3.1 Physical chemistry — revision question pack

40 specification points · notes, questions, answers and worked methods

Checked against AQA 7405 section 3.1. Review basis: the qualification registry sourced from the AQA A-level Chemistry (7405) specification; registry verification recorded 11 July 2026.

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Answer all questions in the spaces provided.

3.1.1.1 · Fundamental particles

Explanation

Atomic models changed when new experimental evidence could not be explained by an earlier model.
The accepted model has a very small nucleus containing protons and neutrons, with electrons occupying the surrounding space.
A proton has relative charge $+1$ and relative mass $1$ ; a neutron has charge $0$ and relative mass $1$ ; an electron has charge $-1$ and relative mass about $1/1836$ .
A neutral atom has equal numbers of protons and electrons.
An ion forms only by losing or gaining electrons, so its nucleus and element identity remain unchanged.

Worked example

An ion contains $17$ protons, $18$ electrons and $20$ neutrons. State its overall charge and explain whether it is still chlorine.

1.Overall charge $=17-18=-1$ because neutrons have no charge.
2.The element is fixed by its $17$ protons, so the ion remains chlorine.

Answer: The particle is a $\mathrm{Cl^-}$ ion.

Common mistakes

Don't include neutrons when calculating the overall charge, although neutrons are uncharged.
Don't change the number of protons when forming an ion, which would create a different element.

Exam tip

For a particle-comparison question, state both relative mass and relative charge using the values in the data table.

Tier 1 · Easy

State the relative charge and relative mass of an electron.
[2 marks]
Total for this question: 2
A neutral atom loses two electrons. Explain why its mass is unchanged to the precision normally used for relative atomic masses.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A particle contains 15 protons, 16 neutrons and 18 electrons. State its overall charge and identify which particles are in its nucleus.
[3 marks]
Total for this question: 3
A student claims that an atom forms a $1-$ ion when one proton in its nucleus changes into a neutron. Identify two errors in this model and give the correct change that forms a $1-$ ion.
[3 marks]
Total for this question: 3
A data table gives these entries: proton — relative charge $+1$ , relative mass $1$ , outside the nucleus; neutron — relative charge $0$ , relative mass $1$ , in the nucleus; electron — relative charge $-1$ , relative mass $1$ , in the nucleus. Identify and correct the three incorrect entries.
[3 marks]
Total for this question: 3

Tier 3 · Hard

The nucleus of an atom has charge $2.08\times10^{-18}\,\mathrm{C}$ . Use the proton charge $1.60\times10^{-19}\,\mathrm{C}$ to determine the number of protons. The atom gains two electrons; state the resulting ion charge and its number of electrons.
[4 marks]
Total for this question: 4
Each beam contains one type of sub-atomic particle. Beams P, Q and R enter the same electric field at the same speed. P is undeflected. Q bends towards the positive plate and R bends towards the negative plate. Q bends much more than R. The particles have equal charge magnitudes where charged. Identify P, Q and R and explain the difference in deflection.
[4 marks]
Total for this question: 4
In an alpha-scattering experiment, most alpha particles passed through a thin metal foil without deflection, but a very small proportion were deflected through large angles. Explain how these observations support the nuclear model rather than the plum-pudding model, and give one feature of the modern model, other than the nucleus, that the plum-pudding model lacks.
[4 marks]
Total for this question: 4
An atom is modelled as a sphere of radius $1.2\times10^{-10}\,\mathrm{m}$ containing a spherical nucleus of radius $4.8\times10^{-15}\,\mathrm{m}$ . Calculate the nucleus-to-atom volume ratio. Explain how this result and the locations of the fundamental particles account for both the atom's volume and almost all of its mass.
[4 marks]
Total for this question: 4
Cathode-ray particles are deflected towards a positively charged plate. The same charge-to-mass ratio is obtained when the gas in the tube and the metal used for the cathode are changed. Explain what each observation reveals about the particles and why the combined evidence required a model containing a sub-atomic particle.
[4 marks]
Total for this question: 4

3.1.1.2 · Mass number and isotopes

Explanation

Atomic number $Z$ is the number of protons; mass number $A$ is the total number of protons and neutrons, so neutrons $=A-Z$ .
Isotopes are atoms of one element with equal proton numbers but different neutron numbers.
In time-of-flight mass spectrometry, gaseous particles are ionised, accelerated so all ions have the same kinetic energy, separated by flight time and detected.
For mononuclear $1+$ ions, the mass-to-charge value identifies relative isotopic mass and signal size gives abundance.
The weighted mean $A_r$ can identify an element, while a molecular-ion peak can give $M_r$ .

Worked example

Chlorine has two isotopes: $^{35}\mathrm{Cl}$ at $75.8\%$ abundance and $^{37}\mathrm{Cl}$ at $24.2\%$ abundance. Calculate $A_r$ .

1.Multiply each isotopic mass by its percentage abundance: $(35\times75.8)+(37\times24.2)$ .
2.Divide the total by $100$ : $A_r=\dfrac{3548.4}{100}$ .

Answer: $A_r=35.484$ , usually reported as $35.5$ .

Common mistakes

Don't use mass number as though it were the relative atomic mass of the element.
Don't add percentage-weighted products without dividing the result by $100$ .

Exam tip

In a calculation from a mass spectrum, show the abundance weighting before giving the final weighted mean.

Tier 1 · Easy

For the ion $^{81}_{35}\mathrm{Br^-}$ , give its proton count, neutron count and electron count.
[3 marks]
Total for this question: 3
Three neutral particles have these compositions. P: 18 protons and 22 neutrons. Q: 19 protons and 21 neutrons. R: 18 protons and 24 neutrons. Identify the pair that are isotopes and explain your choice.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An element has two isotopes, $^{50}\mathrm{Q}$ with abundance $73.0\%$ and $^{52}\mathrm{Q}$ with abundance $27.0\%$ . Calculate the relative atomic mass of Q.
[3 marks]
Total for this question: 3
Two samples of element X give mass-spectrum peaks only at $m/z=63$ and $m/z=65$ . The peak-area ratio $63:65$ is $7:3$ for sample A and $4:6$ for sample B. Deduce which sample is enriched in the heavier isotope and compare the relative atomic masses of the samples.
[3 marks]
Total for this question: 3
A student describes TOF mass spectrometry as follows: neutral particles are accelerated; the ion with the largest mass reaches the detector first; the detector current is unrelated to abundance. Identify and correct the three errors and explain why ionisation must occur before acceleration.
[4 marks]
Total for this question: 4

Tier 3 · Hard

In a TOF mass spectrometer, singly charged $^{64}\mathrm{X^+}$ ions take $1.280\times10^{-5}\,\mathrm{s}$ to reach the detector. Another singly charged isotope of X takes $1.339\times10^{-5}\,\mathrm{s}$ . All ions receive the same kinetic energy. Use $t\propto\sqrt{m}$ to deduce the mass number of the second isotope.
[4 marks]
Total for this question: 4
All ions are accelerated to the same kinetic energy. A $1+$ ion of $^{24}\mathrm{Mg}$ reaches the detector in $1.20\times10^{-5}\,\mathrm{s}$ over a $1.00\,\mathrm{m}$ drift tube. Calculate the flight time of a $1+$ $^{26}\mathrm{Mg}$ ion, then use the abundances Mg-24 $79.0\%$ , Mg-25 $10.0\%$ and Mg-26 $11.0\%$ to calculate $A_r$ .
[4 marks]
Total for this question: 4
Element X has isotopes of mass numbers 60 and 62. Its mass spectrum contains both $\mathrm{X^+}$ and $\mathrm{X^{2+}}$ ions, giving peaks at $m/z=30$ , 31, 60 and 62. Identify the ion responsible for each peak. An ion at $m/z=30$ has flight time $8.00\times10^{-7}\,\mathrm{s}$ ; calculate the flight time of the ion at $m/z=31$ using the constant-kinetic-energy model, $E_k=\tfrac12mv^2$ .
[5 marks]
Total for this question: 5
Element X has two isotopes, $^{10}\mathrm{X}$ and $^{11}\mathrm{X}$ , whose abundances are $80.0\%$ and $20.0\%$ . A molecular-ion region in a mass spectrum contains only singly charged $\mathrm{X_2^+}$ ions formed by random pairing of X atoms. Determine every $m/z$ value in this region, identify the isotopic composition responsible for each value and calculate the expected peak-area ratio.
[5 marks]
Total for this question: 5
A sample contains only $^{68}\mathrm{X}$ and $^{70}\mathrm{X}$ . The same fixed proportion of both isotopes forms $\mathrm{X^{2+}}$ ions; the remainder forms $\mathrm{X^+}$ ions. The peak currents are $1.20\,\mathrm{nA}$ at $m/z=68$ , $2.40\,\mathrm{nA}$ at $m/z=34$ , $0.450\,\mathrm{nA}$ at $m/z=70$ and $0.900\,\mathrm{nA}$ at $m/z=35$ . Every arriving ion transfers its full positive charge. Determine the total ion-population ratio of the isotopes, the percentage abundance of $^{70}\mathrm{X}$ and the relative atomic mass.
[5 marks]
Total for this question: 5

3.1.1.3 · Electron configuration

Explanation

For atoms up to $Z=36$ , electrons fill $1s$ , $2s$ , $2p$ , $3s$ , $3p$ , $4s$ , $3d$ , then $4p$ . Chromium is $\mathrm{[Ar]3d^5\,4s^1}$ and copper is $\mathrm{[Ar]3d^{10}\,4s^1}$ ; transition-metal ions lose $4s$ electrons before $3d$ .
First ionisation energy removes one electron from each atom in one mole of gaseous atoms.
Across Period 3, increasing nuclear charge generally raises first ionisation energy, with sub-shell and electron-pairing anomalies.
Down Group 2, increasing distance and shielding lower it.
A large successive-ionisation jump shows removal from a shell closer to the nucleus and provides evidence for shell structure.

Sub-shell filling order through 3d; 4s fills before 3d in neutral atoms.

Worked example

Write the electron configuration of an iron atom and an $\mathrm{Fe^{2+}}$ ion.

1.Iron has $26$ electrons: $\mathrm{[Ar]3d^6\,4s^2}$ .
2.Remove the two $4s$ electrons before any $3d$ electrons.

Answer: $\mathrm{Fe:[Ar]3d^6\,4s^2}$ and $\mathrm{Fe^{2+}:[Ar]3d^6}$ .

Common mistakes

Don't remove $3d$ electrons before $4s$ electrons when writing a transition-metal ion.
Don't write a successive ionisation equation without gaseous species or without one electron as a product.
Don't explain a Period 3 anomaly using nuclear charge alone while ignoring sub-shell energy or electron pairing.

Exam tip

For a successive-ionisation graph, locate the large jump first and link it explicitly to a change of electron shell.

Tier 1 · Easy

Write the electron configuration, in sub-shell notation, of $\mathrm{S^{2-}}$ . Sulfur has atomic number 16.
[1 mark]
Total for this question: 1
The occupied sub-shells of a neutral atom contain 2, 2, 6, 2 and 3 electrons in $1s$ , $2s$ , $2p$ , $3s$ and $3p$ respectively. Identify the element and state its atomic number.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Define first ionisation energy and write an equation, including state symbols, for the first ionisation of magnesium.
[3 marks]
Total for this question: 3
An atom has electron configuration $\mathrm{[Ar]3d^{10}\,4s^2}$ . One of its ions has configuration $\mathrm{[Ar]3d^{10}}$ . Deduce the charge on the ion and explain which electrons are removed.
[3 marks]
Total for this question: 3
An atom and two ions made from argon, potassium and calcium are isoelectronic and each contains $18$ electrons. Identify the three species, write their common electron configuration in sub-shell notation, and explain why they are not the same element.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Element X is in Period 3. Its first five ionisation energies, in $\mathrm{kJ\,mol^{-1}}$ , are 780, 1580, 3260, 4380 and 16050. Deduce the group and identity of X, write its outer electron configuration, and write an equation for its fifth ionisation.
[5 marks]
Total for this question: 5
A student writes chromium as $\mathrm{[Ar]3d^4\,4s^2}$ and $\mathrm{Cr^{3+}}$ as $\mathrm{[Ar]3d^1\,4s^2}$ . Identify both errors and give the correct electron configurations.
[4 marks]
Total for this question: 4
The third ionisation energy of magnesium is much greater than the third ionisation energy of aluminium. Write the electron configurations of $\mathrm{Mg^{2+}}$ and $\mathrm{Al^{2+}}$ , identify the sub-shell from which the third electron is removed in each case, and explain the difference.
[5 marks]
Total for this question: 5
Copper has electron configuration $\mathrm{[Ar]3d^{10}\,4s^1}$ . Write equations, including state symbols, for its first and second ionisations. Give the electron configuration after each ionisation and explain why the second ionisation requires much more energy than the first.
[6 marks]
Total for this question: 6
The first ionisation energy of Ne, the second ionisation energy of Na and the third ionisation energy of Mg remove an electron from $\mathrm{Ne}$ , $\mathrm{Na^+}$ and $\mathrm{Mg^{2+}}$ respectively. Write the electron configuration of each species immediately before removal, deduce the order of the three ionisation energies from lowest to highest and explain the order.
[5 marks]
Total for this question: 5

3.1.2.1 · Relative atomic mass and relative molecular mass

Explanation

Relative atomic mass, $A_r$ , is the weighted mean mass of an atom of an element relative to one twelfth of the mass of a carbon-12 atom. Relative molecular mass, $M_r$ , is the mean mass of a molecule on the same scale.
Both are ratios and therefore have no units.
Calculate $M_r$ by multiplying each element's $A_r$ by the number of its atoms and summing.
For an ionic substance, the equivalent quantity is relative formula mass because the solid contains a lattice rather than separate molecules.
Brackets and waters of crystallisation must be counted fully.

Worked example

Calculate the relative formula mass of $\mathrm{CuSO_4\cdot5H_2O}$ using $A_r$ : Cu $=63.5$ , S $=32.1$ , O $=16.0$ , H $=1.0$ .

1.Count atoms, including the water: one Cu, one S, nine O and ten H.
2. $M_r=63.5+32.1+(9\times16.0)+(10\times1.0)$ .

Answer: The relative formula mass is $249.6$ with no units.

Common mistakes

Don't count only the oxygen and hydrogen inside one water molecule instead of all five waters.
Don't attach $\mathrm{g\,mol^{-1}}$ to $M_r$ , although that unit belongs to molar mass.

Exam tip

Expand every bracketed or hydrated group into atom counts before substituting relative atomic masses.

Tier 1 · Easy

Calculate the relative molecular mass of urea, $\mathrm{CO(NH_2)_2}$ . Use $A_r$ : H $=1.0$ , C $=12.0$ , N $=14.0$ , O $=16.0$ .
[2 marks]
Total for this question: 2
Give the correct relative-mass term for each: the mean mass of a neon atom, the mean mass of a carbon dioxide molecule, and the mass of one magnesium chloride formula unit, each relative to one twelfth of carbon-12.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Calculate the relative formula mass of $\mathrm{MgSO_4\cdot7H_2O}$ . Use $A_r$ : H $=1.0$ , O $=16.0$ , Mg $=24.3$ , S $=32.1$ .
[3 marks]
Total for this question: 3
Two records describe water as $M_r=18.0\,\mathrm{g\,mol^{-1}}$ and as having a molar mass of 18.0 with no units. Compare the quantities, identify both unit errors and give the correct statements.
[3 marks]
Total for this question: 3
The mass of one carbon-12 atom is $1.9926\times10^{-26}\,\mathrm{kg}$ . One molecule of compound Q has mass $7.9704\times10^{-26}\,\mathrm{kg}$ . Use the definition of relative molecular mass to calculate $M_r$ of Q and state whether $M_r$ has units.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An ionic oxide has formula $\mathrm{X_2O_3}$ and relative formula mass 144.1. Element X contains only isotopes $^{47}\mathrm{X}$ and $^{49}\mathrm{X}$ . Calculate $A_r(\mathrm{X})$ and the percentage abundance of $^{47}\mathrm{X}$ . Use $A_r(\mathrm{O})=16.0$ .
[4 marks]
Total for this question: 4
A hydrated salt has formula $\mathrm{Na_2CO_3\cdot xH_2O}$ and relative formula mass 286.0. The anhydrous salt has relative formula mass 106.0 and water has $M_r=18.0$ . Deduce $x$ .
[3 marks]
Total for this question: 3
An element X has $A_r=52.0$ . Compare the mass percentages of oxygen in $\mathrm{X_2O_3}$ and $\mathrm{XO_2}$ , then calculate the mass of each compound that contains $8.00\,\mathrm{g}$ of oxygen. Use $A_r(\mathrm{O})=16.0$ and identify which compound supplies this oxygen using less material.
[5 marks]
Total for this question: 5
A fertiliser contains $\mathrm{NH_4NO_3}$ and $\mathrm{(NH_4)_2SO_4}$ in a $3:1$ mole ratio. Calculate the mean relative formula mass per formula unit in the mixture. Use $A_r$ : H $=1.0$ , N $=14.0$ , O $=16.0$ , S $=32.1$ .
[4 marks]
Total for this question: 4
A hydrated sample has formula $\mathrm{Na_2S_2O_3\cdot xH_2O}$ and is $36.3\%$ water by mass. Use $A_r$ : H $=1.0$ , O $=16.0$ , Na $=23.0$ , S $=32.1$ . Determine $x$ and calculate the relative formula mass of the hydrate.
[5 marks]
Total for this question: 5

3.1.2.2 · The mole and the Avogadro constant

Explanation

One mole contains the Avogadro constant, $N_A$ , of specified entities, such as atoms, molecules, ions, electrons or formula units.
Amount of substance links measurable quantities to particle number: $n=m/M$ , $N=nN_A$ , and for a solution $n=cV$ when $V$ is in $\mathrm{dm^3}$ .
A secure calculation first converts the given quantity to moles, applies the chemical formula or equation ratio, then converts to the requested quantity.
The named entity matters: one mole of $\mathrm{MgCl_2}$ formula units contains two moles of chloride ions.

Worked example

Calculate the number of chloride ions in $25.0\,\mathrm{cm^3}$ of $0.200\,\mathrm{mol\,dm^{-3}}$ $\mathrm{MgCl_2(aq)}$ . Use $N_A=6.02\times10^{23}\,\mathrm{mol^{-1}}$ .

1.Convert volume: $25.0\,\mathrm{cm^3}=0.0250\,\mathrm{dm^3}$ .
2. $n(\mathrm{MgCl_2})=cV=0.200\times0.0250=5.00\times10^{-3}\,\mathrm{mol}$ .
3. $n(\mathrm{Cl^-})=2\times5.00\times10^{-3}=1.00\times10^{-2}\,\mathrm{mol}$ .
4. $N=1.00\times10^{-2}\times6.02\times10^{23}$ .

Answer: $6.02\times10^{21}$ chloride ions.

Common mistakes

Don't use a volume in $\mathrm{cm^3}$ directly in $n=cV$ .
Don't count formula units of $\mathrm{MgCl_2}$ but omit the factor of two for chloride ions.

Exam tip

Write the entity beside every mole value so that formula-unit multipliers are not lost.

Tier 1 · Easy

Calculate the amount, in moles, in $5.30\,\mathrm{g}$ of $\mathrm{Na_2CO_3}$ . Use $M_r=106.0$ .
[2 marks]
Total for this question: 2
A sample contains $3.01\times10^{22}$ formula units. Calculate the amount of substance in the sample. Use $N_A=6.02\times10^{23}\,\mathrm{mol^{-1}}$ .
[2 marks]
Total for this question: 2

Tier 2 · Standard

A sample contains $0.0125\,\mathrm{mol}$ of oxygen molecules. Calculate the number of $\mathrm{O_2}$ molecules. Use $N_A=6.02\times10^{23}\,\mathrm{mol^{-1}}$ .
[2 marks]
Total for this question: 2
In an RP₁ preparation, the final sodium carbonate solution must have volume $250.0\,\mathrm{cm^3}$ and concentration $0.0800\,\mathrm{mol\,dm^{-3}}$ . Calculate the mass of anhydrous $\mathrm{Na_2CO_3}$ required and identify the apparatus used to set the final volume accurately. Use $M_r(\mathrm{Na_2CO_3})=106.0$ .
[3 marks]
Total for this question: 3
One molecule of an unnamed gas has mass $7.31\times10^{-23}\,\mathrm{g}$ . Calculate its molar mass and the number of molecules in a $0.220\,\mathrm{g}$ sample. Use $N_A=6.02\times10^{23}\,\mathrm{mol^{-1}}$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

$18.0\,\mathrm{cm^3}$ of $0.250\,\mathrm{mol\,dm^{-3}}$ aluminium sulfate solution contains fully dissociated $\mathrm{Al_2(SO_4)_3}$ . Calculate the number of sulfate ions present. Use $N_A=6.02\times10^{23}\,\mathrm{mol^{-1}}$ .
[4 marks]
Total for this question: 4
A vessel contains $15.0\,\mathrm{cm^3}$ of $0.200\,\mathrm{mol\,dm^{-3}}$ $\mathrm{AlCl_3(aq)}$ . A student adds $20.0\,\mathrm{cm^3}$ of $0.250\,\mathrm{mol\,dm^{-3}}$ $\mathrm{AgNO_3(aq)}$ . Silver ions precipitate chloride ions in a $1:1$ ratio. Calculate the number of chloride ions remaining in solution. Use $N_A=6.02\times10^{23}\,\mathrm{mol^{-1}}$ .
[5 marks]
Total for this question: 5
$0.585\,\mathrm{g}$ of sodium chloride contains $\mathrm{Na^+}$ and $\mathrm{Cl^-}$ ions. Calculate the total number of electrons in all the ions in the sample. Use $M_r(\mathrm{NaCl})=58.5$ , $N_A=6.02\times10^{23}\,\mathrm{mol^{-1}}$ , and atomic numbers Na $=11$ and Cl $=17$ .
[5 marks]
Total for this question: 5
$0.5619\,\mathrm{g}$ of $\mathrm{Fe_2(SO_4)_3\cdot9H_2O}$ is sampled. Its molar mass is $561.9\,\mathrm{g\,mol^{-1}}$ . Calculate the total number of oxygen atoms in the sample. Use $N_A=6.022\times10^{23}\,\mathrm{mol^{-1}}$ .
[4 marks]
Total for this question: 4
A sample containing exactly $2.50\times10^{20}$ atoms of element X has mass $0.0264\,\mathrm{g}$ . The molar mass of X is $63.5\,\mathrm{g\,mol^{-1}}$ . Use these measurements to determine an experimental value of the Avogadro constant and calculate its percentage difference from the accepted value $6.022\times10^{23}\,\mathrm{mol^{-1}}$ .
[5 marks]
Total for this question: 5

3.1.2.3 · The ideal gas equation

Explanation

The ideal gas equation is $pV=nRT$ .
The gas constant $R$ has the value $8.31$ in units of $\mathrm{J\,K^{-1}\,mol^{-1}}$ , so pressure must be in $\mathrm{Pa}$ , volume in $\mathrm{m^3}$ , temperature in $\mathrm{K}$ and amount in $\mathrm{mol}$ .
Convert $\mathrm{kPa}$ to $\mathrm{Pa}$ by multiplying by $10^3$ , $\mathrm{cm^3}$ to $\mathrm{m^3}$ by multiplying by $10^{-6}$ , and $\mathrm{dm^3}$ to $\mathrm{m^3}$ by multiplying by $10^{-3}$ .
Convert Celsius using $T/\mathrm{K}=\theta/^{\circ}\mathrm{C}+273$ .
Rearranging the equation before substitution reduces unit and algebra errors in multi-step calculations.

Worked example

A gas occupies $240\,\mathrm{cm^3}$ at $100\,\mathrm{kPa}$ and $298\,\mathrm{K}$ . Calculate the amount of gas.

1. $V=240\times10^{-6}=2.40\times10^{-4}\,\mathrm{m^3}$ and $p=1.00\times10^5\,\mathrm{Pa}$ .
2.Rearrange: $n=\dfrac{pV}{RT}$ .
3. $n=\dfrac{(1.00\times10^5)(2.40\times10^{-4})}{8.31\times298}$ .

Answer: $n=9.69\times10^{-3}\,\mathrm{mol}$ to three significant figures.

Common mistakes

Don't substitute temperature in degrees Celsius rather than kelvin.
Don't combine pressure in kilopascals with volume in cubic centimetres.

Exam tip

Show every SI conversion on a separate line before substituting into $pV=nRT$ .

Tier 1 · Easy

$0.100\,\mathrm{mol}$ of gas occupies $2.49\times10^{-3}\,\mathrm{m^3}$ at $300\,\mathrm{K}$ . Calculate its pressure. Take $R=8.31$ .
[3 marks]
Total for this question: 3
A substitution line for $pV=nRT$ with $R=8.31$ uses $p=96.0$ , $V=250$ and $T=22$ for a gas at $96.0\,\mathrm{kPa}$ , $250\,\mathrm{cm^3}$ and $22\,^{\circ}\mathrm{C}$ . Identify the three unit errors and give the correct values.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Calculate the volume occupied by $0.0350\,\mathrm{mol}$ of an ideal gas at $315\,\mathrm{K}$ and $105\,\mathrm{kPa}$ . Give the answer in $\mathrm{dm^3}$ . Take $R=8.31$ .
[4 marks]
Total for this question: 4
$0.0200\,\mathrm{mol}$ of an ideal gas occupies $410\,\mathrm{cm^3}$ at $120\,\mathrm{kPa}$ . Calculate the temperature of the gas using $R=8.31$ in SI units.
[4 marks]
Total for this question: 4
An experiment uses $0.0100\,\mathrm{mol}$ of an ideal gas at $291\,\mathrm{K}$ . Its pressure is $98.5\,\mathrm{kPa}$ and its volume is $245\,\mathrm{cm^3}$ . Use the measurements to calculate an experimental value of the gas constant $R$ in $\mathrm{J\,K^{-1}\,mol^{-1}}$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

A $0.318\,\mathrm{g}$ sample of a volatile liquid forms $85.0\,\mathrm{cm^3}$ of vapour at $98.0\,\mathrm{kPa}$ and $373\,\mathrm{K}$ . Calculate its molar mass. Take $R=8.31$ .
[5 marks]
Total for this question: 5
Magnesium reacts with hydrochloric acid according to $\mathrm{Mg+2HCl\rightarrow MgCl_2+H_2}$ . A student mixes $0.240\,\mathrm{g}$ Mg with $20.0\,\mathrm{cm^3}$ of $0.800\,\mathrm{mol\,dm^{-3}}$ HCl. Identify the limiting reactant and calculate the maximum hydrogen volume at $305\,\mathrm{K}$ and $102\,\mathrm{kPa}$ . Use $A_r(\mathrm{Mg})=24.3$ and take $R=8.31$ in SI units.
[6 marks]
Total for this question: 6
A vessel fitted with a movable piston and a seal that may leak initially contains gas at $112\,\mathrm{kPa}$ , $1.80\,\mathrm{dm^3}$ and $320\,\mathrm{K}$ . It is changed to $1.50\,\mathrm{dm^3}$ and $300\,\mathrm{K}$ . Calculate the pressure expected if no gas escapes. The measured pressure is only $92.0\,\mathrm{kPa}$ ; calculate the percentage of the original gas that escaped, assuming ideal behaviour.
[5 marks]
Total for this question: 5
A $0.250\,\mathrm{g}$ sample of impure calcium carbonate reacts with excess acid. The carbon dioxide formed occupies $52.5\,\mathrm{cm^3}$ at $100\,\mathrm{kPa}$ and $298\,\mathrm{K}$ . The impurity does not produce a gas. Calculate the percentage purity of the sample. Use $M_r(\mathrm{CaCO_3})=100.1$ and take $R=8.31$ in SI units.
[5 marks]
Total for this question: 5
At $310\,\mathrm{K}$ and $102\,\mathrm{kPa}$ , an ideal gas has density $1.742\,\mathrm{g\,dm^{-3}}$ . Calculate its molar mass and identify it from $\mathrm{N_2}$ , $\mathrm{CO_2}$ and $\mathrm{SO_2}$ . Explain how a measured volume that is $2\%$ too small would affect the calculated molar mass if the measured gas mass were correct. Use $R=8.31$ .
[5 marks]
Total for this question: 5

3.1.2.4 · Empirical and molecular formula

Explanation

An empirical formula is the simplest whole-number ratio of atoms of each element; a molecular formula gives the actual numbers of atoms in one molecule.
From masses or percentages, divide each value by the element's $A_r$ to obtain relative moles, divide every result by the smallest, then multiply all ratios by the same small integer if necessary.
A ratio near $1.5$ requires doubling, not rounding.
To obtain a molecular formula, calculate the empirical formula mass and find the integer $M_r/(\text{empirical formula mass})$ .
Combustion data must be converted from product moles to constituent atom moles.

Worked example

A compound contains $40.0\%$ carbon, $6.7\%$ hydrogen and $53.3\%$ oxygen by mass. Its $M_r$ is $180$ . Determine its empirical and molecular formulae.

1.Relative moles: C $=40.0/12.0=3.33$ , H $=6.7/1.0=6.7$ , O $=53.3/16.0=3.33$ .
2.Divide by $3.33$ to obtain approximately $1:2:1$ , so the empirical formula is $\mathrm{CH_2O}$ .
3.Empirical formula mass $=30$ ; multiplier $=180/30=6$ .

Answer: Empirical formula $\mathrm{CH_2O}$ ; molecular formula $\mathrm{C_6H_{12}O_6}$ .

Common mistakes

Don't use mass percentages directly as atom ratios without dividing by relative atomic masses.
Don't round a half-integer mole ratio instead of multiplying every ratio by two.

Exam tip

Keep unrounded mole values until the common whole-number multiplier has been identified.

Tier 1 · Easy

A compound contains $2.70\,\mathrm{g}$ of aluminium and $2.40\,\mathrm{g}$ of oxygen. Determine its empirical formula. Use $A_r$ : Al $=27.0$ , O $=16.0$ .
[3 marks]
Total for this question: 3
A molecule has formula $\mathrm{C_2H_4O_2}$ . Give its empirical formula and explain why $\mathrm{C_2H_4O_2}$ is not an empirical formula.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A compound is $54.5\%$ carbon, $9.1\%$ hydrogen and $36.4\%$ oxygen by mass. Its $M_r$ is 88. Determine its empirical and molecular formulae. Use $A_r$ : H $=1.0$ , C $=12.0$ , O $=16.0$ .
[5 marks]
Total for this question: 5
An analysis gives $0.0200\,\mathrm{mol}$ of element X and $0.0300\,\mathrm{mol}$ of element Y. A student rounds the ratio $1:1.5$ to $1:2$ and proposes $\mathrm{XY_2}$ . Identify the error and give the correct empirical formula.
[3 marks]
Total for this question: 3
$2.40\,\mathrm{g}$ of element X is heated in oxygen and forms $4.00\,\mathrm{g}$ of an oxide. Determine the empirical formula of the oxide. Use $A_r(\mathrm{X})=48.0$ and $A_r(\mathrm{O})=16.0$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

Complete combustion of $0.900\,\mathrm{g}$ of a compound containing only carbon, hydrogen and oxygen produces $1.320\,\mathrm{g}$ of $\mathrm{CO_2}$ and $0.540\,\mathrm{g}$ of $\mathrm{H_2O}$ . The compound has $M_r=150$ . Determine its molecular formula. Use $A_r$ : H $=1.0$ , C $=12.0$ , O $=16.0$ .
[7 marks]
Total for this question: 7
A gaseous compound contains $85.7\%$ carbon and $14.3\%$ hydrogen by mass. A $0.280\,\mathrm{g}$ sample occupies $124.7\,\mathrm{cm^3}$ at $100\,\mathrm{kPa}$ and $300\,\mathrm{K}$ . Determine its molecular formula. Use $A_r$ : H $=1.0$ , C $=12.0$ , and take $R=8.31$ in SI units.
[7 marks]
Total for this question: 7
An elemental analysis gives $15.7\%$ nitrogen and $36.0\%$ oxygen by mass. Two candidate empirical formulae are $\mathrm{C_3H_7NO_2}$ and $\mathrm{C_2H_5NO}$ . Calculate the theoretical nitrogen and oxygen percentages for each candidate and state which is consistent with the analysis. Use $A_r$ : H $=1.0$ , C $=12.0$ , N $=14.0$ , O $=16.0$ .
[5 marks]
Total for this question: 5
A $10.0\,\mathrm{cm^3}$ sample of a gaseous hydrocarbon is burned completely with $50.0\,\mathrm{cm^3}$ of oxygen. All gas volumes are compared at the same temperature and pressure. After the water formed has condensed, $40.0\,\mathrm{cm^3}$ of gas remains. Passing this gas through aqueous alkali removes all the carbon dioxide and leaves $10.0\,\mathrm{cm^3}$ of oxygen. Determine the molecular formula of the hydrocarbon.
[5 marks]
Total for this question: 5
A molecular oxide contains only nitrogen and oxygen. Nitrogen accounts for $30.4\%$ of its mass, and each molecule contains two nitrogen atoms. Determine the molecular formula and $M_r$ . Use $A_r$ : N $=14.0$ , O $=16.0$ .
[5 marks]
Total for this question: 5

3.1.2.5 · Balanced equations and associated calculations

Explanation

Balanced full and ionic equations conserve every atom and the total charge; only coefficients may be changed.
Quantitative work follows a mole route: convert the known quantity to moles, apply the balanced-equation ratio, identify any limiting reactant, then convert to the requested mass, volume or concentration.
Percentage yield is $100\times\text{actual yield}/\text{theoretical yield}$ .
Atom economy is $100\times M_r(\text{desired product})/\sum M_r(\text{reactants})$ , including stoichiometric coefficients.
A high atom economy reduces waste and raw-material use, with economic, environmental and ethical benefits, but it does not guarantee a high experimental yield.

Worked example

$2\mathrm{Al}+3\mathrm{Cl_2}\rightarrow2\mathrm{AlCl_3}$ . Calculate the theoretical mass of $\mathrm{AlCl_3}$ from $5.40\,\mathrm{g}$ Al with excess chlorine. $M_r(\mathrm{AlCl_3})=133.5$ .

1. $n(\mathrm{Al})=5.40/27.0=0.200\,\mathrm{mol}$ .
2.The equation ratio Al: $\mathrm{AlCl_3}$ is $2:2$ , so $n(\mathrm{AlCl_3})=0.200\,\mathrm{mol}$ .
3. $m=nM=0.200\times133.5$ .

Answer: The theoretical mass is $26.7\,\mathrm{g}$ .

Common mistakes

Don't use a mass ratio from the equation instead of converting masses to moles.
Don't put the actual yield in the denominator when calculating percentage yield.
Don't omit equation coefficients when calculating atom economy.

Exam tip

For a multi-stage calculation, label the moles of each substance and show the equation ratio explicitly.

Tier 1 · Easy

Balance the equation $\mathrm{C_3H_8+O_2\rightarrow CO_2+H_2O}$ using the smallest whole-number coefficients.
[1 mark]
Total for this question: 1
A student writes $\mathrm{Fe_2O_3+2CO\rightarrow2Fe+2CO_2}$ . Identify why this equation is flawed and write the balanced equation using the smallest whole-number coefficients.
[2 marks]
Total for this question: 2

Tier 2 · Standard

$25.0\,\mathrm{cm^3}$ of $0.150\,\mathrm{mol\,dm^{-3}}$ hydrochloric acid reacts with excess calcium carbonate: $\mathrm{CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O}$ . Calculate the maximum mass of calcium carbonate that can react. Use $M_r(\mathrm{CaCO_3})=100.1$ .
[4 marks]
Total for this question: 4
In an RP₁ titration, $25.00\,\mathrm{cm^3}$ of sodium hydroxide is titrated with $0.1000\,\mathrm{mol\,dm^{-3}}$ sulfuric acid. The titres are $18.35$ , $18.45$ and $19.10\,\mathrm{cm^3}$ . Use the concordant titres to calculate the concentration of sodium hydroxide. The equation is $\mathrm{H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O}$ .
[5 marks]
Total for this question: 5
Aqueous calcium nitrate is mixed with aqueous sodium carbonate and calcium carbonate precipitates. Write the balanced full equation and the net ionic equation, both with state symbols, and identify the spectator ions.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Urea is made by $\mathrm{2NH_3+CO_2\rightarrow CO(NH_2)_2+H_2O}$ . A process uses $85.0\,\mathrm{kg}$ of ammonia and $132\,\mathrm{kg}$ of carbon dioxide. Determine the limiting reactant, the maximum mass of urea, the percentage atom economy for urea and the percentage yield if $126\,\mathrm{kg}$ is obtained. Use molar masses in $\mathrm{kg\,kmol^{-1}}$ : $\mathrm{NH_3}=17.0$ , $\mathrm{CO_2}=44.0$ , urea $=60.0$ , $\mathrm{H_2O}=18.0$ .
[7 marks]
Total for this question: 7
$0.270\,\mathrm{g}$ of a metal M with molar mass $27.0\,\mathrm{g\,mol^{-1}}$ reacts completely with excess acid. The hydrogen produced occupies $371.5\,\mathrm{cm^3}$ at $100\,\mathrm{kPa}$ and $298\,\mathrm{K}$ . Use the data to deduce the charge on the metal ion and write the balanced ionic equation, taking $R=8.31$ in SI units.
[6 marks]
Total for this question: 6
An unnamed metal carbonate, $\mathrm{MCO_3}$ , has $M_r=100.0$ and decomposes on heating according to $\mathrm{MCO_3(s)\rightarrow MO(s)+CO_2(g)}$ . Use $M_r(\mathrm{CO_2})=44.0$ . A $4.00\,\mathrm{g}$ sample is heated for a limited time and the solid residue has mass $3.12\,\mathrm{g}$ . Calculate the percentage of the original carbonate that decomposed and the masses of $\mathrm{MCO_3}$ and $\mathrm{MO}$ in the residue.
[5 marks]
Total for this question: 5
An unnamed reactant A forms intermediate B in a $1:1$ mole ratio with a yield of $80.0\%$ . In a second reaction, two moles of B form one mole of product C with a yield of $75.0\%$ . Determine the minimum mass of A needed to obtain $12.0\,\mathrm{g}$ of C. Use molar masses A $=50.0\,\mathrm{g\,mol^{-1}}$ and C $=120\,\mathrm{g\,mol^{-1}}$ .
[5 marks]
Total for this question: 5
Two routes make desired product P. Route A uses $150\,\mathrm{g}$ of A and $40.0\,\mathrm{g}$ of B per stoichiometric batch, has a theoretical P mass of $120\,\mathrm{g}$ and a $92.0\%$ yield; it runs at $320\,\mathrm{K}$ and its only by-product is water. Route B uses $150\,\mathrm{g}$ of C, has a theoretical P mass of $120\,\mathrm{g}$ and a $72.0\%$ yield; it runs at $650\,\mathrm{K}$ and produces a toxic gaseous by-product. For each route, calculate the atom economy and the actual mass of P obtained as a percentage of the total stoichiometric reactant mass. Evaluate which route is preferable.
[6 marks]
Total for this question: 6

3.1.3.1 · Ionic bonding

Explanation

Ionic bonding is the strong electrostatic attraction between oppositely charged ions throughout a giant lattice. Electron transfer forms the ions but is not itself the bond.
Simple-ion charges can be predicted from Periodic Table groups, and the ionic formula uses the smallest whole-number ratio giving zero overall charge.
Compound ions remain intact; for example, two $\mathrm{Al^{3+}}$ ions balance three $\mathrm{SO_4^{2-}}$ ions to give $\mathrm{Al_2(SO_4)_3}$ .
Brackets are required when more than one compound ion is present.
Explanations of high melting point must refer to strong attractions acting in many directions through the lattice.

A two-dimensional slice through an ionic lattice with alternating charges.

Worked example

Deduce the formula of the compound formed from $\mathrm{Ca^{2+}}$ and $\mathrm{NO_3^-}$ ions.

1.One calcium ion has charge $+2$ , so two nitrate ions are needed for charge $-2$ .
2.Keep nitrate intact and place it in brackets because two are present.

Answer: $\mathrm{Ca(NO_3)_2}$ .

Common mistakes

Don't define ionic bonding as electron transfer rather than electrostatic attraction between ions.
Don't write a formula whose ionic charges do not sum to zero.

Exam tip

A bonding-definition question requires both electrostatic attraction and oppositely charged ions.

Tier 1 · Easy

Write the formula of aluminium sulfate from $\mathrm{Al^{3+}}$ and $\mathrm{SO_4^{2-}}$ ions.
[1 mark]
Total for this question: 1
Compare these statements: "electron transfer forms ions" and "electron transfer is the ionic bond". Identify the error in the second statement and give the correct definition of an ionic bond.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Explain the nature of ionic bonding in magnesium oxide.
[3 marks]
Total for this question: 3
Draw a dot-and-cross diagram for lithium nitride, $\mathrm{Li_3N}$ . Show outer-shell electrons only, distinguish the electrons transferred from lithium, and include brackets and charges around every ion.
[4 marks]
Total for this question: 4
Element M is a Group 2 metal and element X is a Group 6 non-metal. Deduce the charge on each simple ion and construct the formula of the compound formed between M and X. The ammonium ion is $\mathrm{NH_4^+}$ ; construct the formula of the compound formed between ammonium ions and the ion of X.
[4 marks]
Total for this question: 4

Tier 3 · Hard

An element X forms the ionic compound $\mathrm{X_2(CO_3)_3}$ . Deduce the charge on the X ion, then write the formulae of its nitrate and hydroxide. Nitrate is $\mathrm{NO_3^-}$ and hydroxide is $\mathrm{OH^-}$ .
[4 marks]
Total for this question: 4
Magnesium oxide melts at about $2850\,^{\circ}\mathrm{C}$ , whereas sodium fluoride melts at about $990\,^{\circ}\mathrm{C}$ . Both are giant ionic lattices. Explain the difference in melting point using the ions present and their electrostatic attractions.
[4 marks]
Total for this question: 4
A neutral representative portion of an ionic lattice contains exactly $20$ ions. The cations are $\mathrm{M^{3+}}$ and the anions are $\mathrm{X^{2-}}$ . Determine the number of each ion in the portion and construct the formula of the compound.
[5 marks]
Total for this question: 5
Sodium forms $\mathrm{Na^+}$ , magnesium forms $\mathrm{Mg^{2+}}$ and aluminium forms $\mathrm{Al^{3+}}$ . The compounds sodium phosphate and magnesium phosphate have formulae $\mathrm{Na_3PO_4}$ and $\mathrm{Mg_3(PO_4)_2}$ . Use both formulae to deduce the charge on the phosphate ion, then construct the formula of aluminium phosphate.
[4 marks]
Total for this question: 4
An ionic solid contains only $\mathrm{Mg^{2+}}$ , $\mathrm{OH^-}$ and $\mathrm{Cl^-}$ ions. It contains equal numbers of hydroxide and chloride ions. Deduce the simplest formula of the solid and show that your formula is electrically neutral. Then give the formula of the compound containing twice as many chloride as hydroxide ions.
[4 marks]
Total for this question: 4

3.1.3.2 · Nature of covalent and dative covalent bonds

Explanation

A covalent bond is a shared pair of electrons. Double and triple bonds contain two and three shared pairs.
In an ordinary covalent bond, each bonded atom contributes one electron to the shared pair. In a co-ordinate, or dative covalent, bond, both electrons in the shared pair are supplied by one atom.
The displayed-formula arrow starts at a lone pair on the electron-pair donor and points towards the electron-deficient acceptor.
Once formed, the dative bond behaves like any other covalent bond.
Dot-and-cross diagrams must show outer-shell electrons, charges and the origin of the bonding pair accurately.

Direction of a dative-bond arrow from a nitrogen lone pair to an electron-pair acceptor.

Worked example

Explain the formation of the dative covalent bond when $\mathrm{NH_3}$ reacts with $\mathrm{H^+}$ .

1.Nitrogen in $\mathrm{NH_3}$ has a lone pair and acts as the electron-pair donor.
2. $\mathrm{H^+}$ accepts both electrons; the arrow runs from the nitrogen lone pair to hydrogen.

Answer: A dative bond forms in $\mathrm{NH_4^+}$ from $\mathrm{N: \rightarrow H^+}$ .

Common mistakes

Don't point the dative-bond arrow towards the atom donating the lone pair.
Don't describe a double bond as one shared pair rather than two shared pairs.

Exam tip

In a dative-bond diagram, place the tail of the arrow exactly on the donating lone pair.

Tier 1 · Easy

State what is meant by a single covalent bond.
[1 mark]
Total for this question: 1
A bonding diagram shows four electrons shared between two carbon atoms. Deduce the type of covalent bond and the number of shared electron pairs.
[2 marks]
Total for this question: 2

Tier 2 · Standard

$\mathrm{BCl_3}$ accepts a lone pair from $\mathrm{NH_3}$ . Represent the dative covalent bond in the product and state which atom donates the electron pair.
[3 marks]
Total for this question: 3
$\mathrm{Cl^-}$ reacts with electron-deficient $\mathrm{AlCl_3}$ to form $\mathrm{AlCl_4^-}$ . A student draws the dative-bond arrow from Al towards Cl. Identify the error and give the correct origin and direction of the arrow.
[3 marks]
Total for this question: 3
Draw a dot-and-cross diagram for $\mathrm{N_2}$ . Show outer-shell electrons only, distinguish the electrons from the two nitrogen atoms, and state the number of shared pairs and lone pairs in the molecule.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Ammonia reacts with a proton to form ammonium. Write an equation for the reaction, show the direction of dative-bond formation, and explain why all four N-H bonds in the ammonium ion are equivalent after formation.
[4 marks]
Total for this question: 4
The complex ion $\mathrm{[Ag(NH_3)_2]^+}$ forms from $\mathrm{Ag^+}$ and two ammonia molecules. Write the formation equation, state the direction of each displayed dative-bond arrow and deduce the total number of electrons donated to silver.
[4 marks]
Total for this question: 4
$\mathrm{CO_2}$ has $16$ outer-shell electrons in total. A student proposes the line formula O-C-O with only single bonds. Use electron counting and the octet rule to reject this model, give the correct line formula, and state the total numbers of shared pairs and lone pairs in the molecule.
[5 marks]
Total for this question: 5
Carbon monoxide has ten outer-shell electrons. Draw a dot-and-cross diagram, state the number of shared pairs between C and O and identify the origin of the electron pair in the dative component of the bond.
[5 marks]
Total for this question: 5
Two electron-deficient $\mathrm{AlCl_3}$ units join to form $\mathrm{Al_2Cl_6}$ with two bridging chlorine atoms. Draw the dimer showing all terminal and bridging Al-Cl links and both dative-bond arrows. Explain the arrow directions and determine the total number of electrons donated in forming the dimer.
[5 marks]
Total for this question: 5

3.1.3.3 · Metallic bonding

Explanation

Metallic bonding is the strong electrostatic attraction between a lattice of positive metal ions and delocalised electrons.
The electrons are not attached to a particular ion and can move through the structure, so a metal conducts electricity when solid or liquid.
Metallic attraction is non-directional, allowing layers of ions to slide while remaining attracted to the delocalised electrons; this explains malleability and ductility.
Bond strength generally increases with greater ionic charge, more delocalised electrons and smaller ionic radius because attraction between the ions and electron density is stronger.
Metals do not consist of separate molecules.

Positive metal ions surrounded by mobile delocalised electrons.

Worked example

Explain why magnesium has a higher melting point than sodium.

1.A magnesium atom supplies more delocalised electrons and forms a $\mathrm{Mg^{2+}}$ ion rather than a $\mathrm{Na^+}$ ion.
2.The higher charge and smaller ion produce stronger electrostatic attraction to the delocalised electrons.

Answer: Magnesium has stronger metallic bonding, so more energy is needed to melt it.

Common mistakes

Don't describe metallic bonding as attraction between neutral atoms.
Don't place fixed electron pairs between particular metal ions instead of showing delocalised electrons.

Exam tip

A conductivity explanation must name the charged particles and state that the delocalised electrons are free to move.

Tier 1 · Easy

Complete the definition of metallic bonding.
[2 marks]
Total for this question: 2
An aluminium atom has outer electron configuration $3s^2\,3p^1$ . Deduce the charge on the metal ion in the lattice and the number of delocalised electrons supplied per aluminium atom.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Explain why a metal conducts electricity when solid and remains conductive when molten.
[3 marks]
Total for this question: 3
Explain why a pure metal can be hammered into a thin sheet without its metallic bonding being completely broken.
[3 marks]
Total for this question: 3
The melting points of the Group 1 metals decrease from lithium to caesium. Explain this trend in terms of metallic bonding.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Magnesium has a higher melting point than sodium. Explain this difference using metallic bonding and the electron configurations $\mathrm{Na:[Ne]3s^1}$ and $\mathrm{Mg:[Ne]3s^2}$ .
[4 marks]
Total for this question: 4
A proposed model of a metal shows neutral atoms joined to fixed neighbours by localised electron pairs. Use the facts that the solid conducts electricity and can deform without shattering to identify two errors and give the correct metallic-bonding model.
[5 marks]
Total for this question: 5
An idealised mixed-metal lattice region contains six positive ion cores. Each P atom contributes one delocalised electron and leaves a $\mathrm{P^+}$ core; each Q atom contributes two and leaves a $\mathrm{Q^{2+}}$ core. The region contains nine delocalised electrons. Determine the numbers of P and Q cores and explain how the electron population makes the region neutral while maintaining metallic bonding.
[5 marks]
Total for this question: 5
Aluminium has density $2.70\,\mathrm{g\,cm^{-3}}$ and supplies three delocalised electrons per atom. Magnesium has density $1.74\,\mathrm{g\,cm^{-3}}$ and supplies two delocalised electrons per atom. Calculate the amount of delocalised electrons, in moles, per $\mathrm{cm^3}$ of each metal and use the results to explain which metal has stronger metallic bonding. Use $A_r(\mathrm{Al})=27.0$ and $A_r(\mathrm{Mg})=24.3$ .
[5 marks]
Total for this question: 5
Sodium melts at $98\,^{\circ}\mathrm{C}$ , whereas sodium chloride melts at $801\,^{\circ}\mathrm{C}$ . Explain the difference by identifying the type of electrostatic attraction in each lattice and comparing their relative strengths.
[4 marks]
Total for this question: 4

3.1.3.4 · Bonding and physical properties

Explanation

The four crystal types are ionic, metallic, macromolecular and molecular. Sodium chloride, magnesium, diamond or graphite, and iodine or ice are respective examples.
Ionic and metallic solids contain giant lattices; macromolecular substances have many covalent bonds; molecular solids contain discrete molecules held by intermolecular forces. Melting a molecular solid overcomes intermolecular attractions, not covalent bonds inside molecules.
Ionic substances conduct only when molten or dissolved because ions can move. Metals and graphite conduct through delocalised electrons.
Structure is inferred from combined melting-point, solubility and conductivity evidence.
A requested diagram must show the specified number and type of particles.

Schematic comparison of ionic, metallic, macromolecular and molecular crystal structures.

Worked example

A solid has a high melting point, does not conduct electricity when solid, and conducts when molten. Deduce its crystal type.

1.The high melting point indicates strong attractions in a giant structure.
2.Conduction only when molten shows that charged ions become mobile on melting.

Answer: The substance has an ionic crystal structure.

Common mistakes

Don't state that boiling a simple molecular substance breaks covalent bonds inside its molecules.
Don't claim that solid ionic compounds conduct because they contain charged ions, without noting that the ions are fixed.
Don't attribute graphite conductivity to moving carbon atoms rather than delocalised electrons.

Exam tip

When deducing crystal type, link each observed property to the particles, attractions and available mobile charge carriers.

Tier 1 · Easy

Iodine forms crystals containing discrete $\mathrm{I_2}$ molecules. State the type of crystal structure and the attractions overcome when iodine melts.
[2 marks]
Total for this question: 2
Solid magnesium chloride does not conduct electricity. The claim that the solid conducts because it contains charged ions conflicts with this evidence. Identify the error and give the condition needed for magnesium chloride to conduct.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Graphite has a high melting point and conducts electricity parallel to its layers. Explain both properties in terms of its structure and bonding.
[4 marks]
Total for this question: 4
Two colourless solids are both soluble in water and neither reacts with water. One is ionic and the other is a molecular non-electrolyte. Suggest the shortest electrical test sequence that distinguishes them, including the observation expected for each structure type.
[3 marks]
Total for this question: 3
Graphite is soft and slippery but diamond is hard, although both are macromolecular carbon. Explain the difference.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Three crystalline solids have these properties. A: high melting point, does not conduct when solid, conducts when molten. B: low melting point, never conducts. C: conducts when solid and is malleable. Deduce the structure type of A, B and C and justify each choice.
[6 marks]
Total for this question: 6
Silicon dioxide has a melting point above $1600\,^{\circ}\mathrm{C}$ , whereas carbon dioxide sublimes at about $-78\,^{\circ}\mathrm{C}$ . Both contain covalent bonds. Deduce the structure type of each substance and explain the large difference in temperature.
[4 marks]
Total for this question: 4
Draw one connected structure diagram for a portion of ice containing exactly four $\mathrm{H_2O}$ molecules. Show the shape and covalent bonds within every molecule, partial charges, an acceptor lone pair for each hydrogen bond, and approximately linear O-H $\cdots$ O hydrogen-bond arrangements. Explain the energy change when this portion of ice melts.
[6 marks]
Total for this question: 6
The enthalpy of fusion of iodine is $15.5\,\mathrm{kJ\,mol^{-1}}$ , whereas the I-I bond enthalpy is $151\,\mathrm{kJ\,mol^{-1}}$ . Use these data to deduce which attractions are overcome when iodine melts and explain why the two values differ by about an order of magnitude.
[4 marks]
Total for this question: 4
Draw two particle diagrams for a compound $\mathrm{M_2X}$ formed from $\mathrm{M^+}$ and $\mathrm{X^{2-}}$ : one solid and one molten. Each diagram must contain exactly six ions in the formula ratio. Explain the change in electrical conductivity on melting without implying that ionic bonds become covalent.
[5 marks]
Total for this question: 5

3.1.3.5 · Shapes of simple molecules and ions

Explanation

Electron charge clouds repel and arrange as far apart as possible around a central atom. Count each bonding region, including a multiple bond as one region, and each lone pair.
Repulsion decreases in the order lone pair–lone pair $>$ lone pair–bond pair $>$ bond pair–bond pair, so lone pairs compress adjacent bond angles.
Base geometries are linear $180^{\circ}$ , trigonal planar $120^{\circ}$ , tetrahedral $109.5^{\circ}$ , trigonal bipyramidal $90^{\circ}$ and $120^{\circ}$ , and octahedral $90^{\circ}$ .
In a trigonal bipyramid, lone pairs occupy equatorial positions first.
Shape names describe atom positions, not lone pairs.

Three-dimensional bond conventions comparing common electron-pair-derived shapes and their bond angles.

Worked example

Predict the shape and bond angle of $\mathrm{NH_3}$ .

1.Nitrogen has three bonding pairs and one lone pair: four electron charge clouds.
2.The electron-pair arrangement is tetrahedral, but the atom positions form a trigonal pyramid.
3.Lone pair–bond pair repulsion compresses the angle below $109.5^{\circ}$ .

Answer: Trigonal pyramidal with a bond angle of about $107^{\circ}$ .

Common mistakes

Don't count a double or triple bond as more than one electron charge cloud.
Don't name the electron-pair arrangement instead of the molecular shape when lone pairs are present.
Don't use $109.5^{\circ}$ unchanged even though a lone pair compresses the bond angle.

Exam tip

A full shape answer states the number of bonding and lone pairs, the repulsion argument, the shape and the bond angle.

Tier 1 · Easy

State the shape and bond angle of $\mathrm{BF_3}$ around the boron atom.
[2 marks]
Total for this question: 2
A student counts each C=O double bond in $\mathrm{CO_2}$ as two electron charge clouds and predicts a tetrahedral molecule. Identify the error and give the correct shape and bond angle.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Use electron-pair repulsion theory to explain the shape of $\mathrm{NH_3}$ and its H-N-H bond angle of $107^{\circ}$ .
[4 marks]
Total for this question: 4
An unfamiliar molecule $\mathrm{XH_3}$ is observed to be trigonal pyramidal. Deduce the number of bonding pairs and lone pairs around X, state the electron-pair arrangement, and explain why its H-X-H angle is less than $109.5^{\circ}$ .
[3 marks]
Total for this question: 3
Predict the shape of $\mathrm{PCl_5}$ and state all principal Cl-P-Cl bond angles. State which bonds are axial and which are equatorial in your explanation.
[4 marks]
Total for this question: 4

Tier 3 · Hard

The ion $\mathrm{BrF_4^-}$ has four Br-F bonding pairs. Deduce the number of lone pairs on Br, the arrangement of all electron pairs, the molecular shape and the F-Br-F bond angles.
[5 marks]
Total for this question: 5
$\mathrm{SF_4}$ and $\mathrm{XeF_2}$ each have five electron charge clouds around the central atom. Deduce the number and preferred positions of the lone pairs in each species, then deduce both molecular shapes and the principal bond angles.
[6 marks]
Total for this question: 6
A student draws $\mathrm{SF_6}$ as a flat regular hexagon with F-S-F angles of $60^{\circ}$ . Use electron-pair repulsion theory to correct the model. State the number of bonding pairs and lone pairs, name the shape, give its bond angles, and describe a correct three-dimensional drawing.
[5 marks]
Total for this question: 5
$\mathrm{IF_5}$ has six electron charge clouds around the iodine atom, five of them bonding pairs. Deduce the number of lone pairs, the molecular shape and the F-I-F bond angles, and explain why the angles between the axial and basal fluorine atoms are not exactly $90^{\circ}$ .
[5 marks]
Total for this question: 5
Sulfur dioxide has two S-O bonding regions and one lone pair around sulfur; each S=O bond counts as one electron charge cloud. Deduce the electron-pair arrangement, molecular shape and approximate O-S-O angle. Explain the difference from the O-C-O angle in carbon dioxide.
[5 marks]
Total for this question: 5

3.1.3.6 · Bond polarity

Explanation

Electronegativity is the power of an atom to attract the bonding pair of electrons in a covalent bond.
A difference in electronegativity produces an unequal electron distribution: the more electronegative atom is $\delta^-$ and the other is $\delta^+$ .
A bond dipole has both magnitude and direction.
To decide whether a whole molecule has a permanent dipole, identify every polar bond and use the molecular shape to combine their dipoles.
Equal bond dipoles arranged symmetrically cancel, so a molecule can contain polar bonds but have no overall permanent dipole.

Worked example

Explain why $\mathrm{CO_2}$ has polar bonds but no permanent dipole.

1.Oxygen is more electronegative than carbon, so each $\mathrm{C=O}$ bond is polar.
2. $\mathrm{CO_2}$ is linear, so its equal bond dipoles act in opposite directions and cancel.

Answer: $\mathrm{CO_2}$ has no resultant permanent dipole.

Common mistakes

Don't assume that any molecule containing a polar bond must be polar overall.
Don't draw $\delta^-$ on the less electronegative atom.

Exam tip

A polarity explanation must combine electronegativity difference with the molecule's three-dimensional symmetry.

Tier 1 · Easy

The electronegativities of hydrogen and chlorine are $2.2$ and $3.2$ , respectively. Add the correct partial charge to each atom in an $\mathrm{H-Cl}$ bond.
[1 mark]
Total for this question: 1
The electronegativities of H, C, N and O are $2.2$ , $2.5$ , $3.0$ and $3.5$ , respectively. Deduce the order of increasing polarity for the $\mathrm{C-H}$ , $\mathrm{N-H}$ and $\mathrm{O-H}$ bonds and state the electronegativity difference for each bond.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Both $\mathrm{CO_2}$ and $\mathrm{SO_2}$ contain polar bonds. Explain why only one of these molecules has a permanent dipole.
[4 marks]
Total for this question: 4
A student draws trigonal planar $\mathrm{BF_3}$ with $\delta^-$ on boron, $\delta^+$ on each fluorine atom and a permanent dipole. Identify the two errors and give the correct polarity of the molecule.
[4 marks]
Total for this question: 4
The electronegativities of H, O and F are $2.2$ , $3.5$ and $4.0$ , respectively. For bent $\mathrm{HOF}$ , state the direction of the $\mathrm{H-O}$ and $\mathrm{O-F}$ bond dipoles and explain why the molecule has a permanent dipole.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Fluorine is more electronegative than carbon. Compare the polarity of tetrahedral $\mathrm{CF_4}$ and tetrahedral $\mathrm{CH_3F}$ , referring to bond dipoles and molecular shape.
[5 marks]
Total for this question: 5
Two planar isomers of 1,2-dichloroethene are tested. In isomer X, both $\mathrm{C-Cl}$ bonds project towards one side of the double-bond axis; its dipole moment is non-zero. In isomer Y, the $\mathrm{C-Cl}$ bonds project towards different sides; its dipole moment is zero. Explain these results in terms of bond dipoles and identify X and Y as the E or Z isomer.
[5 marks]
Total for this question: 5
Molecules M and N both have formula $\mathrm{XCl_2}$ and contain two identical polar $\mathrm{X-Cl}$ bonds. M has zero dipole moment, whereas N has a non-zero dipole moment. Deduce the shape of each molecule and what the result for N implies about lone pairs on X.
[5 marks]
Total for this question: 5
A linear molecule A-B-C has electronegativities A $=3.5$ , B $=2.5$ and C $=3.0$ . The A-B and B-C bond-dipole magnitudes are $1.4$ and $0.6$ arbitrary units, respectively. Determine the partial charges in both bonds and the magnitude and direction of the resultant molecular dipole.
[4 marks]
Total for this question: 4
Bond-polarity measurements show $\mathrm{X^{\delta+}-Y^{\delta-}}$ and $\mathrm{Y^{\delta+}-Z^{\delta-}}$ . Molecule XYZ is bent and the two bond dipoles have different magnitudes. Deduce the electronegativity order and determine whether XYZ has a permanent dipole.
[4 marks]
Total for this question: 4

3.1.3.7 · Forces between molecules

Explanation

Induced dipole–dipole forces act between all atoms and molecules. Their strength generally increases with electron number and with molecular surface contact.
Permanent dipole–dipole attractions also act between polar molecules.
Hydrogen bonding requires hydrogen covalently bonded to nitrogen, oxygen or fluorine and a lone pair on nitrogen, oxygen or fluorine in another molecule.
Boiling-point comparisons must identify the forces present and compare their overall strength, not simply molecular mass.
In ice, hydrogen bonds hold water molecules in an open arrangement, making ice less dense than liquid water; the bonds are not described as inherently stronger.

Worked example

Explain why ethanol has a higher boiling point than ethane.

1.Both molecules experience induced dipole–dipole forces.
2.Ethanol also forms hydrogen bonds because it contains an $\mathrm{O-H}$ bond and oxygen lone pairs.
3.More energy is needed to overcome the stronger intermolecular attractions in ethanol.

Answer: Ethanol has the higher boiling point because it forms hydrogen bonds.

Common mistakes

Don't call a covalent $\mathrm{O-H}$ bond a hydrogen bond.
Don't explain the low density of ice by claiming that its hydrogen bonds are stronger than those in water.

Exam tip

For a boiling-point comparison, name every force present and then identify the strongest difference between the substances.

Tier 1 · Easy

State the strongest type of intermolecular force between $\mathrm{CH_3Cl}$ molecules.
[1 mark]
Total for this question: 1
Choose the molecule that can form hydrogen bonds with other molecules of the same substance: $\mathrm{CH_4}$ , $\mathrm{CH_3F}$ or $\mathrm{CH_3OH}$ . Identify the two structural features that allow this hydrogen bonding.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Explain why ethane, $\mathrm{C_2H_6}$ , has a higher boiling point than methane, $\mathrm{CH_4}$ .
[3 marks]
Total for this question: 3
The boiling points of pentane, 2-methylbutane and 2,2-dimethylpropane are $36.1\,^{\circ}\mathrm{C}$ , $27.8\,^{\circ}\mathrm{C}$ and $9.5\,^{\circ}\mathrm{C}$ , respectively. Explain the trend even though the molecules have the same molecular formula.
[4 marks]
Total for this question: 4
Describe how to draw one hydrogen bond from a water molecule to an ammonia molecule, using water as the hydrogen-bond donor. Include the relevant partial charges, covalent bond, lone pair and intermolecular line.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Explain why water has a much higher boiling point than $\mathrm{H_2S}$ and why solid water is less dense than liquid water.
[5 marks]
Total for this question: 5
Propanone, $M_r=58.0$ , boils at $56\,^{\circ}\mathrm{C}$ , whereas propan-1-ol, $M_r=60.0$ , boils at $97\,^{\circ}\mathrm{C}$ . A student claims that their boiling points should be almost equal because their relative molecular masses are similar. Evaluate this claim.
[5 marks]
Total for this question: 5
A student claims that methoxymethane, $\mathrm{CH_3OCH_3}$ , cannot form hydrogen bonds in any mixture because it has no $\mathrm{O-H}$ bond. Evaluate the claim for pure methoxymethane and for methoxymethane mixed with water.
[5 marks]
Total for this question: 5
Propan-1-ol, propylamine and butane each contain 34 electrons. Their boiling points are 370 K, 321 K and 273 K. Explain the order in terms of the forces between their molecules.
[5 marks]
Total for this question: 5
HF has a much higher boiling point than HCl, but the boiling points then rise from HCl to HBr to HI. Explain both features of this trend.
[5 marks]
Total for this question: 5

3.1.4.1 · Enthalpy change

Explanation

Enthalpy change, $\Delta H$ , is the heat-energy change at constant pressure.
An exothermic process transfers heat to the surroundings and has $\Delta H<0$ ; an endothermic process absorbs heat and has $\Delta H>0$ .
A standard enthalpy change uses substances in their standard states at $100\,\mathrm{kPa}$ and a stated temperature.
Standard enthalpy of combustion is the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions.
Standard enthalpy of formation is the change when one mole of a compound forms from its constituent elements in their standard states.

Worked example

Write the formation equation to which $\Delta_\mathrm{f}H^\circ$ for liquid water refers.

1.Use elements in their standard states: $\mathrm{H_2(g)}$ and $\mathrm{O_2(g)}$ .
2.Form exactly one mole of $\mathrm{H_2O(l)}$ , using a fractional oxygen coefficient if needed.

Answer: $\mathrm{H_2(g)+\tfrac12O_2(g)\rightarrow H_2O(l)}$ .

Common mistakes

Don't define formation enthalpy for producing more than one mole of compound.
Don't omit standard states or use monatomic gaseous elements instead of their standard states.

Exam tip

In an enthalpy-definition question, the one-mole condition, standard states and standard conditions are separate marking points.

Tier 1 · Easy

A reaction transfers $38\,\text{kJ}$ of heat from the reacting chemicals to the surroundings. State whether the reaction is exothermic or endothermic and give the sign of $\Delta H$ .
[2 marks]
Total for this question: 2
The products of a reaction are $42\,\text{kJ mol}^{-1}$ lower in enthalpy than the reactants. A student records $\Delta H=+42\,\text{kJ mol}^{-1}$ . Identify the error and give the correct value.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Define standard enthalpy of formation and standard enthalpy of combustion. Include the amount of substance and the required conditions in each definition.
[4 marks]
Total for this question: 4
Identify which equation represents a standard enthalpy of formation, which represents a standard enthalpy of combustion and which represents neither as written: A $\mathrm{2Na(s)+Cl_2(g)\rightarrow2NaCl(s)}$ ; B $\mathrm{Na(s)+\tfrac12Cl_2(g)\rightarrow NaCl(s)}$ ; C $\mathrm{CO(g)+\tfrac12O_2(g)\rightarrow CO_2(g)}$ . Explain each choice.
[5 marks]
Total for this question: 5
At $298\,\text{K}$ and $100\,\text{kPa}$ , element M is a solid, element Q exists as $\mathrm{Q_2(g)}$ , and compound $\mathrm{MQ_2}$ has melting point $210\,\text{K}$ and boiling point $330\,\text{K}$ . Write the equation, including state symbols, for the standard enthalpy of formation of $\mathrm{MQ_2}$ at $298\,\text{K}$ and justify the product state.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Complete combustion of $0.250\,\text{mol}$ of a liquid releases $222\,\text{kJ}$ under standard conditions. Calculate its standard enthalpy of combustion.
[3 marks]
Total for this question: 3
A reaction with $\Delta H=-55.0\,\text{kJ mol}^{-1}$ releases $2.20\,\text{kJ}$ when a sample of X, $M_\mathrm{r}=46.0$ , reacts completely. Calculate the mass of X that reacted.
[3 marks]
Total for this question: 3
For the thermochemical equation $\mathrm{2X(g)\rightarrow Y(g)}$ , $\Delta H=-164\,\text{kJ mol}^{-1}$ as written. A $12.0\,\text{g}$ sample of Y, $M_\mathrm{r}=48.0$ , decomposes completely to X. Calculate the heat transferred and state whether it is absorbed or released.
[4 marks]
Total for this question: 4
The equation $\mathrm{2A+3B\rightarrow C}$ has $\Delta H=-240\,\text{kJ mol}^{-1}$ as written. A mixture contains $0.500\,\text{mol}$ of A and $0.900\,\text{mol}$ of B and reacts completely as far as possible. Determine the limiting reactant, the heat transferred, the amount of C formed and the amount of excess reactant remaining.
[5 marks]
Total for this question: 5
State the thermochemical equation, with state symbols, that defines the standard enthalpy of combustion of ethane. Explain why the fractional oxygen coefficient in this equation must not be replaced by whole-number coefficients.
[4 marks]
Total for this question: 4

3.1.4.2 · Calorimetry

Explanation

Calorimetry estimates heat transfer using $q=mc\Delta T$, where $m$ is the mass undergoing the temperature change, $c$ is its specific heat capacity and $\Delta T$ is the measured temperature change.
In solution calorimetry, total solution volume and density commonly estimate mass, while the limiting reagent supplies the reacting amount.
Convert heat to a molar enthalpy using $\Delta H=-q/n$ when a temperature rise shows that the reaction released heat.
Required practical 2 measures an enthalpy change, for example by dissolution, neutralisation, displacement or alcohol combustion.
Heat loss, incomplete combustion, evaporation and heating the apparatus create systematic differences that must be evaluated.

Worked example

$50.0\,\mathrm{g}$ of solution warms by $6.20\,\mathrm{K}$ . Calculate $q$ using $c=4.18\,\mathrm{J\,g^{-1}\,K^{-1}}$ .

1.Substitute into $q=mc\Delta T$ .
2. $q=50.0\times4.18\times6.20=1295.8\,\mathrm{J}$ .

Answer: $q=1.30\,\mathrm{kJ}$ to three significant figures.

Common mistakes

Don't use only the solute mass rather than the mass of solution that changes temperature.
Don't report a positive reaction enthalpy after a temperature rise without applying $\Delta H=-q/n$ .

Exam tip

For an uncertainty evaluation, state the direction in which the named heat-transfer error changes the calculated $\Delta H$ .

Tier 1 · Easy

A $100\,\text{g}$ sample of water warms by $6.5\,\text{K}$ . Use $c=4.18\,\text{J g}^{-1}\text{K}^{-1}$ to calculate the heat gained by the water.
[2 marks]
Total for this question: 2
A student compares four cup-and-lid designs for measuring an enthalpy change by mixing an acid with an alkali. State the dependent variable and two variables that should be controlled.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Equal $50.0\,\text{cm}^3$ portions of hydrochloric acid and sodium hydroxide are combined. Each solution has concentration $1.00\,\text{mol dm}^{-3}$ , and the temperature rises by $6.80\,\text{K}$ . Assume density $=1.00\,\text{g cm}^{-3}$ and $c=4.18\,\text{J g}^{-1}\text{K}^{-1}$ . Calculate the enthalpy change per mole of water formed.
[5 marks]
Total for this question: 5
Four repeats of a calorimetry experiment give temperature rises of $6.3$ , $6.4$ , $8.9$ and $6.2\,\text{K}$ . Identify the anomalous result, calculate a suitable mean temperature rise and explain why repeats improve the result.
[4 marks]
Total for this question: 4
An endothermic solid-dissolving experiment uses $45.0\,\text{g}$ of water and $5.0\,\text{g}$ of a solid that dissolves completely. A student uses only $45.0\,\text{g}$ as $m$ in $q=mc\Delta T$ . Assume the solution has the same specific heat capacity as water. Explain how this affects the calculated $q$ and molar enthalpy of solution.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Burning $0.720\,\text{g}$ of propan-1-ol, $\mathrm{C_3H_8O}$ , heats $250\,\text{g}$ of water by $18.4\,\text{K}$ . Use $c=4.18\,\text{J g}^{-1}\text{K}^{-1}$ and $M_r(\mathrm{C_3H_8O})=60.0$ . Calculate the experimental enthalpy of combustion and suggest one reason its magnitude is lower than the accepted value.
[6 marks]
Total for this question: 6
In an RP₂ calorimetry experiment, solutions are mixed at $3.0\,\text{min}$ . Before mixing, the temperatures at $0.0$ , $1.0$ and $2.0\,\text{min}$ are $20.30$ , $20.20$ and $20.10\,^{\circ}\text{C}$ . After mixing, the temperatures at $4.0$ , $5.0$ and $6.0\,\text{min}$ are $24.20$ , $24.00$ and $23.80\,^{\circ}\text{C}$ . (a) Extrapolate both linear trends to $3.0\,\text{min}$ and use a solution mass of $100.0\,\text{g}$ , $c=4.18\,\text{J g}^{-1}\text{K}^{-1}$ and an amount reacted of $0.0400\,\text{mol}$ to calculate the corrected molar enthalpy change. (b) Explain why the trends are extrapolated.
[7 marks]
Total for this question: 7
In a solution calorimetry experiment, $80.0\,\text{g}$ of solution warms by $5.60\,\text{K}$ . The solution has specific heat capacity $4.18\,\text{J g}^{-1}\text{K}^{-1}$ , and the cup and temperature probe together have heat capacity $32.0\,\text{J K}^{-1}$ . If $0.0400\,\text{mol}$ reacts, calculate the molar enthalpy change after allowing for energy absorbed by the apparatus. Calculate the value obtained if the apparatus is ignored and evaluate the bias.
[6 marks]
Total for this question: 6
$40.0\,\text{g}$ of one solution at $18.00\,^{\circ}\text{C}$ is mixed with $60.0\,\text{g}$ of another solution at $22.00\,^{\circ}\text{C}$ . The solutions have the same specific heat capacity, $4.18\,\text{J g}^{-1}\text{K}^{-1}$ , and their reaction forms $0.0500\,\text{mol}$ of product. The maximum temperature is $26.80\,^{\circ}\text{C}$ . Calculate the molar enthalpy change, allowing for the different initial temperatures. Assume no heat is lost to the surroundings and that the apparatus absorbs a negligible amount of heat.
[4 marks]
Total for this question: 4
A calorimetry result uses $100.0\,\text{g}$ of solution, $c=4.18\,\text{J g}^{-1}\text{K}^{-1}$ , $0.0400\,\text{mol}$ reacted and a measured rise of $5.00\pm0.10\,\text{K}$ . An independent reference value is $-50.0\,\text{kJ mol}^{-1}$ . Calculate the range of molar enthalpies allowed by the stated temperature uncertainty and evaluate whether that uncertainty explains the difference.
[5 marks]
Total for this question: 5

3.1.4.3 · Applications of Hess's law

Explanation

Hess's law states that the enthalpy change for a reaction is independent of the route, provided the initial and final states are identical. A Hess cycle replaces an unknown route with known enthalpy changes.
With formation enthalpies, $\Delta_\mathrm{r}H^\circ=\sum\Delta_\mathrm{f}H^\circ(\text{products})-\sum\Delta_\mathrm{f}H^\circ(\text{reactants})$ .
With combustion enthalpies leading to common products, use the total for reactants minus the total for products.
Every equation coefficient multiplies its enthalpy value.
Reversing a chemical equation reverses the sign of $\Delta H$ ; the algebra must follow the arrows in the cycle.

Worked example

Calculate $\Delta_\mathrm{r}H^\circ$ for $\mathrm{C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)}$ from formation enthalpies $+52$ , $0$ and $-85\,\mathrm{kJ\,mol^{-1}}$ respectively.

1.Use products minus reactants.
2. $\Delta_\mathrm{r}H^\circ=-85-[52+0]$ .

Answer: $\Delta_\mathrm{r}H^\circ=-137\,\mathrm{kJ\,mol^{-1}}$ .

Common mistakes

Don't use reactants minus products with formation enthalpies.
Don't reverse a chemical equation without reversing the sign of its enthalpy change.

Exam tip

Write the complete products-minus-reactants or reactants-minus-products expression before entering numerical values.

Tier 1 · Easy

For the changes $\mathrm{A\rightarrow B}$ and $\mathrm{B\rightarrow C}$ , the enthalpy changes are $+25\,\text{kJ mol}^{-1}$ and $-80\,\text{kJ mol}^{-1}$ . Calculate the enthalpy change for $\mathrm{A\rightarrow C}$ .
[1 mark]
Total for this question: 1
For $\mathrm{A\rightarrow B}$ , $\Delta H=+48\,\text{kJ mol}^{-1}$ . A Hess cycle uses the step $\mathrm{B\rightarrow A}$ but labels it $+48\,\text{kJ mol}^{-1}$ . Identify the error and give the correct label.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Use $\Delta_\mathrm{f}H^\circ[\mathrm{SO_2(g)}]=-297\,\text{kJ mol}^{-1}$ and $\Delta_\mathrm{f}H^\circ[\mathrm{SO_3(g)}]=-396\,\text{kJ mol}^{-1}$ to calculate $\Delta_\mathrm{r}H^\circ$ for $\mathrm{SO_2(g)+\tfrac12O_2(g)\rightarrow SO_3(g)}$ .
[2 marks]
Total for this question: 2
Use the equations $\mathrm{C(s)+O_2(g)\rightarrow CO_2(g)}$ , $\Delta H=-394\,\text{kJ mol}^{-1}$ , and $\mathrm{CO(g)+\tfrac12O_2(g)\rightarrow CO_2(g)}$ , $\Delta H=-283\,\text{kJ mol}^{-1}$ , to calculate $\Delta H$ for $\mathrm{C(s)+\tfrac12O_2(g)\rightarrow CO(g)}$ . Make the direction of each Hess step explicit.
[4 marks]
Total for this question: 4
For $\mathrm{2H_2S(g)+3O_2(g)\rightarrow2H_2O(l)+2SO_2(g)}$ , the standard enthalpies of formation of $\mathrm{H_2S(g)}$ , $\mathrm{H_2O(l)}$ and $\mathrm{SO_2(g)}$ are $-21$ , $-286$ and $-297\,\text{kJ mol}^{-1}$ , respectively. Calculate the standard enthalpy change of reaction.
[4 marks]
Total for this question: 4

Tier 3 · Hard

The standard enthalpies of combustion of $\mathrm{C_3H_6(g)}$ , $\mathrm{H_2(g)}$ and $\mathrm{C_3H_8(g)}$ are $-2058$ , $-286$ and $-2220\,\text{kJ mol}^{-1}$ , respectively. Use Hess's law to calculate the enthalpy change for $\mathrm{C_3H_6(g)+H_2(g)\rightarrow C_3H_8(g)}$ .
[3 marks]
Total for this question: 3
For $\mathrm{CaCO_3(s)\rightarrow CaO(s)+CO_2(g)}$ , $\Delta_\mathrm{r}H^\circ=+178\,\text{kJ mol}^{-1}$ . The standard enthalpies of formation of $\mathrm{CaO(s)}$ and $\mathrm{CO_2(g)}$ are $-635$ and $-394\,\text{kJ mol}^{-1}$ , respectively. Calculate the standard enthalpy of formation of $\mathrm{CaCO_3(s)}$ .
[4 marks]
Total for this question: 4
For $\mathrm{X(g)+H_2(g)\rightarrow Y(g)}$ , $\Delta H=-92\,\text{kJ mol}^{-1}$ . The standard enthalpies of combustion of $\mathrm{H_2(g)}$ and $\mathrm{Y(g)}$ are $-286$ and $-1510\,\text{kJ mol}^{-1}$ , respectively. Use a combustion Hess cycle to calculate the standard enthalpy of combustion of X.
[4 marks]
Total for this question: 4
Determine $\Delta H$ for $\mathrm{2A+2B\rightarrow D}$ from $\mathrm{A+B\rightarrow C}$ , $\Delta H=-72\,\text{kJ mol}^{-1}$ , and $\mathrm{2C\rightarrow D}$ , $\Delta H=+40\,\text{kJ mol}^{-1}$ . Write the equation combination.
[4 marks]
Total for this question: 4
A Hess network contains these changes: $\mathrm{A\rightarrow B}$ , $\Delta H=+35\,\text{kJ mol}^{-1}$ ; $\mathrm{B\rightarrow C}$ , $\Delta H=-82\,\text{kJ mol}^{-1}$ ; and $\mathrm{A\rightarrow D}$ , $\Delta H=-20\,\text{kJ mol}^{-1}$ . Determine $\Delta H$ for $\mathrm{D\rightarrow C}$ and for $\mathrm{C\rightarrow A}$ .
[5 marks]
Total for this question: 5

3.1.4.4 · Bond enthalpies

Explanation

Mean bond enthalpy is the mean energy required to break one mole of a specified covalent bond in gaseous molecules.
Bond breaking is endothermic and bond formation is exothermic, so estimate a gaseous reaction enthalpy using $\Delta H=\sum E(\text{bonds broken})-\sum E(\text{bonds formed})$ .
Count every bond in the displayed structures, include equation coefficients, and cancel unchanged bonds only after the full count is secure.
The result is approximate because a mean bond enthalpy averages that bond across different molecular environments; it is not the exact energy for one particular molecule.

Worked example

Estimate $\Delta H$ for $\mathrm{H_2+Cl_2\rightarrow2HCl}$ using $E(\mathrm{H-H})=436$ , $E(\mathrm{Cl-Cl})=243$ and $E(\mathrm{H-Cl})=432\,\mathrm{kJ\,mol^{-1}}$ .

1.Bonds broken: one H–H and one Cl–Cl, giving $436+243=679$ .
2.Bonds formed: two H–Cl, giving $2\times432=864$ .
3. $\Delta H=679-864$ .

Answer: $\Delta H=-185\,\mathrm{kJ\,mol^{-1}}$ .

Common mistakes

Don't add bond formation energies instead of subtracting them.
Don't count one product H–Cl bond although the equation forms two moles of HCl.

Exam tip

Draw or inspect every displayed bond and present separate totals for bonds broken and bonds formed.

Tier 1 · Easy

Define the term mean bond enthalpy.
[2 marks]
Total for this question: 2
A student assigns $-413\,\text{kJ mol}^{-1}$ to breaking a mole of $\mathrm{C-H}$ bonds and $+413\,\text{kJ mol}^{-1}$ to forming them. Identify both sign errors.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Use the mean bond enthalpies $E(\mathrm{H-H})=436$ , $E(\mathrm{Cl-Cl})=243$ and $E(\mathrm{H-Cl})=431\,\text{kJ mol}^{-1}$ to estimate $\Delta H$ for $\mathrm{H_2(g)+Cl_2(g)\rightarrow2HCl(g)}$ .
[3 marks]
Total for this question: 3
For $\mathrm{CH_4(g)+Cl_2(g)\rightarrow CH_3Cl(g)+HCl(g)}$ , $\Delta H=-114\,\text{kJ mol}^{-1}$ . Use $E(\mathrm{C-H})=413$ , $E(\mathrm{Cl-Cl})=243$ and $E(\mathrm{H-Cl})=432\,\text{kJ mol}^{-1}$ to calculate the mean $\mathrm{C-Cl}$ bond enthalpy.
[4 marks]
Total for this question: 4
$8.50\,\text{g}$ of gaseous ammonia, $\mathrm{NH_3}$ , contains $0.500\,\text{mol}$ of molecules. Estimate the energy required to break every $\mathrm{N-H}$ bond in the sample using $E(\mathrm{N-H})=388\,\text{kJ mol}^{-1}$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

Estimate the enthalpy change for $\mathrm{C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)}$ using $E(\mathrm{C{=}C})=612$ , $E(\mathrm{H-H})=436$ , $E(\mathrm{C-C})=348$ and $E(\mathrm{C-H})=413\,\text{kJ mol}^{-1}$ . Explain why a value obtained from standard formation enthalpies may differ.
[4 marks]
Total for this question: 4
A student estimates $\Delta H$ for adding $\mathrm{HBr}$ to propene. The student breaks one $\mathrm{C{=}C}$ and one $\mathrm{H-Br}$ bond, then forms one $\mathrm{C-C}$ and one $\mathrm{C-Br}$ bond. Identify the omitted bond and correct the estimate using $E(\mathrm{C{=}C})=612$ , $E(\mathrm{H-Br})=366$ , $E(\mathrm{C-C})=348$ , $E(\mathrm{C-H})=413$ and $E(\mathrm{C-Br})=276\,\text{kJ mol}^{-1}$ .
[5 marks]
Total for this question: 5
The structures of butane and 2-methylpropane are $\mathrm{CH_3CH_2CH_2CH_3}$ and $\mathrm{(CH_3)_3CH}$ , respectively. Deduce the enthalpy change that mean bond enthalpies predict for $\mathrm{butane(g)\rightarrow2\text{-}methylpropane(g)}$ , then explain why accurate standard enthalpy data need not give exactly this value.
[4 marks]
Total for this question: 4
Estimate the enthalpy change for the gas-phase cracking reaction $\mathrm{C_4H_{10}(g)\rightarrow C_2H_6(g)+C_2H_4(g)}$ using $E(\mathrm{C-C})=348$ , $E(\mathrm{C=C})=612$ and $E(\mathrm{C-H})=413\,\text{kJ mol}^{-1}$ . State why the method would need an additional term if a liquid species were used.
[5 marks]
Total for this question: 5

3.1.5.1 · Collision theory

Explanation

Particles must collide before they can react. A collision is successful only when the particles have energy at least equal to the activation energy.
Activation energy is the minimum energy required for a reaction to occur following a collision between particles. Collision theory separates collision frequency from the fraction of collisions that are successful.
Concentration and pressure mainly alter collision frequency; temperature strongly alters the fraction above the activation energy.
Most collisions are unsuccessful because they lack sufficient energy, not because the particles failed to meet.
Explanations should identify which factor changes and how this affects successful collisions per second.

Worked example

Explain why powdered calcium carbonate reacts faster with acid than equal-mass marble chips at the same temperature.

1.Powder has a larger surface area exposed to the acid.
2.Acid particles collide with carbonate particles more frequently.
3.More successful collisions occur per second, so the reaction rate is greater.

Answer: The powdered solid reacts faster because its greater surface area increases collision frequency.

Common mistakes

Don't state that every collision causes a reaction.
Don't claim that larger surface area gives particles more energy.

Exam tip

A rate explanation should finish with the change in successful collisions per second.

Tier 1 · Easy

Define activation energy.
[2 marks]
Total for this question: 2
A reaction has activation energy $80\,\text{kJ mol}^{-1}$ . Collisions P, Q and R have energies of $72$ , $91$ and $84\,\text{kJ mol}^{-1}$ , respectively. Identify which collisions have sufficient energy to react and explain your choice.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Explain why most collisions between reactant particles do not produce a reaction.
[3 marks]
Total for this question: 3
For a reacting gas, the collision frequency rises from $5.0\times10^8$ to $1.0\times10^9\,\text{s}^{-1}$ while the fraction of collisions with energy at least equal to $E_\mathrm{a}$ is unchanged. Temperature and reaction route are unchanged. Deduce the type of condition change and predict the rate change.
[4 marks]
Total for this question: 4
Reactions P and Q have activation energies $50\,\text{kJ mol}^{-1}$ and $150\,\text{kJ mol}^{-1}$ at the same temperature and collision frequency. Deduce which reaction is faster and explain your answer in terms of the proportion of collisions with sufficient energy.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A reacting mixture undergoes $2.0\times10^{10}$ molecular collisions each second. A fraction $4.0\times10^{-6}$ have at least the activation energy. Assuming collisions with sufficient energy react, calculate the numbers of successful and unsuccessful collisions per second and explain the role of the activation energy.
[4 marks]
Total for this question: 4
At constant temperature, doubling a reactant concentration doubles the collision frequency. A results table also shows the fraction of collisions with energy at least equal to $E_\mathrm{a}$ increasing from $0.015$ to $0.030$ . Evaluate the consistency of the table and state the expected effect on successful collisions.
[4 marks]
Total for this question: 4
At temperature $T_1$ , a mixture undergoes $4.0\times10^9$ collisions each second and $1.20\times10^8$ are successful. The cumulative energy data show that the fractions of collisions with energy at least $40$ , $60$ , $80$ and $100\,\text{kJ mol}^{-1}$ are $0.16$ , $0.080$ , $0.030$ and $0.0090$ , respectively. Assume collision energy is the only success criterion. (a) Deduce the activation energy. (b) At a higher temperature $T_2$ , the fraction above this activation energy is $0.072$ and the collision frequency is $5.0\times10^9\,\text{s}^{-1}$ . Calculate the factor by which the successful-collision frequency increases.
[5 marks]
Total for this question: 5
At the same temperature, reaction R has $6.0\times10^9$ collisions per second and $1.2\%$ of them have sufficient energy. Reaction S has $3.0\times10^9$ collisions per second and $3.0\%$ have sufficient energy. Assume collision energy is the only success criterion. Determine which reaction is faster and calculate the factor between their successful-collision frequencies.
[4 marks]
Total for this question: 4

3.1.5.2 · Maxwell–Boltzmann distribution

Explanation

A Maxwell–Boltzmann distribution for a gas plots number of molecules against molecular energy. The curve starts at the origin, rises to a most-probable-energy peak, then approaches the energy axis without touching it because there is no maximum molecular energy.
The area under the curve represents the total number of molecules.
Equal samples at different temperatures therefore have equal total areas.
At higher temperature the peak becomes lower and moves right, the curve broadens, and a greater area lies beyond a fixed activation energy.
A sketch must preserve these features and label both axes and $E_\mathrm{a}$ .

Maxwell–Boltzmann distributions at lower and higher temperature for equal numbers of molecules.

Worked example

State three changes when a gas sample is heated while the number of molecules remains constant.

1.The distribution broadens and its peak becomes lower.
2.The most probable energy moves to a higher value.
3.The area beyond a fixed $E_\mathrm{a}$ increases while total area remains constant.

Answer: Lower, right-shifted peak; broader curve; greater fraction above $E_\mathrm{a}$ .

Common mistakes

Don't draw the high-energy tail meeting or crossing the energy axis.
Don't change the total area when comparing equal numbers of molecules.

Exam tip

On a distribution sketch, label the higher-temperature curve and shade or describe the area beyond $E_\mathrm{a}$ .

Tier 1 · Easy

State the quantity represented on each axis of a Maxwell–Boltzmann distribution for a gas.
[2 marks]
Total for this question: 2
A proposed Maxwell–Boltzmann curve starts above the origin and its high-energy tail crosses the energy axis. Identify the two corrections needed.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Describe how to add a higher-temperature curve to a Maxwell–Boltzmann distribution while keeping the number of gas molecules constant.
[4 marks]
Total for this question: 4
A student draws a broader, lower Maxwell–Boltzmann curve for the same gas sample at a higher temperature, but makes its total area $15\%$ larger than the original area. Explain the error and state how the curve should be corrected.
[3 marks]
Total for this question: 3
Two samples of the same gas are at the same temperature, but sample B contains twice as many molecules as sample A. The distributions are plotted as number of molecules against energy. Compare their Maxwell–Boltzmann curves, including peak position, shape and area.
[3 marks]
Total for this question: 3
A student states that the most probable energy on a Maxwell–Boltzmann distribution is given by the height of the peak. Identify the error and state how the most probable energy changes when the same sample is heated.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Two equal samples of the same gas are at temperatures $T_1$ and $T_2$ , where $T_2>T_1$ . A fixed energy $E^*$ lies in the high-energy tail, beyond the point where the two Maxwell–Boltzmann curves cross. Explain why the curves must cross and compare the fractions of molecules with energy greater than $E^*$ .
[5 marks]
Total for this question: 5
Two equal samples of the same gas give these energy-population data for a fixed high-energy threshold $E_\mathrm{a}$ . Sample A has $99\,000$ molecules below $E_\mathrm{a}$ and $1\,000$ at or above it. Sample B has $97\,000$ below $E_\mathrm{a}$ and $3\,000$ at or above it. Deduce which sample is hotter and describe how its full Maxwell–Boltzmann curve differs.
[4 marks]
Total for this question: 4
On one Maxwell–Boltzmann distribution, $86\%$ of molecules lie below a lower activation energy $E_{\mathrm{a,P}}$ , $9\%$ lie between $E_{\mathrm{a,P}}$ and a higher activation energy $E_{\mathrm{a,Q}}$ , and $5\%$ lie above $E_{\mathrm{a,Q}}$ . Calculate the fractions energetic enough for routes P and Q and compare their successful-collision frequencies if the collision frequency is otherwise identical.
[5 marks]
Total for this question: 5
For a sample of $200\,000$ molecules at $T_1$ , $72\%$ have energy below $E_1$ , $18\%$ lie from $E_1$ to below $E_2$ , and $10\%$ have energy at least $E_2$ . At $T_2$ , where the sample size is unchanged, the percentages are $60\%$ , $21\%$ and $19\%$ , respectively. Calculate the increase factors for the populations energetic enough to clear thresholds $E_1$ and $E_2$ , then determine which threshold population is proportionally more temperature-sensitive.
[6 marks]
Total for this question: 6

3.1.5.3 · Effect of temperature on reaction rate

Explanation

Rate of reaction is the change in concentration of a reactant or product per unit time, although another measured quantity may track reaction progress. Increasing temperature raises mean molecular energy and collision frequency.
More importantly, the Maxwell–Boltzmann distribution changes so a much greater fraction of molecules has energy at least equal to $E_\mathrm{a}$ .
Consequently, a much larger proportion of collisions is successful and the rate rises.
For an uncatalysed route, activation energy remains unchanged.
Required practical 3 investigates how reaction rate changes with temperature, commonly using an initial-rate method with sodium thiosulfate and hydrochloric acid, controlled quantities and repeat measurements.

Worked example

Explain, using a Maxwell–Boltzmann distribution, why a reaction is much faster at $40^{\circ}\mathrm{C}$ than at $20^{\circ}\mathrm{C}$ .

1.At the higher temperature the distribution is broader with a lower, right-shifted peak.
2.The area beyond the unchanged $E_\mathrm{a}$ is larger.
3.A greater fraction of collisions is successful, so rate increases.

Answer: Higher temperature increases the fraction of molecules with $E\geq E_\mathrm{a}$ .

Common mistakes

Don't state that heating lowers the activation energy.
Don't attribute the whole rate increase only to more frequent collisions.

Exam tip

Use the phrase 'greater proportion of molecules with energy at least equal to $E_\mathrm{a}$ ' in a temperature-rate explanation.

Tier 1 · Easy

State the effect of increasing temperature on the rate of a reaction, with all other conditions unchanged.
[1 mark]
Total for this question: 1
In RP₃, a student investigates the reaction of sodium thiosulfate with hydrochloric acid at different temperatures by timing the disappearance of a cross. State the independent variable, the measured dependent variable and two variables that should be controlled.
[4 marks]
Total for this question: 4

Tier 2 · Standard

Use a Maxwell–Boltzmann distribution to explain why a small temperature increase can cause a large increase in reaction rate.
[4 marks]
Total for this question: 4
An RP₃ experiment gives cross-disappearance times of $80.0\,\text{s}$ at $20\,^{\circ}\text{C}$ , $50.0\,\text{s}$ at $30\,^{\circ}\text{C}$ and $25.0\,\text{s}$ at $40\,^{\circ}\text{C}$ . Calculate $1/t$ at each temperature and describe the relationship shown by these data.
[4 marks]
Total for this question: 4
The same mass of a solid reacts with excess acid at two temperatures. The gas syringe is maintained at the same measurement temperature and pressure in both runs. Describe how the product-volume–time curve at the higher reaction temperature differs when both reactions produce the same final amount of gas.
[3 marks]
Total for this question: 3

Tier 3 · Hard

In the first $10.0\,\text{s}$ of a reaction, $8.4\,\text{cm}^3$ of gas forms at $25\,^{\circ}\text{C}$ and $14.1\,\text{cm}^3$ forms at $35\,^{\circ}\text{C}$ . Calculate the factor by which the average rate over this interval increases, then explain the change using molecular energies.
[5 marks]
Total for this question: 5
Magnesium reacts with excess hydrochloric acid at $20\,^{\circ}\text{C}$ and $35\,^{\circ}\text{C}$ . Each run uses $0.450\,\text{g}$ of magnesium ribbon. The acid volume is $50.0\,\text{cm}^3$ and its concentration is $1.00\,\text{mol dm}^{-3}$ . Method A records hydrogen volume against time with a $500\,\text{cm}^3$ gas syringe graduated every $2\,\text{cm}^3$ . Method B records mass against time on a balance reading to $0.001\,\text{g}$ . Using $A_\mathrm{r}(\mathrm{Mg})=24.3$ , $M_\mathrm{r}(\mathrm{H_2})=2.00$ and a molar gas volume of $24.0\,\text{dm}^3\,\text{mol}^{-1}$ , calculate the expected final hydrogen volume and the expected total mass loss, then evaluate both methods and recommend one for comparing initial rates at the two temperatures.
[6 marks]
Total for this question: 6
In an RP₃ comparison at $50.0\,^{\circ}\text{C}$ , a student places room-temperature reactants in a water bath, mixes them immediately and starts the timer. The mixture reaches $50.0\,^{\circ}\text{C}$ only after part of the reaction has occurred. Explain the direction of the error in the measured initial rate and give an improved procedure.
[5 marks]
Total for this question: 5
An RP₃ endpoint experiment gives $1/t$ values of $0.0196$ , $0.0200$ and $0.0204\,\text{s}^{-1}$ at $30\,^{\circ}\text{C}$ , and $0.0268$ , $0.0270$ and $0.0272\,\text{s}^{-1}$ at $35\,^{\circ}\text{C}$ . Calculate the mean rate measure at each temperature and the factor increase, then evaluate from the repeat ranges whether the observed increase is resolved by the data.
[5 marks]
Total for this question: 5
A student compares a sodium thiosulfate reaction at two temperatures. The cooler mixture is prepared from $30.0\,\text{cm}^3$ of $0.120\,\text{mol dm}^{-3}$ thiosulfate, $20.0\,\text{cm}^3$ of water and $12.0\,\text{cm}^3$ of $0.800\,\text{mol dm}^{-3}$ acid. For the hotter run, only the water portion is changed, to $8.0\,\text{cm}^3$ . Calculate both reactant concentrations after mixing in each run, evaluate the comparison and give a valid improvement.
[6 marks]
Total for this question: 6

3.1.5.4 · Effect of concentration and pressure

Explanation

Increasing solution concentration places more reactant particles in each unit volume, so particles collide more frequently.
Increasing the pressure of reacting gases by decreasing their volume likewise raises the number of particles per unit volume and collision frequency.
Provided temperature is constant, neither change alters the Maxwell–Boltzmann energy distribution, the particles' mean energy or the activation energy.
The fraction of collisions with sufficient energy therefore remains the same, but more collisions occur each second and hence more successful collisions occur each second.
Pressure arguments apply to gaseous reactants; changing pressure has little direct effect on the concentration of a solid.

Worked example

Explain why compressing a mixture of reacting gases at constant temperature increases reaction rate.

1.Compression places the same number of gas particles in a smaller volume.
2.Collision frequency increases while the energy distribution remains unchanged.
3.More successful collisions occur per second.

Answer: The rate increases because gas-particle collision frequency increases.

Common mistakes

Don't claim that higher pressure makes gas particles move faster at constant temperature.
Don't state that increased concentration lowers activation energy.

Exam tip

When temperature is fixed, explain concentration or pressure effects through particle density and collision frequency.

Tier 1 · Easy

State why increasing the concentration of a dissolved reactant usually increases reaction rate.
[2 marks]
Total for this question: 2
Choose the two changes that increase the number of reacting particles per unit volume without changing the temperature: increasing the concentration of a solution, compressing reacting gases, adding a catalyst, or heating the mixture.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A gaseous reacting mixture is compressed to half its original volume at constant temperature. Explain why its reaction rate increases.
[3 marks]
Total for this question: 3
At constant temperature, experiments with reactant X give the following data while every other concentration is fixed: $[\mathrm{X}]=0.10$ , $0.20$ and $0.40\,\text{mol dm}^{-3}$ ; initial rates $=0.80\times10^{-3}$ , $1.60\times10^{-3}$ and $3.20\times10^{-3}\,\text{mol dm}^{-3}\,\text{s}^{-1}$ . Describe the pattern and explain it using collision theory. Do not infer a rate equation.
[4 marks]
Total for this question: 4
$25.0\,\text{cm}^3$ of a $0.800\,\text{mol dm}^{-3}$ reactant solution is diluted with water to $100.0\,\text{cm}^3$ at constant temperature. Calculate the new concentration and explain the expected effect on its collision frequency with a second dissolved reactant whose concentration is unchanged.
[4 marks]
Total for this question: 4

Tier 3 · Hard

The concentration of one aqueous reactant is increased while temperature and all other concentrations are fixed. Explain why the rate changes, distinguishing the effect on collision frequency from any effect on activation energy or particle energy.
[5 marks]
Total for this question: 5
A reacting gas mixture is at constant temperature. In experiment A, helium is added at constant volume, increasing the total pressure. In experiment B, the original reacting mixture is compressed to half its volume. Compare the effects on reactant particle concentrations, collision frequency and reaction rate.
[5 marks]
Total for this question: 5
Helium is added to a reacting gas mixture in a cylinder fitted with a freely moving piston. Temperature and total pressure remain constant, so the piston moves outwards. Evaluate the claim that the reaction rate is unchanged simply because helium is inert.
[5 marks]
Total for this question: 5
A gas X reacts at the surface of a fixed mass of solid Y. At constant temperature, the same amount of X is compressed from $2.40$ to $1.50\,\text{dm}^3$ while the exposed area of Y is unchanged. Calculate the factor change in the concentration of X and explain the expected rate change.
[4 marks]
Total for this question: 4
Vessels P and Q each have a volume of $0.500\,\text{dm}^3$ and are at the same temperature. P contains $0.0400\,\text{mol}$ of X and $0.0600\,\text{mol}$ of Y. Q contains $0.0600\,\text{mol}$ of X and $0.0300\,\text{mol}$ of Y. Calculate the four concentrations. Then state the effect on the frequency of collisions between X and Y of the change in $[\mathrm{X}]$ alone and of the change in $[\mathrm{Y}]$ alone.
[5 marks]
Total for this question: 5

3.1.5.5 · Catalysts

Explanation

A catalyst increases reaction rate without being changed in chemical composition or amount by the overall reaction. It provides an alternative route with a lower activation energy.
At fixed temperature, the Maxwell–Boltzmann distribution is unchanged, but the lower $E_\mathrm{a}$ threshold leaves a larger area of molecules able to react, so a greater fraction of collisions is successful.
A catalyst does not change reactant or product enthalpies, $\Delta H$ , the equilibrium constant or the equilibrium position.
It accelerates both forward and reverse reactions and therefore allows equilibrium to be reached sooner.
An intermediate may form and later be regenerated.

Energy profiles for catalysed and uncatalysed routes with unchanged reactant and product enthalpies.

Worked example

Explain the effect of a catalyst on the energy profile and equilibrium of a reversible reaction.

1.The catalysed pathway has a lower activation energy in both directions.
2.Reactant and product energy levels, so $\Delta H$ , are unchanged.
3.Both rates increase equally at equilibrium, so the equilibrium position is unchanged.

Answer: Equilibrium is reached faster but its position and $K$ are unchanged.

Common mistakes

Don't state that a catalyst lowers the energy of every molecule.
Don't claim that a catalyst changes $\Delta H$ or moves the equilibrium position.

Exam tip

On a catalysed profile, keep reactant and product levels fixed and lower only the pathway peak.

Tier 1 · Easy

Define a catalyst.
[2 marks]
Total for this question: 2
A reaction occurs in two steps: $\mathrm{X+Y\rightarrow XY}$ and $\mathrm{XY+Z\rightarrow P+Y}$ . Identify the catalyst and the intermediate.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Use a Maxwell–Boltzmann distribution to explain how a catalyst increases the rate of a gas-phase reaction at constant temperature.
[4 marks]
Total for this question: 4
An energy-profile table gives reactants at $40\,\text{kJ mol}^{-1}$ and products at $10\,\text{kJ mol}^{-1}$ . Path P has a maximum at $155\,\text{kJ mol}^{-1}$ and path Q at $95\,\text{kJ mol}^{-1}$ . Identify the catalysed path and calculate its forward activation energy and $\Delta H$ .
[4 marks]
Total for this question: 4
For the same fixed amount of product, a reaction takes $240\,\text{s}$ without a solid additive and $120\,\text{s}$ with it. The additive has mass $1.250\,\text{g}$ before use and $1.249\,\text{g}$ after it is recovered, dried and reweighed on a balance with uncertainty $\pm0.001\,\text{g}$ . Its chemical composition is unchanged. Evaluate the claim that the additive cannot be a catalyst because its measured mass decreased.
[4 marks]
Total for this question: 4

Tier 3 · Hard

An exothermic reaction has $\Delta H=-35\,\text{kJ mol}^{-1}$ and an uncatalysed forward activation energy of $145\,\text{kJ mol}^{-1}$ . A catalyst lowers the forward activation energy to $82\,\text{kJ mol}^{-1}$ . Calculate the reverse activation energy for each route and state why $\Delta H$ is unchanged.
[5 marks]
Total for this question: 5
Two identical starting mixtures are kept at the same temperature. The uncatalysed reversible reaction reaches a constant product concentration of $0.640\,\text{mol dm}^{-3}$ after $300\,\text{s}$ . With a catalyst it reaches $0.640\,\text{mol dm}^{-3}$ after $80\,\text{s}$ . A student claims that the catalyst increases $K_\mathrm{c}$ . Evaluate the claim using the data.
[4 marks]
Total for this question: 4
An uncatalysed profile has reactants at $25$ , products at $70$ and a maximum at $185\,\text{kJ mol}^{-1}$ . Proposal Q keeps the two levels but lowers the maximum to $120\,\text{kJ mol}^{-1}$ . Proposal R lowers the maximum to $120$ but also moves the product level to $40\,\text{kJ mol}^{-1}$ . Calculate the forward and reverse activation energies for Q and evaluate which proposal can represent a catalyst.
[6 marks]
Total for this question: 6
Distinguish the changes shown on a Maxwell–Boltzmann distribution when a catalyst is added at constant temperature and when the same uncatalysed mixture is heated. State what changes, what stays fixed and why each change increases the fraction of molecules able to react.
[5 marks]
Total for this question: 5

3.1.6.1 · Chemical equilibria and Le Chatelier's principle

Explanation

At dynamic equilibrium in a closed system, forward and reverse reactions continue at equal rates, so reactant and product concentrations remain constant. Le Chatelier's principle predicts that the equilibrium position shifts to oppose a change.
Adding a species favours the direction that removes it. Higher pressure favours the side with fewer moles of gas and has no positional effect when gaseous mole totals are equal.
Higher temperature favours the endothermic direction.
A catalyst changes neither position nor equilibrium yield.
Industrial conditions balance equilibrium yield with rate, energy cost, pressure cost, safety and catalyst performance rather than maximising one factor.

Worked example

For $\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}$ , the forward reaction is exothermic. Predict the effects of higher pressure and higher temperature on equilibrium yield.

1.Higher pressure favours the side with fewer gas moles: two moles of ammonia rather than four reactant moles.
2.Higher temperature favours the endothermic reverse direction.

Answer: Higher pressure increases ammonia yield; higher temperature decreases it.

Common mistakes

Don't say equilibrium means both reactions have stopped.
Don't predict a pressure effect by counting solid or liquid species as gas moles.
Don't claim that a catalyst increases equilibrium yield.

Exam tip

For an industrial-condition question, separate the equilibrium-yield effect from the rate, cost and safety compromise.

Tier 1 · Easy

State two features of a reversible reaction at dynamic equilibrium.
[2 marks]
Total for this question: 2
For $\mathrm{A\rightleftharpoons B}$ in a closed vessel, the concentrations $[\mathrm{A}]$ and $[\mathrm{B}]$ are $0.55$ and $0.45\,\text{mol dm}^{-3}$ at $40\,\text{s}$ , then $0.50$ and $0.50\,\text{mol dm}^{-3}$ at both $60$ and $80\,\text{s}$ . State the earliest time at which the equilibrium composition is recorded. Explain why the equilibrium is dynamic.
[3 marks]
Total for this question: 3

Tier 2 · Standard

For $\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}$ , the forward reaction is exothermic. Predict the effect on the equilibrium yield of ammonia of increasing pressure, increasing temperature and adding a catalyst.
[3 marks]
Total for this question: 3
An industrial plant considers the exothermic hydration equilibrium $\mathrm{C_2H_4(g)+H_2O(g)\rightleftharpoons C_2H_5OH(g)}$ under otherwise identical feed conditions. Choice A uses $570\,\text{K}$ and $2.0\times10^4\,\text{kPa}$ , giving $18\%$ equilibrium conversion but a lower rate and high compression cost. Choice B uses $640\,\text{K}$ and $8.0\times10^3\,\text{kPa}$ , giving $8\%$ equilibrium conversion but a higher rate and lower compression cost. Compare the choices and justify an industrial recommendation.
[6 marks]
Total for this question: 6
A closed system containing $\mathrm{A(g)\rightleftharpoons B(g)}$ is at equilibrium. Some B is removed suddenly without changing the temperature. Compare the forward and reverse rates immediately after removal, predict the shift, and state the rate relationship at the new equilibrium.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Sulfur trioxide is made by the exothermic equilibrium $\mathrm{2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)}$ . Explain why an industrial process may use a moderate temperature, a pressure that is not extremely high, and a catalyst.
[6 marks]
Total for this question: 6
For $\mathrm{A(g)\rightleftharpoons B(g)}$ at a fixed temperature, one equilibrium mixture has $[\mathrm{A}]=0.80$ and $[\mathrm{B}]=0.40\,\text{mol dm}^{-3}$ . Some B is added and the system re-establishes equilibrium with $[\mathrm{A}]=1.00$ and $[\mathrm{B}]=0.50\,\text{mol dm}^{-3}$ . Use the data to distinguish the change in equilibrium position from the value of $K_\mathrm{c}$ .
[5 marks]
Total for this question: 5
For the endothermic equilibrium $\mathrm{A(g)\rightleftharpoons2B(g)}$ , temperature is increased and pressure is increased by compression (a decrease in volume) at the same time. A student states that the equilibrium must shift right. Evaluate the conclusion and distinguish the effects on equilibrium position from any effect on the equilibrium constant.
[5 marks]
Total for this question: 5
For a gaseous equilibrium, separate experiments show that compression at constant temperature increases the equilibrium yield of product, while heating at constant pressure decreases it. Adding a catalyst changes only the time taken to reach the final composition. Deduce what can be concluded about gaseous mole numbers, the forward enthalpy change and the catalyst's effect.
[5 marks]
Total for this question: 5
The closed equilibrium $\mathrm{A_2(g)+B_2(g)\rightleftharpoons2AB(g)}$ has constant macroscopic concentrations. In one tracer experiment, a tiny amount of isotopically labelled $\mathrm{A_2}$ is introduced and labelled AB later appears. In a second experiment begun at the same equilibrium, labelled AB is introduced and labelled $\mathrm{A_2}$ later appears. Explain what the combined evidence demonstrates.
[5 marks]
Total for this question: 5

3.1.6.2 · Equilibrium constant Kc for homogeneous systems

Explanation

For $aA+bB\rightleftharpoons cC+dD$ , $K_\mathrm{c}=[C]^c[D]^d/([A]^a[B]^b)$ using equilibrium concentrations and equation coefficients as powers.
Calculate equilibrium amounts through the stoichiometric change, then divide by volume in $\mathrm{dm^3}$ before substitution.
Derive units from the net concentration powers; they can cancel.
At fixed temperature, changing concentration, pressure or adding a catalyst does not change $K_\mathrm{c}$ , although the equilibrium position may move.
Temperature changes $K_\mathrm{c}$ : heating increases it for an endothermic forward reaction and decreases it for an exothermic forward reaction.

Worked example

For $\mathrm{H_2+I_2\rightleftharpoons2HI}$ , equilibrium concentrations are $[\mathrm{H_2}]=0.20$ , $[\mathrm{I_2}]=0.30$ and $[\mathrm{HI}]=0.60\,\mathrm{mol\,dm^{-3}}$ . Calculate $K_\mathrm{c}$ .

1. $K_\mathrm{c}=\dfrac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}$ .
2. $K_\mathrm{c}=\dfrac{0.60^2}{0.20\times0.30}$ .

Answer: $K_\mathrm{c}=6.0$ with no units for this equation.

Common mistakes

Don't use initial concentrations in the $K_\mathrm{c}$ expression instead of equilibrium concentrations.
Don't omit equation coefficients as powers in the expression.
Don't state that a catalyst changes $K_\mathrm{c}$ .

Exam tip

Write the symbolic $K_\mathrm{c}$ expression and its units before substituting equilibrium values.

Tier 1 · Easy

Write the expression for $K_\mathrm{c}$ for $\mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)}$ .
[1 mark]
Total for this question: 1
For $\mathrm{2NO_2(g)\rightleftharpoons N_2O_4(g)}$ , a student writes $K_\mathrm{c}=[\mathrm{N_2O_4}]/2[\mathrm{NO_2}]$ . Identify the error and give the correct expression.
[3 marks]
Total for this question: 3

Tier 2 · Standard

At equilibrium, $[\mathrm{H_2}]=0.200\,\text{mol dm}^{-3}$ , $[\mathrm{I_2}]=0.300\,\text{mol dm}^{-3}$ and $[\mathrm{HI}]=1.20\,\text{mol dm}^{-3}$ for $\mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)}$ . Calculate $K_\mathrm{c}$ and state its units.
[3 marks]
Total for this question: 3
For $\mathrm{N_2O_4(g)\rightleftharpoons2NO_2(g)}$ , $K_\mathrm{c}=0.360\,\text{mol dm}^{-3}$ and the equilibrium concentration of $\mathrm{N_2O_4}$ is $0.250\,\text{mol dm}^{-3}$ . Calculate the equilibrium concentration of $\mathrm{NO_2}$ .
[3 marks]
Total for this question: 3
For $\mathrm{2A(g)+B(g)\rightleftharpoons C(g)}$ , write the expression for $K_\mathrm{c}$ and deduce its units from concentration units of $\text{mol dm}^{-3}$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

For the exothermic equilibrium $\mathrm{A(g)+B(g)\rightleftharpoons2C(g)}$ , a $2.00\,\text{dm}^3$ vessel initially contains $0.800\,\text{mol}$ of A, $0.600\,\text{mol}$ of B and no C. At equilibrium it contains $0.500\,\text{mol}$ of C. Calculate $K_\mathrm{c}$ and predict the effect of increasing temperature on its value.
[6 marks]
Total for this question: 6
At a fixed temperature, $K_\mathrm{c}=4.00$ for $\mathrm{CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)}$ . Initially, $[\mathrm{CO}]=[\mathrm{H_2O}]=0.600\,\text{mol dm}^{-3}$ and no products are present. Calculate all four equilibrium concentrations and state the units of $K_\mathrm{c}$ .
[6 marks]
Total for this question: 6
For $\mathrm{2P(g)+Q(g)\rightleftharpoons 2R(g)}$ , the equilibrium amounts in a $0.500\,\text{dm}^3$ vessel are $0.250\,\text{mol}$ of P, $0.200\,\text{mol}$ of Q and $0.400\,\text{mol}$ of R. Calculate $K_\mathrm{c}$ and state its units.
[4 marks]
Total for this question: 4
For $\mathrm{A(g)+B(g)\rightleftharpoons C(g)}$ , the equilibrium amounts are $0.200\,\text{mol}$ of A, $0.300\,\text{mol}$ of B and $0.120\,\text{mol}$ of C. At this temperature, $K_\mathrm{c}=4.00\,\text{mol}^{-1}\text{ dm}^3$ . Determine the vessel volume and the three equilibrium concentrations.
[5 marks]
Total for this question: 5
At a fixed temperature, $K_\mathrm{c}=0.0400\,\text{mol}^{-1}\text{ dm}^3$ for the exothermic equilibrium $\mathrm{2A(g)\rightleftharpoons B(g)}$ . Determine $K_\mathrm{c}$ and its units for the reverse equation and for the equation with every coefficient doubled. State how heating affects the constant for the original and reverse equations.
[6 marks]
Total for this question: 6

3.1.7 · Oxidation, reduction and redox equations

Explanation

Redox reactions transfer electrons from the reducing agent to the oxidising agent.
Oxidation is electron loss, so a reducing agent donates electrons and is oxidised; reduction is electron gain, so an oxidising agent accepts electrons and is reduced.
Oxidation states identify which element changes: values sum to zero in a neutral compound and to the overall charge in an ion, while an uncombined element has oxidation state zero.
A redox equation is constructed from separate oxidation and reduction half-equations, multiplied until electron numbers match, then added with electrons cancelled.
Both atoms and total charge must balance.

Worked example

Combine $\mathrm{Fe^{2+}\rightarrow Fe^{3+}+e^-}$ and $\mathrm{Cl_2+2e^-\rightarrow2Cl^-}$ into an overall redox equation.

1.Multiply the iron half-equation by $2$ so that it releases $2e^-$ .
2.Add the half-equations and cancel the two electrons.

Answer: $\mathrm{2Fe^{2+}+Cl_2\rightarrow2Fe^{3+}+2Cl^-}$ .

Common mistakes

Don't call the electron donor the oxidising agent, although donating electrons makes it the reducing agent.
Don't balance atoms in two half-equations but add them without first making the numbers of transferred electrons equal.
Don't make oxidation states in an ion sum to zero instead of to the ion's overall charge.

Exam tip

For a 'write the overall redox equation' question, show balanced half-equations and the electron multiplier before cancellation.

Tier 1 · Easy

Determine the oxidation state of manganese in $\mathrm{MnO_4^-}$ .
[2 marks]
Total for this question: 2
In the reaction $\mathrm{Cl_2+2Br^-\rightarrow2Cl^-+Br_2}$ , identify the oxidising agent and the reducing agent.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Combine the half-equations $\mathrm{Fe^{2+}\rightarrow Fe^{3+}+e^-}$ and $\mathrm{Cl_2+2e^-\rightarrow2Cl^-}$ to give the overall redox equation.
[3 marks]
Total for this question: 3
Complete the reduction half-equation $\mathrm{NO_3^-\rightarrow NO}$ in acidic solution. Include electrons, $\mathrm{H^+}$ and $\mathrm{H_2O}$ as required.
[3 marks]
Total for this question: 3
The overall reaction is $\mathrm{2Fe^{3+}+H_2O_2\rightarrow2Fe^{2+}+O_2+2H^+}$ . The oxidation half-equation is $\mathrm{H_2O_2\rightarrow O_2+2H^++2e^-}$ . Deduce the reduction half-equation and identify the oxidising agent.
[3 marks]
Total for this question: 3

Tier 3 · Hard

In acidic solution, dichromate(VI) ions oxidise $\mathrm{Sn^{2+}}$ ions to $\mathrm{Sn^{4+}}$ . Construct the overall ionic equation.
[5 marks]
Total for this question: 5
A student proposes $\mathrm{2MnO_4^-+5H_2O_2+16H^+\rightarrow2Mn^{2+}+5O_2+8H_2O}$ for the oxidation of hydrogen peroxide by acidified manganate(VII). Show that the equation is not balanced and write the corrected overall ionic equation.
[5 marks]
Total for this question: 5
In acidic solution, iodate ions, $\mathrm{IO_3^-}$ , react with iodide ions to form iodine. Use oxidation states to deduce the electron-transfer ratio, then construct the balanced overall ionic equation.
[5 marks]
Total for this question: 5
An oxide, $\mathrm{M_3O_4}$ , contains M only in the $+2$ and $+3$ oxidation states. In acid solution it oxidises iodide ions while every M atom forms $\mathrm{M^{2+}}$ . Deduce the numbers of $\mathrm{M^{2+}}$ and $\mathrm{M^{3+}}$ ions represented in one formula unit, then write the balanced overall ionic equation.
[4 marks]
Total for this question: 4
In acid solution, one mole of $\mathrm{XO_4^-}$ reacts with $2.50$ moles of $\mathrm{SO_3^{2-}}$ . Sulfur forms $\mathrm{SO_4^{2-}}$ , and X forms a single monatomic ion. Deduce the charge on the X ion and construct the balanced overall ionic equation.
[5 marks]
Total for this question: 5

3.1.8.1 · Born–Haber cycles (A-level only)

Explanation

Lattice enthalpy may mean formation, when one mole of ionic solid forms from gaseous ions, or dissociation, the equal-magnitude opposite-sign process.
A Born–Haber cycle applies Hess's law to enthalpy of formation, atomisation, bond enthalpy, ionisation energy, electron affinity and lattice enthalpy; any one unknown can be calculated.
Comparing the experimental cycle value with a perfect ionic-model value tests bonding: a substantial difference is evidence of covalent character.
A solution cycle combines lattice dissociation with hydration enthalpies, defined for forming one mole of aqueous ions from gaseous ions, to obtain enthalpy of solution.

A Born–Haber cycle links elements, gaseous ions and the ionic solid through Hess's law.

Worked example

For $\mathrm{NaCl(s)}$ , $\Delta_fH^\circ=-411\,\mathrm{kJ\,mol^{-1}}$ . Atomising Na and half a mole of chlorine requires $108$ and $121\,\mathrm{kJ\,mol^{-1}}$ ; $IE_1(\mathrm{Na})=496$ and $EA_1(\mathrm{Cl})=-349\,\mathrm{kJ\,mol^{-1}}$ . Calculate the lattice enthalpy of formation.

1.Apply Hess's law: $-411=108+121+496-349+\Delta_{latt}H^\circ$ .
2.The non-lattice terms total $376\,\mathrm{kJ\,mol^{-1}}$ .
3.Rearrange: $\Delta_{latt}H^\circ=-411-376$ .

Answer: $\Delta_{latt}H^\circ=-787\,\mathrm{kJ\,mol^{-1}}$ .

Common mistakes

Don't use lattice dissociation with the negative sign of lattice formation instead of reversing the sign.
Don't omit a second electron-affinity or hydration term when the ionic formula contains two identical anions.
Don't describe disagreement with the perfect ionic model as experimental error rather than evidence of covalent character.

Exam tip

Label every gaseous, aqueous and solid state in a Born–Haber cycle because each enthalpy definition depends on state.

Tier 1 · Easy

Define the lattice enthalpy of formation of $\mathrm{MgCl_2}$ .
[2 marks]
Total for this question: 2
The lattice enthalpy of dissociation of $\mathrm{CaF_2}$ is $+2630\,\mathrm{kJ\,mol^{-1}}$ . State its lattice enthalpy of formation and write the equation for lattice formation.
[3 marks]
Total for this question: 3

Tier 2 · Standard

For $\mathrm{NaCl(s)}$ , $\Delta_fH^\circ=-411\,\mathrm{kJ\,mol^{-1}}$ . The atomisation enthalpy of sodium is $+108\,\mathrm{kJ\,mol^{-1}}$ , half the chlorine bond enthalpy is $+121\,\mathrm{kJ\,mol^{-1}}$ , the first ionisation energy of sodium is $+496\,\mathrm{kJ\,mol^{-1}}$ and the first electron affinity of chlorine is $-349\,\mathrm{kJ\,mol^{-1}}$ . Calculate the lattice enthalpy of formation of $\mathrm{NaCl}$ .
[4 marks]
Total for this question: 4
When one mole of $\mathrm{KBr}$ dissolves, $\Delta H=+19.9\,\mathrm{kJ\,mol^{-1}}$ . Converting $\mathrm{K^+(g)}$ to $\mathrm{K^+(aq)}$ has $\Delta H=-322\,\mathrm{kJ\,mol^{-1}}$ , while converting $\mathrm{Br^-(g)}$ to $\mathrm{Br^-(aq)}$ has $\Delta H=-335\,\mathrm{kJ\,mol^{-1}}$ . Calculate the lattice enthalpy of dissociation of $\mathrm{KBr}$ and explain why its sign is positive.
[4 marks]
Total for this question: 4
Calculate the missing hydration-enthalpy value for $\mathrm{M^+}$ in the solution cycle of an unnamed ionic solid, $\mathrm{M_2X}$ . Its lattice enthalpy of formation is $-2350\,\mathrm{kJ\,mol^{-1}}$ , its enthalpy of solution is $-90\,\mathrm{kJ\,mol^{-1}}$ and the hydration enthalpy of $\mathrm{X^{2-}}$ is $-1480\,\mathrm{kJ\,mol^{-1}}$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

The lattice enthalpy of formation of $\mathrm{MgCl_2}$ is $-2526\,\mathrm{kJ\,mol^{-1}}$ . The hydration enthalpies of $\mathrm{Mg^{2+}}$ and $\mathrm{Cl^-}$ are $-1920$ and $-364\,\mathrm{kJ\,mol^{-1}}$ respectively. Calculate the enthalpy of solution of $\mathrm{MgCl_2}$ .
[4 marks]
Total for this question: 4
A Born–Haber cycle for $\mathrm{MgO(s)}$ uses these data in $\mathrm{kJ\,mol^{-1}}$ : $\Delta_fH^\circ=-602$ ; atomisation of Mg $=+150$ ; atomisation of O $=+249$ ; $IE_1(\mathrm{Mg})=+738$ ; $EA_1(\mathrm{O})=-141$ ; $EA_2(\mathrm{O})=+742$ ; lattice enthalpy of formation $=-3791$ . Calculate $IE_2(\mathrm{Mg})$ . A perfect-ionic model predicts a lattice enthalpy of formation of $-3650\,\mathrm{kJ\,mol^{-1}}$ . Explain the difference between this theoretical value and the experimental Born–Haber value.
[6 marks]
Total for this question: 6
For an unnamed ionic solid $\mathrm{MX_2(s)}$ , $\Delta_fH^\circ=-690\,\mathrm{kJ\,mol^{-1}}$ . Use these data in $\mathrm{kJ\,mol^{-1}}$ : atomisation of M, $+180$ ; $IE_1(\mathrm{M})=+600$ ; $IE_2(\mathrm{M})=+1200$ ; electron affinity of X, $-350$ ; lattice enthalpy of formation, $-2200$ . The atomisation enthalpy of X is half the $\mathrm{X-X}$ bond dissociation enthalpy. Calculate the bond dissociation enthalpy of $\mathrm{X_2}$ . Then write, with state symbols, the equation whose enthalpy change is the first electron affinity of X.
[6 marks]
Total for this question: 6
For an unnamed ionic solid $\mathrm{MX(s)}$ , $\Delta_fH^\circ=-412\,\mathrm{kJ\,mol^{-1}}$ . A Born–Haber cycle uses these data in $\mathrm{kJ\,mol^{-1}}$ : atomisation of M, $+126$ ; atomisation of X, $+104$ ; first ionisation energy of M, $+511$ ; first electron affinity of X, $-328$ . Calculate the lattice enthalpy of formation. The hydration enthalpies of $\mathrm{M^+}$ and $\mathrm{X^-}$ are $-304$ and $-372\,\mathrm{kJ\,mol^{-1}}$ respectively. Hence calculate the enthalpy of solution of MX. Write, with state symbols, the equation defining the hydration enthalpy of the cation.
[6 marks]
Total for this question: 6

3.1.8.2 · Gibbs free-energy change, ΔG, and entropy change, ΔS (A-level only)

Explanation

Entropy change, $\Delta S$ , describes changing disorder and accounts for feasible physical and chemical changes that enthalpy alone cannot explain.
Calculate $\Delta S^\circ=\sum S^\circ(\text{products})-\sum S^\circ(\text{reactants})$, including stoichiometric coefficients.
Feasibility follows from $\Delta G=\Delta H-T\Delta S$: a reaction is feasible under the stated conditions when $\Delta G$ is zero or negative.
Temperature must be in kelvin and $\Delta H$ and $T\Delta S$ must use consistent units.
On a graph of $\Delta G$ against $T$ , the intercept is $\Delta H$ and the gradient is $-\Delta S$ ; the zero crossing gives the feasibility threshold.

For positive entropy change, the Gibbs free-energy line falls with temperature and crosses into the feasible region.

Worked example

A reaction has $\Delta H=+42.0\,\mathrm{kJ\,mol^{-1}}$ and $\Delta S=+135\,\mathrm{J\,K^{-1}\,mol^{-1}}$ . Calculate $\Delta G$ at $350\,\mathrm{K}$ and state whether the reaction is feasible.

1.Convert entropy to $0.135\,\mathrm{kJ\,K^{-1}\,mol^{-1}}$ .
2.Substitute: $\Delta G=42.0-350(0.135)$ .
3.Evaluate the sign of the result against the feasibility condition.

Answer: $\Delta G=-5.25\,\mathrm{kJ\,mol^{-1}}$ ; the reaction is feasible at $350\,\mathrm{K}$ .

Common mistakes

Don't substitute entropy in joules while enthalpy is in kilojoules, making $T\Delta S$ one thousand times too large.
Don't state that a positive $\Delta G$ means feasible, reversing the zero-or-negative criterion.
Don't read the gradient of a $\Delta G$ against $T$ graph as $+\Delta S$ instead of $-\Delta S$ .

Exam tip

In a temperature-threshold calculation, set $\Delta G=0$ explicitly before rearranging for $T$ .

Tier 1 · Easy

For a reaction, the total standard entropy of the products is $465\,\mathrm{J\,K^{-1}\,mol^{-1}}$ and that of the reactants is $392\,\mathrm{J\,K^{-1}\,mol^{-1}}$ . Calculate $\Delta S^\circ$ .
[2 marks]
Total for this question: 2
A reaction has $\Delta G=-18\,\mathrm{kJ\,mol^{-1}}$ at a stated temperature. A student concludes that the reaction must be fast. Explain why this conclusion is not valid.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A reaction has $\Delta H=+42.0\,\mathrm{kJ\,mol^{-1}}$ and $\Delta S=+135\,\mathrm{J\,K^{-1}\,mol^{-1}}$ . Calculate $\Delta G$ at $350\,\mathrm{K}$ and state whether the reaction is feasible at this temperature.
[4 marks]
Total for this question: 4
For $\mathrm{N_2(g)+3H_2(g)\rightarrow2NH_3(g)}$ , molar entropy data are $192$ for $\mathrm{N_2}$ , $131$ for $\mathrm{H_2}$ and $193$ for $\mathrm{NH_3}$ , all in $\mathrm{J\,K^{-1}\,mol^{-1}}$ . The reaction enthalpy is $-92.4\,\mathrm{kJ\,mol^{-1}}$ . Calculate $\Delta S^\circ$ and $\Delta G^\circ$ at $298\,\mathrm{K}$ , then state whether the reaction is feasible at this temperature.
[5 marks]
Total for this question: 5
Reaction P has $\Delta H=-45\,\mathrm{kJ\,mol^{-1}}$ and $\Delta S=+80\,\mathrm{J\,K^{-1}\,mol^{-1}}$ . Reaction Q has $\Delta H=+60\,\mathrm{kJ\,mol^{-1}}$ and $\Delta S=-120\,\mathrm{J\,K^{-1}\,mol^{-1}}$ . Without calculating a threshold temperature, state the feasibility pattern of each reaction as temperature changes and justify each answer using $\Delta G=\Delta H-T\Delta S$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

For a reaction, $\Delta G=+18.0\,\mathrm{kJ\,mol^{-1}}$ at $300\,\mathrm{K}$ and $\Delta G=-12.0\,\mathrm{kJ\,mol^{-1}}$ at $500\,\mathrm{K}$ . Assume $\Delta H$ and $\Delta S$ are constant. Determine $\Delta H$ , $\Delta S$ and the temperature at which the reaction first becomes feasible.
[5 marks]
Total for this question: 5
A reaction has $\Delta H=-95.0\,\mathrm{kJ\,mol^{-1}}$ and $\Delta S=-180\,\mathrm{J\,K^{-1}\,mol^{-1}}$ . At $600\,\mathrm{K}$ a student writes $\Delta G=-95.0-600(-180)=+107905\,\mathrm{kJ\,mol^{-1}}$ . Identify the unit error, calculate the correct $\Delta G$ , and state what can and cannot be concluded.
[5 marks]
Total for this question: 5
For $\mathrm{A(g)+B(g)\rightarrow C(g)}$ , $\Delta H^\circ=-72.0\,\mathrm{kJ\,mol^{-1}}$ and $\Delta G^\circ=-25.2\,\mathrm{kJ\,mol^{-1}}$ at $400\,\mathrm{K}$ . The standard molar entropies of A and B are $180$ and $210\,\mathrm{J\,K^{-1}\,mol^{-1}}$ . Calculate $\Delta S^\circ$ , the standard molar entropy of C, and the highest temperature at which the reaction is thermodynamically feasible, assuming $\Delta H^\circ$ and $\Delta S^\circ$ are constant.
[6 marks]
Total for this question: 6
Reaction P has $\Delta H=+48.0\,\mathrm{kJ\,mol^{-1}}$ and $\Delta S=+120\,\mathrm{J\,K^{-1}\,mol^{-1}}$ . Reaction Q has $\Delta H=-72.0\,\mathrm{kJ\,mol^{-1}}$ and $\Delta S=-150\,\mathrm{J\,K^{-1}\,mol^{-1}}$ . Assume these values are constant. Determine the temperature range over which both reactions are thermodynamically feasible.
[5 marks]
Total for this question: 5
A reaction has $\Delta H=+36.0\,\mathrm{kJ\,mol^{-1}}$ . It is not thermodynamically feasible at $300\,\mathrm{K}$ but is feasible at $450\,\mathrm{K}$ . Assume $\Delta H$ and $\Delta S$ are constant. Determine the allowed range of values for $\Delta S$ in $\mathrm{J\,K^{-1}\,mol^{-1}}$ .
[4 marks]
Total for this question: 4

3.1.9.1 · Rate equations (A-level only)

Explanation

A rate equation relates rate to reactant concentrations: $\text{rate}=k[A]^m[B]^n$ , where experimentally determined orders $m$ and $n$ are restricted here to $0$ , $1$ or $2$ .
The rate constant $k$ has units obtained by rearranging the particular rate equation.
Raising temperature increases $k$ .
The Arrhenius equation is $k=Ae^{-E_a/(RT)}$ , where $A$ is the Arrhenius constant, $E_a$ is activation energy and $T$ is kelvin temperature.
Its linear form, $\ln k=-E_a/(RT)+\ln A$ , makes a plot of $\ln k$ against $1/T$ straight, with gradient $-E_a/R$ and intercept $\ln A$ .

Worked example

For $\text{rate}=k[A][B]^2$ , the rate is $4.80\times10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ when $[A]=0.200$ and $[B]=0.0500\,\mathrm{mol\,dm^{-3}}$ . Calculate $k$ with units.

1.Rearrange: $k=\dfrac{\text{rate}}{[A][B]^2}$ .
2.Substitute: $k=\dfrac{4.80\times10^{-4}}{0.200(0.0500)^2}$ .
3.Divide the rate units by $(\mathrm{mol\,dm^{-3}})^3$ .

Answer: $k=0.960\,\mathrm{dm^6\,mol^{-2}\,s^{-1}}$ .

Common mistakes

Don't copy balancing coefficients into the rate equation even though reaction orders are experimental.
Don't plot $\ln k$ against $T$ rather than against $1/T$ when using the linear Arrhenius form.
Don't give the Arrhenius gradient as $+E_a/R$ , losing the negative sign.

Exam tip

For a rate-constant calculation, show the rearranged equation and derive units from that equation for the method marks.

Tier 1 · Easy

A reaction has rate equation $\text{rate}=k[A]^2[B]$ . State the factor by which the rate changes when $[A]$ is halved and $[B]$ is doubled at constant temperature.
[2 marks]
Total for this question: 2
A reaction has rate equation $\text{rate}=k[A]^m[B]^n$ . The units of $k$ are $\mathrm{dm^3\,mol^{-1}\,s^{-1}}$ , and the reaction is zero order with respect to $A$ . Deduce $n$ and write the rate equation.
[3 marks]
Total for this question: 3

Tier 2 · Standard

For $\text{rate}=k[A][B]^2$ , the rate is $4.80\times10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ when $[A]=0.200\,\mathrm{mol\,dm^{-3}}$ and $[B]=0.0500\,\mathrm{mol\,dm^{-3}}$ . Calculate $k$ and give its units.
[4 marks]
Total for this question: 4
Take the gas constant as $R=8.31$ in its usual SI units. A plot of $\ln k$ against $1/T$ has gradient $-8.20\times10^3\,\mathrm{K}$ . Calculate the activation energy in $\mathrm{kJ\,mol^{-1}}$ .
[3 marks]
Total for this question: 3
An unnamed reaction has rate equation $\text{rate}=k[A]^2[B]$ . At one temperature, $k=0.250\,\mathrm{dm^6\,mol^{-2}\,s^{-1}}$ and $[B]=0.200\,\mathrm{mol\,dm^{-3}}$ . Calculate the value of $[A]$ required for a rate of $4.00\times10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

For this calculation use $R=8.31$ . A reaction has rate constants $1.80\times10^{-3}\,\mathrm{s^{-1}}$ at $298\,\mathrm{K}$ and $7.20\times10^{-3}\,\mathrm{s^{-1}}$ at $318\,\mathrm{K}$ . Calculate $E_a$ in $\mathrm{kJ\,mol^{-1}}$ from $\ln(k_2/k_1)=-(E_a/R)(1/T_2-1/T_1)$ .
[5 marks]
Total for this question: 5
The Arrhenius equation is $k=Ae^{-E_a/(RT)}$ ; take $R$ to be $8.31$ in SI units. A first-order reaction has $k=2.50\times10^{-3}\,\mathrm{s^{-1}}$ at $300\,\mathrm{K}$ and $E_a=52.0\,\mathrm{kJ\,mol^{-1}}$ . Calculate $A$ and then calculate $k$ at $335\,\mathrm{K}$ .
[6 marks]
Total for this question: 6
For an unnamed reaction, $\text{rate}=k[X]^2$ and $E_a=45.0\,\mathrm{kJ\,mol^{-1}}$ . The concentration is $0.180\,\mathrm{mol\,dm^{-3}}$ at $300\,\mathrm{K}$ . Use $R=8.31$ and $k_2/k_1=\exp[(E_a/R)(1/T_1-1/T_2)]$ to calculate the concentration required at $330\,\mathrm{K}$ to keep the rate unchanged.
[6 marks]
Total for this question: 6
For an unnamed reaction, $E_a=62.0\,\mathrm{kJ\,mol^{-1}}$ . Set $R=8.31$ in SI units. Determine the temperature at which its rate constant is eight times its value at $295\,\mathrm{K}$ , assuming the Arrhenius constant is unchanged, and state how the direction of the temperature change confirms your result.
[4 marks]
Total for this question: 4
At $310\,\mathrm{K}$ a catalyst makes an unnamed reaction's rate constant $40.0$ times larger. The uncatalysed activation energy is $78.0\,\mathrm{kJ\,mol^{-1}}$ . Set $R=8.31$ in SI units. Assuming the catalyst does not change the Arrhenius constant, calculate the reduction in activation energy and the catalysed activation energy.
[4 marks]
Total for this question: 4

3.1.9.2 · Determination of rate equation (A-level only)

Explanation

The rate equation is determined experimentally. A concentration–time curve gives rate from a tangent gradient; its initial tangent gives initial rate.
Rate–concentration data distinguish orders: unchanged rate is zero order, proportional change is first order and a squared response is second order.
Orders for all reactants combine to give the rate equation and can constrain the rate-determining step, including species produced in a preceding fast equilibrium.
Required practical 7 measures rate by both an initial-rate method and continuous monitoring.
For zero order, concentration falls linearly and the gradient magnitude equals $k$ ; collected or supplied data require a suitable best-fit curve and tangents at specified times.

A tangent to a concentration–time curve gives the instantaneous rate; the initial tangent gives the initial rate.

Worked example

Doubling $[A]$ multiplies initial rate by four, while doubling $[B]$ leaves it unchanged. The rate is $6.00\times10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ at $[A]=0.200\,\mathrm{mol\,dm^{-3}}$ . Deduce the rate equation and calculate $k$ .

1.A fourfold rate change for doubled $[A]$ gives second order in $A$ .
2.No rate change for doubled $[B]$ gives zero order in $B$ .
3. $\text{rate}=k[A]^2$ and $k=(6.00\times10^{-4})/(0.200)^2$ .

Answer: $\text{rate}=k[A]^2$ and $k=1.50\times10^{-2}\,\mathrm{dm^3\,mol^{-1}\,s^{-1}}$ .

Common mistakes

Don't use the gradient of a chord between two points as the instantaneous rate instead of drawing a tangent.
Don't call an unchanged rate first order when concentration changes, rather than zero order.
Don't claim that a matching rate equation proves a mechanism, although it only provides supporting information about the limiting step.

Exam tip

When deducing order from trials, compare a pair in which only one reactant concentration changes and state both concentration and rate factors.

Tier 1 · Easy

At constant temperature, doubling $[X]$ while all other concentrations remain unchanged doubles the initial rate. Deduce the order with respect to $X$ .
[1 mark]
Total for this question: 1
A reaction produces carbon dioxide but no other gas. State a continuous-monitoring measurement suitable for required practical 7 and state one variable that must be controlled when rates are compared.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Three invented kinetic trials gave these values. Trial 1: $[A]=0.100$ , $[B]=0.200$ , rate $=1.50\times10^{-4}$ ; trial 2: $[A]=0.200$ , $[B]=0.200$ , rate $=6.00\times10^{-4}$ ; trial 3: $[A]=0.200$ , $[B]=0.400$ , rate $=6.00\times10^{-4}$ . Concentrations are in $\mathrm{mol\,dm^{-3}}$ and rates in $\mathrm{mol\,dm^{-3}\,s^{-1}}$ . Deduce the rate equation and calculate $k$ with units.
[5 marks]
Total for this question: 5
Three initial-rate trials were carried out at one temperature. Trial 1: $[A]=0.100$ , $[B]=0.100$ , rate $=1.00\times10^{-4}$ . Trial 2: $[A]=0.200$ , $[B]=0.0500$ , rate $=2.00\times10^{-4}$ . Trial 3: $[A]=0.0500$ , $[B]=0.400$ , rate $=1.00\times10^{-4}$ . Concentrations are in $\mathrm{mol\,dm^{-3}}$ and rates in $\mathrm{mol\,dm^{-3}\,s^{-1}}$ . Deduce the orders, write the rate equation and calculate $k$ with units.
[6 marks]
Total for this question: 6
For an unnamed reaction, these initial-rate data were obtained at one temperature: $[X]=0.100$ , $0.200$ and $0.300\,\mathrm{mol\,dm^{-3}}$ give rates $2.00\times10^{-3}$ , $8.00\times10^{-3}$ and $1.80\times10^{-2}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ respectively. State which plot gives a straight line through the origin, deduce the order in $X$ and calculate $k$ with units.
[4 marks]
Total for this question: 4

Tier 3 · Hard

The experimental rate equation for a reaction is $\text{rate}=k[X][Y]^2$ . Mechanism I has a single slow step $\mathrm{X+Y\rightarrow products}$ . Mechanism II has a fast equilibrium $\mathrm{X+Y\rightleftharpoons I}$ followed by the slow step $\mathrm{I+Y\rightarrow products}$ . Determine which mechanism is consistent with the rate equation and explain your choice.
[4 marks]
Total for this question: 4
Reactant $X$ has coefficient $1$ in the reaction equation. Continuous monitoring gives a concentration–time curve for $X$ . When $[X]=0.0800\,\mathrm{mol\,dm^{-3}}$ , the tangent gradient is $-1.60\times10^{-3}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ . When $[X]=0.0400\,\mathrm{mol\,dm^{-3}}$ , it is $-4.00\times10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ . Deduce the order with respect to $X$ and calculate $k$ with units.
[5 marks]
Total for this question: 5
For an unnamed reaction, the concentration of $X$ was measured during one experiment. The data are: at $0$ , $45$ , $90$ , $135$ and $180\,\mathrm{s}$ , $[X]$ is $0.1260$ , $0.1098$ , $0.0936$ , $0.0774$ and $0.0612\,\mathrm{mol\,dm^{-3}}$ respectively. Show that a concentration–time plot is a straight line, deduce the order with respect to $X$ , and calculate $k$ with units.
[6 marks]
Total for this question: 6
For an unnamed reaction, the concentration of X is $0.160$ , $0.0800$ , $0.0400$ and $0.0200\,\mathrm{mol\,dm^{-3}}$ at $0$ , $70$ , $140$ and $210\,\mathrm{s}$ respectively. At $[X]=0.0800\,\mathrm{mol\,dm^{-3}}$ , a tangent to the concentration–time curve has gradient $-7.92\times10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ . Deduce the order in X, calculate $k$ with units, and determine the rate when $[X]=0.0300\,\mathrm{mol\,dm^{-3}}$ .
[5 marks]
Total for this question: 5
Three initial-rate trials for an unnamed reaction gave: trial 1, $[A]=0.100$ , $[B]=0.100$ , rate $=2.00\times10^{-4}$ ; trial 2, $[A]=0.250$ , $[B]=0.100$ , rate $=2.00\times10^{-4}$ ; trial 3, $[A]=0.250$ , $[B]=0.300$ , rate $=6.00\times10^{-4}$ . Concentrations are in $\mathrm{mol\,dm^{-3}}$ and rates in $\mathrm{mol\,dm^{-3}\,s^{-1}}$ . Deduce the rate equation, calculate $k$ with units, and predict the rate when $[A]=0.0400$ and $[B]=0.220\,\mathrm{mol\,dm^{-3}}$ .
[5 marks]
Total for this question: 5

3.1.10 · Equilibrium constant Kp for homogeneous systems (A-level only)

Explanation

For a homogeneous gas equilibrium, partial pressure is $p_i=x_iP_{total}$ with mole fraction $x_i=n_i/n_{total}$ , using equilibrium amounts.
Construct $K_p$ from product partial pressures divided by reactant partial pressures, each raised to its equation coefficient; its units follow from the net pressure power.
Changing pressure at constant temperature may shift the equilibrium position toward fewer gas molecules but does not change $K_p$ .
Temperature can change both position and the value of $K_p$ , with the direction depending on reaction enthalpy.
A catalyst only increases the rate at which equilibrium is reached and changes neither equilibrium position nor $K_p$ .

Worked example

At equilibrium for $\mathrm{N_2O_4(g)\rightleftharpoons2NO_2(g)}$ , amounts are $0.300$ and $0.400\,\mathrm{mol}$ respectively at $200\,\mathrm{kPa}$ . Calculate $K_p$ .

1.Total amount $=0.700\,\mathrm{mol}$ , so $p(\mathrm{N_2O_4})=(0.300/0.700)(200)=85.7\,\mathrm{kPa}$ .
2. $p(\mathrm{NO_2})=(0.400/0.700)(200)=114.3\,\mathrm{kPa}$ .
3. $K_p=\dfrac{p(\mathrm{NO_2})^2}{p(\mathrm{N_2O_4})}$ .

Answer: $K_p=152\,\mathrm{kPa}$ .

Common mistakes

Don't use initial mole amounts instead of equilibrium amounts to calculate partial pressures.
Don't omit the square on $p(\mathrm{NO_2})$ when constructing $K_p$ from the balanced equation.
Don't state that increasing pressure changes $K_p$ , although only temperature changes its value.

Exam tip

In a $K_p$ calculation, write the expression first and keep one pressure unit throughout so the final units follow correctly.

Tier 1 · Easy

A gas mixture contains $2.00\,\mathrm{mol}$ of $A$ and $3.00\,\mathrm{mol}$ of $B$ at a total pressure of $500\,\mathrm{kPa}$ . Calculate the partial pressure of each gas.
[2 marks]
Total for this question: 2
An equilibrium constant $K_p$ has units $\mathrm{kPa^{-2}}$ . Identify which of these equations it belongs to: $\mathrm{2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)}$ or $\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}$ . Justify your answer using the change in the total number of moles of gas.
[3 marks]
Total for this question: 3

Tier 2 · Standard

At equilibrium for $\mathrm{N_2O_4(g)\rightleftharpoons2NO_2(g)}$ , there are $0.300\,\mathrm{mol}$ of $\mathrm{N_2O_4}$ and $0.400\,\mathrm{mol}$ of $\mathrm{NO_2}$ at a total pressure of $2.00\times10^2\,\mathrm{kPa}$ . Calculate $K_p$ , including units.
[5 marks]
Total for this question: 5
For $\mathrm{A(g)\rightleftharpoons2B(g)}$ , a vessel initially contains $1.00\,\mathrm{mol}$ of $A$ only. At equilibrium, $40.0\%$ of $A$ has dissociated and the total pressure is $350\,\mathrm{kPa}$ . Calculate the equilibrium partial pressures and $K_p$ , including units.
[6 marks]
Total for this question: 6
For $\mathrm{2A(g)\rightleftharpoons B(g)}$ , $K_p=0.0800\,\mathrm{kPa^{-1}}$ at a stated temperature. At equilibrium $p_A=50.0\,\mathrm{kPa}$ . Calculate $p_B$ and the total pressure.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Initially, a vessel contains $1.00\,\mathrm{mol}$ of $\mathrm{H_2}$ and $1.00\,\mathrm{mol}$ of $\mathrm{I_2}$ for $\mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)}$ . At equilibrium there are $1.20\,\mathrm{mol}$ of $\mathrm{HI}$ and the total pressure is $250\,\mathrm{kPa}$ . Calculate $K_p$ . State the effect of increasing the total pressure at constant temperature on the equilibrium position and on $K_p$ .
[6 marks]
Total for this question: 6
Phosphorus(V) chloride dissociates as $\mathrm{PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)}$ . A vessel initially contains only $\mathrm{PCl_5}$ . At equilibrium the total pressure is $210\,\mathrm{kPa}$ and $K_p=40.0\,\mathrm{kPa}$ . Calculate the percentage of the original $\mathrm{PCl_5}$ that has dissociated.
[6 marks]
Total for this question: 6
For $\mathrm{A(g)\rightleftharpoons2B(g)}$ , one equilibrium mixture at a fixed temperature has $p_A=120\,\mathrm{kPa}$ and $p_B=60.0\,\mathrm{kPa}$ . A second mixture at the same temperature is reported as $p_A=80.0\,\mathrm{kPa}$ and $p_B=40.0\,\mathrm{kPa}$ . Deduce whether the second report is consistent with the first. If it is not, calculate the corrected value of $p_B$ for the second mixture and explain why changing total pressure cannot be used to justify a different $K_p$ .
[5 marks]
Total for this question: 5
For $\mathrm{A(g)+B(g)\rightleftharpoons2C(g)}$ , the equilibrium amounts of A, B and C are $0.200$ , $0.400$ and $0.800\,\mathrm{mol}$ at $400\,\mathrm{K}$ , and $0.500$ , $0.500$ and $0.400\,\mathrm{mol}$ at $600\,\mathrm{K}$ . Calculate $K_p$ at each temperature from the mole fractions, showing why total pressure cancels. Deduce the sign of the forward reaction enthalpy. State and explain why a catalyst does not change $K_p$ .
[6 marks]
Total for this question: 6
For $\mathrm{A(g)+2B(g)\rightleftharpoons C(g)}$ , an equilibrium mixture has mole fractions $x_A=0.200$ , $x_B=0.500$ and $x_C=0.300$ . At this temperature, $K_p=1.20\times10^{-4}\,\mathrm{kPa^{-2}}$ . Determine the total pressure and all three partial pressures.
[5 marks]
Total for this question: 5

3.1.11.1 · Electrode potentials and cells (A-level only)

Explanation

Electrode half-equations follow the IUPAC reduction convention. Potentials are measured relative to the standard hydrogen electrode; standard $E^\circ$ conditions are $298\,\mathrm{K}$ , $100\,\mathrm{kPa}$ and aqueous ion concentrations of $1.00\,\mathrm{mol\,dm^{-3}}$ .
The more positive couple is reduced, while the other half-equation is reversed for oxidation.
Calculate $E^\circ_{cell}=E^\circ_{positive}-E^\circ_{negative}$ and use its sign to predict a simple redox direction.
Conventional notation places oxidation on the left, reduction on the right, single lines at phase boundaries and a double line for the salt bridge.
Required practical 8 measures cell EMF and can investigate effects of non-standard concentration or temperature.

An electrochemical cell separates oxidation and reduction so electrons flow through the external circuit.

Worked example

Given $E^\circ(\mathrm{Fe^{3+}/Fe^{2+}})=+0.77\,\mathrm{V}$ and $E^\circ(\mathrm{Sn^{4+}/Sn^{2+}})=+0.15\,\mathrm{V}$ , calculate the EMF and write the overall reaction.

1.The iron couple is more positive, so $\mathrm{Fe^{3+}}$ is reduced.
2.Reverse the tin equation and balance electrons: $\mathrm{Sn^{2+}\rightarrow Sn^{4+}+2e^-}$ and $\mathrm{2Fe^{3+}+2e^-\rightarrow2Fe^{2+}}$ .
3. $E^\circ_{cell}=0.77-0.15$ .

Answer: $E^\circ_{cell}=+0.62\,\mathrm{V}$ ; $\mathrm{2Fe^{3+}+Sn^{2+}\rightarrow2Fe^{2+}+Sn^{4+}}$ .

Common mistakes

Don't multiply an electrode potential when multiplying a half-equation to balance electrons.
Don't subtract the more positive potential from the less positive one and predict the reverse reaction.
Don't use a metal electrode in cell notation when both members of a redox couple are aqueous and an inert platinum electrode is required.

Exam tip

For a cell-notation question, identify oxidation and reduction first, then place the oxidation half-cell on the left.

Tier 1 · Easy

Given $E^\circ(\mathrm{Cu^{2+}/Cu})=+0.34\,\mathrm{V}$ and $E^\circ(\mathrm{Zn^{2+}/Zn})=-0.76\,\mathrm{V}$ , calculate $E^\circ_{cell}$ and identify the species reduced.
[3 marks]
Total for this question: 3
A standard $\mathrm{Fe^{3+}/Fe^{2+}}$ half-cell contains both ions in aqueous solution. Explain why a platinum electrode is used.
[2 marks]
Total for this question: 2

Tier 2 · Standard

The standard potentials are $E^\circ(\mathrm{Fe^{3+}/Fe^{2+}})=+0.77\,\mathrm{V}$ and $E^\circ(\mathrm{Sn^{4+}/Sn^{2+}})=+0.15\,\mathrm{V}$ . Calculate the EMF, write the overall reaction and give the conventional cell representation.
[6 marks]
Total for this question: 6
In required practical 8, a student connects a standard $\mathrm{Ag^+/Ag}$ half-cell, for which $E^\circ=+0.80\,\mathrm{V}$ , to a standard $\mathrm{X^{2+}/X}$ half-cell. The measured cell EMF is $1.05\,\mathrm{V}$ and electrons flow from the X electrode to the silver electrode. State how the cell EMF should be measured and deduce $E^\circ(\mathrm{X^{2+}/X})$ .
[6 marks]
Total for this question: 6
The standard reduction potentials are $E^\circ(\mathrm{A^{3+}/A^{2+}})=+0.68\,\mathrm{V}$ and $E^\circ(\mathrm{B^{2+}/B})=-0.22\,\mathrm{V}$ . A student multiplies the first potential by two when balancing electrons and obtains $E^\circ_{cell}=+1.58\,\mathrm{V}$ . Identify the error and calculate the correct value, then write the spontaneous overall equation.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Use $E^\circ(\mathrm{Cu^+/Cu})=+0.52\,\mathrm{V}$ and $E^\circ(\mathrm{Cu^{2+}/Cu^+})=+0.15\,\mathrm{V}$ to determine whether $\mathrm{Cu^+}$ ions can disproportionate under standard conditions. Give the equation and calculate $E^\circ_{cell}$ .
[5 marks]
Total for this question: 5
The standard reduction potentials are $E^\circ(\mathrm{Cl_2/Cl^-})=+1.36\,\mathrm{V}$ , $E^\circ(\mathrm{I_2/I^-})=+0.54\,\mathrm{V}$ and $E^\circ(\mathrm{Zn^{2+}/Zn})=-0.76\,\mathrm{V}$ . Identify the two half-cells that give the largest positive standard EMF. Write the overall reaction, calculate the EMF and give the conventional cell representation.
[7 marks]
Total for this question: 7
For an unnamed element M, $E^\circ(\mathrm{M^{3+}/M^{2+}})=+0.70\,\mathrm{V}$ and $E^\circ(\mathrm{M^{2+}/M})=-0.10\,\mathrm{V}$ . Determine whether $\mathrm{M^{3+}(aq)}$ and $\mathrm{M(s)}$ react together under standard conditions to form only $\mathrm{M^{2+}(aq)}$ , and justify your answer using the electrode potentials. Write the balanced equation and give the conventional cell representation.
[6 marks]
Total for this question: 6
The standard reduction potentials are $E^\circ(\mathrm{X^+/X})=+0.85\,\mathrm{V}$ , $E^\circ(\mathrm{Y^{2+}/Y})=-0.30\,\mathrm{V}$ and $E^\circ(\mathrm{Z^{2+}/Z})=-0.90\,\mathrm{V}$ . Determine what happens when Y metal is added first to a solution of $\mathrm{X^+}$ and then, in a separate experiment, to a solution of $\mathrm{Z^{2+}}$ . For each proposed direction, use the standard EMF and give a balanced equation where a reaction occurs.
[5 marks]
Total for this question: 5
A cell is made from $\mathrm{Ag^+/Ag}$ and $\mathrm{Zn^{2+}/Zn}$ half-cells, whose standard reduction potentials are $+0.80$ and $-0.76\,\mathrm{V}$ respectively. Calculate the standard cell EMF. Deduce and explain separately how the EMF changes when $[\mathrm{Ag^+}]$ is raised and when $[\mathrm{Zn^{2+}}]$ is raised, naming the electrode potential that shifts each time.
[5 marks]
Total for this question: 5

3.1.11.2 · Commercial applications of electrochemical cells (A-level only)

Explanation

Electrochemical cells provide electrical energy and may be non-rechargeable, rechargeable or fuel cells.
Given electrode data can identify each electrode reaction and the cell EMF; oxidation releases electrons at the negative electrode and reduction consumes them at the positive electrode, driving current through the external circuit.
In the simplified lithium cell, lithium is oxidised and $\mathrm{Li^++CoO_2}$ is reduced.
An alkaline hydrogen–oxygen fuel cell receives reactants continuously and is refuelled rather than electrically recharged.
Evaluation must balance point-of-use products and continuous operation against fuel production, storage, safety, materials, lifetime and disposal; benefits and risks depend on the whole system.

Separated electrode reactions in an alkaline hydrogen–oxygen fuel cell force electrons through the external circuit.

Worked example

An alkaline fuel cell uses $\mathrm{H_2+2OH^-\rightarrow2H_2O+2e^-}$ and $\mathrm{O_2+2H_2O+4e^-\rightarrow4OH^-}$ . Deduce the overall equation and explain how current is generated.

1.Multiply the hydrogen equation by $2$ so both half-equations transfer $4e^-$ .
2.Add and cancel electrons, hydroxide ions and two water molecules.
3.State that electron release and electron consumption occur at separate electrodes.

Answer: $\mathrm{2H_2+O_2\rightarrow2H_2O}$ ; electrons travel through the external circuit from the oxidation electrode to the reduction electrode.

Common mistakes

Don't describe a fuel cell as rechargeable even though operation continues by supplying fresh fuel and oxidant.
Don't claim hydrogen fuel cells are unconditionally carbon-free without considering the energy source used to produce hydrogen.
Don't add fuel-cell half-equations without first balancing the number of electrons.

Exam tip

For an 'evaluate' question, give a linked benefit and limitation and distinguish point-of-use products from whole-system impacts.

Tier 1 · Easy

State one difference between a rechargeable cell and a hydrogen–oxygen fuel cell.
[2 marks]
Total for this question: 2
During discharge of the simplified lithium cell, $\mathrm{Li\rightarrow Li^++e^-}$ occurs at one electrode and $\mathrm{Li^++CoO_2+e^-\rightarrow Li^+[CoO_2]^-}$ at the other. Identify which species is oxidised and name the product formed at the positive electrode.
[2 marks]
Total for this question: 2

Tier 2 · Standard

The alkaline fuel-cell half-equations are $\mathrm{H_2+2OH^-\rightarrow2H_2O+2e^-}$ and $\mathrm{O_2+2H_2O+4e^-\rightarrow4OH^-}$ . Deduce the overall equation and explain how the reactions generate a current.
[4 marks]
Total for this question: 4
Use the simplified lithium-cell electrode reactions $\mathrm{Li\rightarrow Li^++e^-}$ and $\mathrm{Li^++CoO_2+e^-\rightarrow Li^+[CoO_2]^-}$ to deduce the overall reaction during discharge and explain why an electric current flows in the external circuit.
[4 marks]
Total for this question: 4
A cell has a positive standard EMF. A student concludes that the cell must therefore be rechargeable. Explain why this conclusion does not follow from the EMF alone, and state the chemical evidence needed before the cell can be classified as rechargeable.
[4 marks]
Total for this question: 4

Tier 3 · Hard

An alkaline hydrogen–oxygen fuel cell uses couples with electrode potentials $+0.40\,\mathrm{V}$ and $-0.83\,\mathrm{V}$ . Calculate its EMF. Then give one benefit and two risks or limitations of using hydrogen fuel cells in vehicles.
[5 marks]
Total for this question: 5
A rechargeable cell uses the reduction couples $\mathrm{P^{3+}+e^-\rightleftharpoons P^{2+}}$ , $E^\circ=+0.60\,\mathrm{V}$ , and $\mathrm{Q^{2+}+2e^-\rightleftharpoons Q(s)}$ , $E^\circ=-0.40\,\mathrm{V}$ . Deduce the electrode reactions and overall reaction during discharge, calculate the standard EMF, and state the overall reaction during charging.
[7 marks]
Total for this question: 7
A delivery fleet returns to a depot every night. The depot has renewable electricity for charging but no hydrogen-production or high-pressure storage equipment. Compare a rechargeable cell with a hydrogen–oxygen fuel cell for this fleet. Justify which system should be used with two linked advantages in this context and two limitations or risks that would still need managing.
[6 marks]
Total for this question: 6
A hydrogen–oxygen fuel-cell unit transfers $12.0\,\mathrm{mol}$ of electrons during a journey. Hydrogen oxidation transfers two electrons per $\mathrm{H_2}$ molecule and oxygen reduction transfers four electrons per $\mathrm{O_2}$ molecule. Determine the amounts and masses of both gases consumed and the amount of water formed. Use $M_r(\mathrm{H_2})=2.00$ and $M_r(\mathrm{O_2})=32.0$ .
[5 marks]
Total for this question: 5
A non-rechargeable cell uses the half-cells $\mathrm{Zn^{2+}/Zn}$ , for which $E^\circ=-0.76\,\mathrm{V}$ , and $\mathrm{2MnO_2+H_2O+2e^-\rightleftharpoons Mn_2O_3+2OH^-}$ , for which $E^\circ=+0.15\,\mathrm{V}$ . Write both electrode reactions in the directions occurring during discharge, construct the overall equation, calculate the standard EMF and explain how the cell generates current.
[5 marks]
Total for this question: 5

3.1.12.1 · Brønsted–Lowry acid–base equilibria in aqueous solution (A-level only)

Explanation

A Brønsted–Lowry acid is a proton donor and a Brønsted–Lowry base is a proton acceptor, so an aqueous acid–base equilibrium transfers $\mathrm{H^+}$ between two species.
Each conjugate acid–base pair differs by exactly one proton: after an acid donates $\mathrm{H^+}$ it becomes its conjugate base, while a base becomes its conjugate acid after accepting $\mathrm{H^+}$ .
Roles must be assigned from the direction shown in the equation because an amphoteric species such as water can donate or accept a proton.
Valid proton-transfer equations balance every atom and the total charge on both sides.

Worked example

In $\mathrm{NH_3+H_2O\rightleftharpoons NH_4^++OH^-}$ , identify the acid, base and conjugate pairs.

1. $\mathrm{H_2O}$ donates $\mathrm{H^+}$ , so it is the acid and $\mathrm{OH^-}$ is its conjugate base.
2. $\mathrm{NH_3}$ accepts $\mathrm{H^+}$ , so it is the base and $\mathrm{NH_4^+}$ is its conjugate acid.

Answer: Acid/base: $\mathrm{H_2O/NH_3}$ ; conjugate pairs: $\mathrm{H_2O/OH^-}$ and $\mathrm{NH_4^+/NH_3}$ .

Common mistakes

Don't define a base as an electron-pair donor instead of using the required Brønsted–Lowry proton definition.
Don't pair two species that differ by more than one $\mathrm{H^+}$ as conjugates.
Don't label water permanently as an acid even though its role depends on the reaction partner.

Exam tip

For an 'identify the conjugate pairs' question, join formulas that differ by one proton and verify the charge changes by one.

Tier 1 · Easy

In $\mathrm{NH_3+H_2O\rightleftharpoons NH_4^++OH^-}$ , identify the acid and the base on the left-hand side.
[2 marks]
Total for this question: 2
State the formula of the conjugate acid of $\mathrm{CO_3^{2-}}$ and the conjugate base of $\mathrm{H_2O}$ .
[2 marks]
Total for this question: 2

Tier 2 · Standard

For $\mathrm{HSO_4^-+H_2O\rightleftharpoons SO_4^{2-}+H_3O^+}$ , identify both conjugate acid–base pairs.
[3 marks]
Total for this question: 3
Complete the proton-transfer equilibrium between sulfurous acid, $\mathrm{H_2SO_3}$ , and ammonia, $\mathrm{NH_3}$ . Identify the acid, the base and both conjugate acid–base pairs.
[5 marks]
Total for this question: 5
A student proposes $\mathrm{H_2O+S^{2-}\rightleftharpoons H_3O^++HS^-}$ as a Brønsted–Lowry proton-transfer equation. Identify the error, write the corrected equation and state the conjugate acid–base pairs.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Use two equations with water to show that $\mathrm{H_2PO_4^-}$ is amphoteric. Identify its role in each equation.
[4 marks]
Total for this question: 4
A student states that $\mathrm{H_2PO_4^-}$ and $\mathrm{PO_4^{3-}}$ are a conjugate acid–base pair. Explain the error and use the missing intermediate species to state two correct conjugate pairs.
[4 marks]
Total for this question: 4
A triprotic acid forms the sequence $\mathrm{H_3A}$ , $\mathrm{H_2A^-}$ , $\mathrm{HA^{2-}}$ and $\mathrm{A^{3-}}$ . State all three conjugate acid–base pairs, identify every amphoteric species in the sequence, and write one equation with water that shows $\mathrm{HA^{2-}}$ acting as a base.
[5 marks]
Total for this question: 5

3.1.12.2 · Definition and determination of pH (A-level only)

Explanation

Hydrogen-ion concentrations span a very wide range, so acidity is expressed logarithmically as $\mathrm{pH}=-\log_{10}[H^+]$ , with concentration in $\mathrm{mol\,dm^{-3}}$ .
Reverse the relationship using $[H^+]=10^{-\mathrm{pH}}$ ; one pH unit represents a tenfold concentration change.
A strong acid is treated as completely dissociated, so a monoprotic acid gives one mole of $\mathrm{H^+}$ per mole after any dilution or mixing calculation.
Precision is reported carefully: the number of decimal places in pH conventionally matches the number of significant figures in $[H^+]$ .
Intermediate values should remain unrounded until the final answer.

Worked example

$25.0\,\mathrm{cm^3}$ of $0.120\,\mathrm{mol\,dm^{-3}}$ hydrochloric acid is diluted to $200\,\mathrm{cm^3}$ . Calculate the pH.

1.Amount of acid $=0.120\times0.0250=3.00\times10^{-3}\,\mathrm{mol}$ .
2.Complete dissociation gives $[H^+]=(3.00\times10^{-3})/0.200=0.0150\,\mathrm{mol\,dm^{-3}}$ .
3. $\mathrm{pH}=-\log_{10}(0.0150)$ .

Answer: $\mathrm{pH}=1.824$ .

Common mistakes

Don't use volume in $\mathrm{cm^3}$ directly in an amount calculation without converting to $\mathrm{dm^3}$ .
Don't calculate $10^{\mathrm{pH}}$ instead of $10^{-\mathrm{pH}}$ when finding hydrogen-ion concentration.
Don't round $[H^+]$ before taking the logarithm and lose accuracy in the final pH.

Exam tip

For a strong-acid pH calculation, show the mole or dilution step before applying the logarithm.

Tier 1 · Easy

Calculate the pH of a solution for which $[H^+]=3.2\times10^{-3}\,\mathrm{mol\,dm^{-3}}$ .
[2 marks]
Total for this question: 2
Solution X has pH $2.30$ and solution Y has pH $4.30$ . Calculate the ratio $[H^+]_X:[H^+]_Y$ .
[2 marks]
Total for this question: 2

Tier 2 · Standard

A solution has pH $3.68$ . Calculate $[H^+]$ in $\mathrm{mol\,dm^{-3}}$ .
[2 marks]
Total for this question: 2
A hydrochloric acid solution has pH $1.30$ . Calculate the dilution factor required to produce a solution of pH $2.45$ . Give your answer to 3 significant figures.
[3 marks]
Total for this question: 3
A strong diprotic acid dissociates completely as $\mathrm{H_2X\rightarrow2H^++X^{2-}}$ . A $150\,\mathrm{cm^3}$ sample has pH $1.75$ . Calculate the concentration of $\mathrm{H_2X}$ and the amount of acid in the sample.
[4 marks]
Total for this question: 4

Tier 3 · Hard

$27.5\,\mathrm{cm^3}$ of $0.145\,\mathrm{mol\,dm^{-3}}$ hydrochloric acid is mixed with $32.5\,\mathrm{cm^3}$ of $0.0730\,\mathrm{mol\,dm^{-3}}$ nitric acid. Calculate the pH of the mixture. Assume volumes are additive.
[5 marks]
Total for this question: 5
A student must prepare $250\,\mathrm{cm^3}$ of hydrochloric acid with pH $2.20$ using $0.150\,\mathrm{mol\,dm^{-3}}$ hydrochloric acid. Calculate the volume of the stock solution required. Assume complete dissociation and additive volumes. Give your answer to 3 significant figures.
[5 marks]
Total for this question: 5
An unnamed strong monoprotic acid solution is $0.365\%$ acid by mass and has density $1.02\,\mathrm{g\,cm^{-3}}$ . The acid has $M_r=36.5$ . Calculate the hydrogen-ion concentration and pH. Carry unrounded values through the calculation.
[5 marks]
Total for this question: 5
$20.0\,\mathrm{cm^3}$ of $0.180\,\mathrm{mol\,dm^{-3}}$ hydrochloric acid is mixed with $30.0\,\mathrm{cm^3}$ of an unknown completely dissociated strong monoprotic acid. The mixture has pH $1.000$ . Determine the concentration of the unknown acid. Assume volumes are additive.
[5 marks]
Total for this question: 5
A $1.46\,\mathrm{g}$ sample of a strong acid, $\mathrm{H_nA}$ , with $M_r=73.0$ is dissolved to make $500\,\mathrm{cm^3}$ of solution. The solution has pH $1.097$ , and every acidic proton dissociates completely. Deduce $n$ .
[4 marks]
Total for this question: 4

3.1.12.3 · The ionic product of water, Kw (A-level only)

Explanation

Water dissociates slightly, giving the ionic product $K_w=[H^+][OH^-]$ , whose value varies with temperature. It is derived from the equilibrium constant for water dissociation by incorporating the effectively constant concentration of liquid water.
For a strong base, complete dissociation first determines $[OH^-]$ , including the number of hydroxide ions in its formula.
Then $[H^+]=K_w/[OH^-]$ and pH follows from $-\log_{10}[H^+]$ .
In neutral water, $[H^+]=[OH^-]=\sqrt{K_w}$ ; neutral pH is therefore $7.00$ only when $K_w=1.00\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}$ .
Appropriate standard form and decimal places must be retained throughout the calculation.

Worked example

At $298\,\mathrm{K}$ , calculate the pH of $0.00350\,\mathrm{mol\,dm^{-3}}$ $\mathrm{Ba(OH)_2}$ . Use $K_w=1.00\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}$ .

1. $\mathrm{Ba(OH)_2}$ gives two hydroxide ions, so $[OH^-]=2(0.00350)=0.00700\,\mathrm{mol\,dm^{-3}}$ .
2. $[H^+]=(1.00\times10^{-14})/0.00700=1.43\times10^{-12}\,\mathrm{mol\,dm^{-3}}$ .
3.Apply $\mathrm{pH}=-\log_{10}[H^+]$ .

Answer: $\mathrm{pH}=11.845$ .

Common mistakes

Don't set $[OH^-]$ equal to the concentration of $\mathrm{Ba(OH)_2}$ and miss the factor of two.
Don't take $-\log_{10}[OH^-]$ and report the result as pH.
Don't assume neutral pH is always $7$ even when the given $K_w$ is for a different temperature.

Exam tip

When a strong-base formula contains more than one $\mathrm{OH^-}$ , state the hydroxide stoichiometric factor before using $K_w$ .

Tier 1 · Easy

At $298\,\mathrm{K}$ , $K_w=1.00\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}$ . Calculate the pH of $0.0250\,\mathrm{mol\,dm^{-3}}$ sodium hydroxide.
[3 marks]
Total for this question: 3
Water dissociates as $\mathrm{H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)}$ . Write the expression for $K_w$ and explain why the concentration of liquid water does not appear in it.
[3 marks]
Total for this question: 3

Tier 2 · Standard

At $298\,\mathrm{K}$ , calculate the pH of $0.00350\,\mathrm{mol\,dm^{-3}}$ barium hydroxide. Use $K_w=1.00\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}$ and assume complete dissociation.
[4 marks]
Total for this question: 4
At $298\,\mathrm{K}$ , an aqueous sodium hydroxide solution has pH $12.30$ . Use $K_w=1.00\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}$ to calculate the concentration of sodium hydroxide.
[4 marks]
Total for this question: 4
At a stated temperature, neutral water has pH $6.42$ . Calculate $K_w$ at this temperature, including units. Deduce whether this temperature is above or below $298\,\mathrm{K}$ and justify your answer.
[4 marks]
Total for this question: 4

Tier 3 · Hard

At a higher temperature, $K_w=4.00\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}$ . Calculate the pH of neutral water and the pH of $0.0200\,\mathrm{mol\,dm^{-3}}$ potassium hydroxide at this temperature.
[5 marks]
Total for this question: 5
The value of $K_w$ is $1.00\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}$ at $298\,\mathrm{K}$ and $5.48\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}$ at $323\,\mathrm{K}$ . Calculate the pH of neutral water at each temperature, explain why neutral water does not have pH $7$ at $323\,\mathrm{K}$ , and deduce whether the dissociation of water is endothermic or exothermic. Give each pH to 2 decimal places and carry unrounded values through the working.
[6 marks]
Total for this question: 6
At a stated temperature, $K_w=2.50\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}$ . A completely dissociated strong hydroxide has formula $\mathrm{M(OH)_n}$ , concentration $0.00600\,\mathrm{mol\,dm^{-3}}$ and pH $11.681$ . Deduce $n$ . The solution is then diluted to four times its original volume; calculate the new pH.
[6 marks]
Total for this question: 6
At a stated temperature, $K_w=2.50\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}$ . A sodium hydroxide portion has volume $25.0\,\mathrm{cm^3}$ and concentration $0.0800\,\mathrm{mol\,dm^{-3}}$ ; a barium hydroxide portion has volume $15.0\,\mathrm{cm^3}$ and concentration $0.0500\,\mathrm{mol\,dm^{-3}}$ . Calculate the pH after mixing them, assuming complete dissociation and additive volumes.
[5 marks]
Total for this question: 5
At an unknown temperature, a $0.00250\,\mathrm{mol\,dm^{-3}}$ solution of a completely dissociated strong hydroxide, $\mathrm{M(OH)_2}$ , has pH $10.893$ . Calculate $K_w$ at this temperature and deduce whether the temperature is above or below $298\,\mathrm{K}$ .
[4 marks]
Total for this question: 4

3.1.12.4 · Weak acids and bases Ka for weak acids (A-level only)

Explanation

Weak acids and weak bases dissociate only slightly in aqueous solution. For $\mathrm{HA\rightleftharpoons H^++A^-}$ , the acid dissociation constant is $K_a=[H^+][A^-]/[HA]$ ; it has units for this expression.
For a weak monoprotic acid of concentration $c$ , small dissociation permits $[HA]\approx c$ and $[H^+]=[A^-]$ , giving $[H^+]\approx\sqrt{K_ac}$ .
If dissociation is not negligible, equilibrium $[HA]$ must be reduced by the dissociated amount.
The logarithmic measure is $pK_a=-\log_{10}K_a$ ; larger $K_a$ and smaller $pK_a$ both indicate a stronger acid.
At half-neutralisation, measured pH can determine $pK_a$ and hence $K_a$ .

Worked example

Calculate the pH of $0.150\,\mathrm{mol\,dm^{-3}}$ weak monoprotic acid with $K_a=6.30\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ , using the small-dissociation approximation.

1. $[H^+]\approx\sqrt{K_a[HA]}=\sqrt{(6.30\times10^{-5})(0.150)}$ .
2. $[H^+]=3.07\times10^{-3}\,\mathrm{mol\,dm^{-3}}$ .
3. $\mathrm{pH}=-\log_{10}(3.07\times10^{-3})$ .

Answer: $\mathrm{pH}=2.512$ .

Common mistakes

Don't use the initial acid concentration as $[H^+]$ , treating a weak acid as completely dissociated.
Don't write $K_a=[HA]/([H^+][A^-])$ , inverting the equilibrium expression.
Don't state that a larger $pK_a$ means a stronger acid, reversing the logarithmic relationship.

Exam tip

For a weak-acid calculation, write the $K_a$ expression before applying any small-dissociation approximation.

Tier 1 · Easy

A weak acid has $K_a=1.74\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ . Calculate its $pK_a$ .
[2 marks]
Total for this question: 2
Weak acid P has $pK_a=3.90$ and weak acid Q has $pK_a=4.80$ . Identify the stronger acid and calculate how many times larger its $K_a$ is.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Calculate the pH of a $0.150\,\mathrm{mol\,dm^{-3}}$ solution of a weak monoprotic acid with $K_a=6.30\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ . Use the small-dissociation approximation.
[4 marks]
Total for this question: 4
A weak monoprotic acid has pH $2.70$ and $K_a=4.00\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ . Use the small-dissociation approximation to calculate its initial concentration. Give your answer to 3 significant figures.
[4 marks]
Total for this question: 4
A $0.200\,\mathrm{mol\,dm^{-3}}$ solution of a weak monoprotic acid has $K_a=5.60\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ . Use the small-dissociation approximation to calculate the percentage of the acid that is dissociated.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A $0.0800\,\mathrm{mol\,dm^{-3}}$ solution of a weak monoprotic acid has pH $2.40$ . Calculate $K_a$ without assuming that the equilibrium acid concentration equals its initial concentration, and hence calculate $pK_a$ .
[5 marks]
Total for this question: 5
A weak monoprotic acid has concentration $0.0100\,\mathrm{mol\,dm^{-3}}$ and $K_a=4.00\times10^{-4}\,\mathrm{mol\,dm^{-3}}$ . Show that the small-dissociation approximation is unsuitable, then use $K_a=x^2/(0.0100-x)$ to calculate the pH.
[6 marks]
Total for this question: 6
An unnamed weak monoprotic acid has mass concentration $61.0\,\mathrm{g\,dm^{-3}}$ , $K_a=1.60\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ and pH $2.500$ . Without using the small-dissociation approximation, calculate its equilibrium acid concentration, its initial concentration and its relative molecular mass.
[6 marks]
Total for this question: 6
A weak monoprotic acid has $K_a=2.50\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ . Its initial concentration is $0.0500\,\mathrm{mol\,dm^{-3}}$ , then a sample is diluted tenfold. Using the small-dissociation approximation, calculate the pH and percentage dissociation before and after dilution, and explain the change in percentage dissociation.
[5 marks]
Total for this question: 5
A weak monoprotic acid has $K_a=1.74\times10^{-4}\,\mathrm{mol\,dm^{-3}}$ and is $8.00\%$ dissociated at equilibrium. Calculate its initial concentration and pH.
[5 marks]
Total for this question: 5

3.1.12.5 · pH curves, titrations and indicators (A-level only)

Explanation

Acid–base titration calculations use experimental amounts and reaction stoichiometry to locate equivalence. Typical pH curves are required for all combinations of weak and strong monoprotic acids and bases.
Curve shape depends on initial strength, any buffer region, the size and position of the rapid pH change, and the final excess titrant.
Equivalence pH is about $7$ only for strong acid–strong base; conjugate-ion hydrolysis shifts weak-acid or weak-base equivalence.
An indicator is suitable only when its transition range lies within the steep curve section.
Required practical 9 investigates pH change for weak acid–strong base and strong acid–weak base titrations.

Typical strong-acid and weak-acid titration curves have different starting and equivalence pH values.

Worked example

$25.0\,\mathrm{cm^3}$ of $0.120\,\mathrm{mol\,dm^{-3}}$ hydrochloric acid is titrated with $0.125\,\mathrm{mol\,dm^{-3}}$ sodium hydroxide. Calculate the equivalence volume and state the approximate equivalence pH.

1.Acid amount $=0.120\times0.0250=0.00300\,\mathrm{mol}$ .
2.The reaction is $1{:}1$ , so $0.00300\,\mathrm{mol}$ of sodium hydroxide is required.
3. $V=0.00300/0.125=0.0240\,\mathrm{dm^3}$ .

Answer: Equivalence volume $=24.0\,\mathrm{cm^3}$ and equivalence pH $\approx7$ .

Common mistakes

Don't assume every acid–base titration has equivalence pH $7$ , ignoring conjugate-ion hydrolysis.
Don't identify an indicator because its range contains pH $7$ rather than because it lies inside the steep section.
Don't treat half-equivalence as equivalence in a weak acid–strong base curve.

Exam tip

For a 'sketch and explain' curve question, label both equivalence volume and the chemically correct equivalence-pH region.

Tier 1 · Easy

A strong acid–weak base titration has a steep pH change from $3.8$ to $6.5$ . Methyl orange changes colour from pH $3.1$ to $4.4$ , while phenolphthalein changes from pH $8.3$ to $10.0$ . Select the suitable indicator and explain your choice.
[2 marks]
Total for this question: 2
A titration curve starts at pH $2.9$ , has a buffer region and has an equivalence point at pH $8.8$ . A base was added from the burette. Identify whether the acid and base are strong or weak. Justify each answer using the curve.
[3 marks]
Total for this question: 3

Tier 2 · Standard

At $298\,\mathrm{K}$ , a flask holds $25.0\,\mathrm{cm^3}$ of $0.120\,\mathrm{mol\,dm^{-3}}$ hydrochloric acid. A burette supplies $0.125\,\mathrm{mol\,dm^{-3}}$ sodium hydroxide. Calculate the equivalence volume and state the approximate pH at equivalence.
[3 marks]
Total for this question: 3
In required practical 9, a student adds base in $5.0\,\mathrm{cm^3}$ portions throughout a titration and records pH with an uncalibrated probe. Explain two changes that would give better evidence for the shape and position of the steep pH change.
[4 marks]
Total for this question: 4
At $298\,\mathrm{K}$ , a measured aliquot of a strong monoprotic acid is titrated with a strong base. Before adding the base, distilled water is added to the flask without any loss of acid. Compare the diluted titration curve with the original curve in terms of initial pH, volume of base at equivalence, pH at equivalence, and pH after the same fixed volume of base has been added beyond equivalence.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A flask contains $20.0\,\mathrm{cm^3}$ of $0.150\,\mathrm{mol\,dm^{-3}}$ weak monoprotic acid with $K_a=2.50\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ . Sodium hydroxide of concentration $0.125\,\mathrm{mol\,dm^{-3}}$ is added from a burette. Calculate the volume at half-equivalence and the pH at that point. State whether the equivalence pH is below, equal to or above $7$ .
[5 marks]
Total for this question: 5
A flask contains $20.0\,\mathrm{cm^3}$ of a monoprotic acid. Sodium hydroxide of concentration $0.150\,\mathrm{mol\,dm^{-3}}$ is added from a burette. A pH curve shows equivalence at $30.0\,\mathrm{cm^3}$ , pH $4.76$ at $15.0\,\mathrm{cm^3}$ , and a rapid pH rise from $7.8$ to $10.2$ around equivalence. Deduce the acid concentration and $K_a$ , identify the acid as strong or weak, and identify either methyl orange $(3.1$ – $4.4)$ or phenolphthalein $(8.3$ – $10.0)$ as a suitable indicator. Give your answer to 3 significant figures.
[7 marks]
Total for this question: 7
At $298\,\mathrm{K}$ , $26.0\,\mathrm{cm^3}$ of a $0.115\,\mathrm{mol\,dm^{-3}}$ weak monoprotic acid, $\mathrm{HA}$ , for which $K_a=2.20\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ , is titrated with $0.0800\,\mathrm{mol\,dm^{-3}}$ sodium hydroxide. Calculate the pH after $9.50\,\mathrm{cm^3}$ of sodium hydroxide has been added and identify the region of the titration curve. Assume the volumes are additive. Around equivalence, the curve rises rapidly from pH $7.7$ to pH $10.5$ . Methyl orange changes colour over pH $3.1$ – $4.4$ and phenolphthalein over pH $8.3$ – $10.0$ . Justify which indicator should be used.
[6 marks]
Total for this question: 6
At $298\,\mathrm{K}$ , $30.0\,\mathrm{cm^3}$ of $0.120\,\mathrm{mol\,dm^{-3}}$ hydrochloric acid is titrated with $0.150\,\mathrm{mol\,dm^{-3}}$ sodium hydroxide. Calculate the equivalence volume and the pH after $20.0\,\mathrm{cm^3}$ of sodium hydroxide has been added. Identify the region of the titration curve.
[5 marks]
Total for this question: 5
At $298\,\mathrm{K}$ , a flask contains $25.0\,\mathrm{cm^3}$ of $0.120\,\mathrm{mol\,dm^{-3}}$ hydrochloric acid. A burette contains a $0.100\,\mathrm{mol\,dm^{-3}}$ weak base, B. Calculate the initial pH and the volume of B required for equivalence. State whether the equivalence pH is below, equal to or above $7$ and explain why. Identify where the buffer region occurs. Methyl orange changes colour over pH $3.1$ – $4.4$ and phenolphthalein over pH $8.3$ – $10.0$ ; select the suitable indicator and explain the rejection of the other indicator.
[6 marks]
Total for this question: 6

3.1.12.6 · Buffer action (A-level only)

Explanation

A buffer maintains approximately constant pH despite dilution or addition of small amounts of acid or base. An acidic buffer contains a weak acid and its salt; its conjugate base removes added $\mathrm{H^+}$ , while the weak acid removes added $\mathrm{OH^-}$ .
A basic buffer contains a weak base and its salt and acts through the equivalent conjugate pair.
Applications depend on resisting harmful pH changes.
For an acidic buffer, $[H^+]=K_a[HA]/[A^-]$ , equivalently $\mathrm{pH}=pK_a+\log_{10}([A^-]/[HA])$ .
After adding acid or base, amounts must first be adjusted stoichiometrically; dilution alone leaves the ratio approximately unchanged.

Worked example

A buffer contains $0.100\,\mathrm{mol}$ of $\mathrm{HA}$ and $0.0800\,\mathrm{mol}$ of $\mathrm{A^-}$ . After adding $0.0100\,\mathrm{mol}$ of HCl, calculate pH if $K_a=1.80\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ .

1.Added $\mathrm{H^+}$ reacts with $\mathrm{A^-}$ , giving $0.0700\,\mathrm{mol}$ of $\mathrm{A^-}$ and $0.110\,\mathrm{mol}$ of $\mathrm{HA}$ .
2. $pK_a=-\log_{10}(1.80\times10^{-5})=4.7447$ .
3. $\mathrm{pH}=4.7447+\log_{10}(0.0700/0.110)$ .

Answer: $\mathrm{pH}=4.548$ .

Common mistakes

Don't substitute the original buffer amounts after added acid has changed both components.
Don't claim dilution destroys buffer action immediately even though both component concentrations change by the same factor.
Don't explain acidic buffer action using only the weak acid and omit removal of added $\mathrm{H^+}$ by the conjugate base.

Exam tip

In a buffer calculation after addition, write the neutralisation reaction and update both mole amounts before using the buffer ratio.

Tier 1 · Easy

An acidic buffer contains equal concentrations of a weak acid and its conjugate base. The acid has $K_a=1.80\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ . Calculate the pH.
[3 marks]
Total for this question: 3
An acidic buffer is diluted with water. State the effect on its pH and on its ability to resist pH change when acid or base is added.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An acidic buffer contains $\mathrm{HA}$ and $\mathrm{A^-}$ . Explain, using equations, how it resists small additions of acid and base.
[4 marks]
Total for this question: 4
A beaker contains $100\,\mathrm{cm^3}$ of $0.200\,\mathrm{mol\,dm^{-3}}$ weak acid $\mathrm{HA}$ . To it are added $50.0\,\mathrm{cm^3}$ of sodium hydroxide whose concentration is $0.100\,\mathrm{mol\,dm^{-3}}$ . For $\mathrm{HA}$ , $K_a=1.80\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ . Calculate the pH of the resulting buffer. Give your answer to 2 decimal places and carry unrounded values through the working.
[5 marks]
Total for this question: 5
An acidic buffer contains $0.0750\,\mathrm{mol}$ of $\mathrm{A^-}$ and $0.0300\,\mathrm{mol}$ of the weak acid $\mathrm{HA}$ in the same solution. Its pH is $5.10$ . Determine $pK_a$ and $K_a$ for $\mathrm{HA}$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

$0.500\,\mathrm{dm^3}$ of a buffer contains $0.100\,\mathrm{mol}$ of $\mathrm{HA}$ and $0.0800\,\mathrm{mol}$ of $\mathrm{A^-}$ . The acid has $K_a=1.80\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ . Calculate the pH after adding $0.0100\,\mathrm{mol}$ of hydrochloric acid. Assume the volume is unchanged.
[5 marks]
Total for this question: 5
$200\,\mathrm{cm^3}$ of a buffer contains $0.0400\,\mathrm{mol}$ of $\mathrm{HA}$ and $0.0600\,\mathrm{mol}$ of $\mathrm{A^-}$ . For $\mathrm{HA}$ , $K_a=1.80\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ . The buffer is diluted to $500\,\mathrm{cm^3}$ , then $10.0\,\mathrm{cm^3}$ of $0.500\,\mathrm{mol\,dm^{-3}}$ sodium hydroxide is added. Calculate the pH after dilution and the pH after adding the sodium hydroxide. Assume volumes are additive. Give each pH to 2 decimal places and carry unrounded values through the working.
[7 marks]
Total for this question: 7
A buffer initially contains $0.0400\,\mathrm{mol}$ HA and $0.0600\,\mathrm{mol}$ $\mathrm{A^-}$ ; $pK_a=4.80$ . Calculate the maximum amount of hydrochloric acid that can be added while keeping pH at or above $4.60$ , and the maximum amount of sodium hydroxide that can be added while keeping pH at or below $5.10$ . Assume volume changes are negligible and compare the two limits.
[6 marks]
Total for this question: 6
A buffer is to contain a total of $0.100\,\mathrm{mol}$ of $\mathrm{HA}$ and $\mathrm{A^-}$ . The weak acid has $pK_a=4.20$ , and the target pH is $3.90$ . Determine the amounts of $\mathrm{HA}$ and $\mathrm{A^-}$ required. Then determine the amount of sodium hydroxide that must be added to $0.100\,\mathrm{mol}$ of HA to prepare this buffer by partial neutralisation.
[5 marks]
Total for this question: 5
Buffer 1 has volume $0.200\,\mathrm{dm^3}$ and contains $0.100\,\mathrm{mol\,dm^{-3}}$ HA and $0.0500\,\mathrm{mol\,dm^{-3}}$ $\mathrm{A^-}$ . Buffer 2 has volume $0.300\,\mathrm{dm^3}$ and contains $0.0400\,\mathrm{mol\,dm^{-3}}$ HA and $0.120\,\mathrm{mol\,dm^{-3}}$ $\mathrm{A^-}$ . Both use the same conjugate pair with $pK_a=4.600$ . Calculate the pH after the buffers are mixed, assuming additive volumes.
[5 marks]
Total for this question: 5

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

3.1.1.1 · Fundamental particles

Tier 1 · Easy

Mark scheme for 3.1.1.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Relative charge $=-1$ Relative mass $\approx1/1836$	Recall the fundamental-particle data: an electron has relative charge $-1$ and a mass much smaller than a nucleon's, approximately $1/1836$ on the relative scale.	2
02.1	The proton and neutron counts, which account for almost all the mass, are unchanged Each electron has a relative mass of only about $1/1836$ , so the lost mass is negligible on this scale	Ion formation changes only the electron count. The nucleus is unchanged, and losing two particles whose individual relative mass is about $1/1836$ produces a negligible change compared with the mass of the protons and neutrons.	2

Tier 2 · Standard

Mark scheme for 3.1.1.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Overall charge $=3-$ The nucleus contains the 15 protons and 16 neutrons	Only protons and electrons contribute to the overall charge. There are three more electrons than protons, so the charge is $3-$ . Protons and neutrons occupy the nucleus; electrons are outside it.	3
02.1	Changing the proton number would change the element Ion formation does not change the nucleus A $1-$ ion forms when the atom gains one electron	Element identity is fixed by proton number, so converting a proton into a neutron would not describe ion formation. A negative ion is produced by gaining electrons; gaining one electron gives charge $1-$ while leaving the nucleus unchanged.	3
03.1	A proton is in the nucleus An electron has relative mass approximately $1/1836$ An electron is outside the nucleus	Check location and relative mass separately. Protons and neutrons occupy the nucleus. Electrons occupy the surrounding space and have relative mass about $1/1836$ , not $1$ ; the three relative charges in the table are already correct.	3

Tier 3 · Hard

Mark scheme for 3.1.1.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	13 protons Ion charge $=2-$ 15 electrons	The proton number is $(2.08\times10^{-18})/(1.60\times10^{-19})=13$ . A neutral atom with 13 protons has 13 electrons. Gaining two electrons gives 15 electrons while leaving 13 protons, so the ion has charge $2-$ .	4
02.1	P is a neutron Q is an electron R is a proton The electron and proton have opposite charges, and the electron is deflected more because its mass is about $1/1836$ of the proton mass	An uncharged neutron is not deflected. A negative electron moves towards the positive plate and a positive proton towards the negative plate. With equal charge magnitudes and the same starting speed, the much smaller electron mass produces the much larger deflection.	4
03.1	Most alpha particles passing through without deflection shows that most of the atom is empty space The few large-angle deflections show that the positive charge and most of the mass are concentrated in a very small, dense nucleus The plum-pudding model spreads positive charge throughout the atom, so it cannot produce large-angle deflections Electrons occupy shells or energy levels around the nucleus (allow: neutrons are present in the nucleus)	If positive charge were diffuse, as in the plum-pudding model, large deflections would not occur. The scattering evidence instead requires an atom that is mainly empty space with its positive charge and most of its mass concentrated in a tiny nucleus. The modern model adds a nucleus of protons and neutrons with electrons in the surrounding space, neither of which the plum-pudding model contains.	4
04.1	The radius ratio is $(4.8\times10^{-15})/(1.2\times10^{-10})=4.0\times10^{-5}$ The volume ratio is $(4.0\times10^{-5})^3=6.4\times10^{-14}$ Electrons occupy the space outside the nucleus, so they determine most of the atomic volume Protons and neutrons are concentrated in the nucleus and each has relative mass about 1, whereas electron mass is negligible, so the nucleus contains almost all the mass	For spheres, the common factor $4\pi/3$ cancels, so the volume ratio is the cube of the radius ratio. Cubing $4.0\times10^{-5}$ gives $6.4\times10^{-14}$ . The tiny nucleus contains the massive nucleons, while the very light electrons occupy the much larger surrounding region.	4
05.1	Deflection towards the positive plate shows that the cathode-ray particles are negatively charged An unchanged charge-to-mass ratio with different gases shows that the same particle is present regardless of the gas An unchanged ratio with different cathode metals shows that the particle is a component of atoms of different elements The particles are electrons, so atoms are divisible and an indivisible-atom model cannot explain the evidence	First infer the sign of charge from the direction of electrostatic attraction. Then use the invariance of charge-to-mass ratio: changing both sources without changing the measured ratio identifies one universal particle rather than a property of one material. This supports electrons as constituents of atoms and rules out an indivisible atom.	4

3.1.1.2 · Mass number and isotopes

Tier 1 · Easy

Mark scheme for 3.1.1.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	35 protons 46 neutrons 36 electrons	The atomic number gives 35 protons. The neutron number is $81-35=46$ . A $1-$ ion has gained one electron, so it has $35+1=36$ electrons.	3
02.1	P and R are isotopes They have the same number of protons but different numbers of neutrons	Isotopes must have the same proton number, which fixes the element, but different neutron numbers. P and R both have 18 protons and have 22 and 24 neutrons respectively.	2

Tier 2 · Standard

Mark scheme for 3.1.1.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$A_r=50.54$	Use the weighted mean: $A_r=(50\times73.0+52\times27.0)/100=(3650+1404)/100=50.54$ .	3
02.1	Sample B is enriched in the heavier isotope Its $m/z=65$ peak has the greater relative area Sample B has the greater relative atomic mass	Peak area represents relative abundance. The heavier isotope contributes $60\%$ of sample B but only $30\%$ of sample A, so the weighted mean mass, and therefore $A_r$ , is greater for sample B.	3
03.1	The sample particles must be ionised before acceleration At the same kinetic energy, the ion with the smaller mass reaches the detector first The detector signal or current is proportional to the number, and hence relative abundance, of ions arriving An electric field exerts a force only on charged particles, so neutral particles would not be accelerated	Ionisation is required because an electric field accelerates charged particles rather than neutral particles. Since all ions receive the same kinetic energy and $E_k=\tfrac12mv^2$ , a lower-mass ion travels faster and arrives sooner. Each arriving ion contributes to the detector current, so a larger current indicates a greater number and relative abundance of ions.	4

Tier 3 · Hard

Mark scheme for 3.1.1.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Mass number $=70$	Both ions have the same charge, so $m_2/m_1=(t_2/t_1)^2$ . Thus $m_2=64(1.339/1.280)^2=70.04$ . An isotope has an integer mass number, so the second isotope is $^{70}\mathrm{X}$ .	4
02.1	Flight time $=1.25\times10^{-5}\,\mathrm{s}$ ; allow $1.249\times10^{-5}\,\mathrm{s}$ $A_r(\mathrm{Mg})=24.32$	At constant kinetic energy, flight time is proportional to the square root of mass for ions with the same charge: $t\propto\sqrt{m}$ . Therefore $t(^{26}\mathrm{Mg^+})=1.20\times10^{-5}\times\sqrt{26/24}=1.249\times10^{-5}\,\mathrm{s}$ . The abundance-weighted mean is $A_r=(24\times79.0+25\times10.0+26\times11.0)/100=24.32$ .	4
03.1	$m/z=30$ : $^{60}\mathrm{X^{2+}}$ ; $m/z=31$ : $^{62}\mathrm{X^{2+}}$ $m/z=60$ : $^{60}\mathrm{X^+}$ ; $m/z=62$ : $^{62}\mathrm{X^+}$ $t=8.00\times10^{-7}\sqrt{62/60}$ $t=8.13\times10^{-7}\,\mathrm{s}$ to three significant figures	For a singly charged ion, $m/z$ equals its mass number; a $2+$ charge halves the value. The reference and target ions are therefore $^{60}\mathrm{X^{2+}}$ and $^{62}\mathrm{X^{2+}}$ . All ions receive the same kinetic energy, so flight time is proportional to the square root of mass. Hence $t_{62}/t_{60}=\sqrt{62/60}$ and $t_{62}=8.00\times10^{-7}\sqrt{62/60}=8.13\times10^{-7}\,\mathrm{s}$ .	5
04.1	$m/z=20$ is produced by $^{10}\mathrm{X}$ - $^{10}\mathrm{X}$ $m/z=21$ is produced by the two equivalent mixed arrangements $^{10}\mathrm{X}$ - $^{11}\mathrm{X}$ and $^{11}\mathrm{X}$ - $^{10}\mathrm{X}$ $m/z=22$ is produced by $^{11}\mathrm{X}$ - $^{11}\mathrm{X}$ The respective probabilities are $0.800^2=0.640$ , $2(0.800)(0.200)=0.320$ and $0.200^2=0.0400$ The expected peak-area ratio for $m/z$ 20:21:22 is $16:8:1$	Add the two isotope mass numbers because every ion contains two X atoms and has charge $1+$ . For the mixed ion, either atom can be the heavier isotope, so its probability contains a factor of two. Squaring and multiplying the fractional abundances gives $0.640:0.320:0.0400$ , which simplifies to $16:8:1$ .	5
05.1	Ion arrival rate is proportional to detector current divided by ionic charge $N(^{68}\mathrm{X})=1.20+(2.40/2)=2.40$ in proportional units $N(^{70}\mathrm{X})=0.450+(0.900/2)=0.900$ , so the isotope-population ratio is $2.40:0.900=8:3$ The percentage abundance of $^{70}\mathrm{X}$ is $(3/11)\times100=27.3\%$ $A_r=[68(8)+70(3)]/11=68.5$ to three significant figures	Detector current measures charge delivered per unit time, not ion count directly. A $2+$ ion delivers twice the charge of a $1+$ ion, so divide both $2+$ currents by 2. Summing the corrected $1+$ and $2+$ contributions for each isotope gives $2.40:0.900=8:3$ . This population ratio supplies both the percentage abundance and the weighted mean.	5

3.1.1.3 · Electron configuration

Tier 1 · Easy

Mark scheme for 3.1.1.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1s^2\,2s^2\,2p^6\,3s^2\,3p^6$	A sulfur atom has 16 electrons and the $2-$ ion has gained two, giving 18. Filling the sub-shells in order gives $1s^2\,2s^2\,2p^6\,3s^2\,3p^6$ .	1
02.1	Phosphorus Atomic number $=15$	The configuration contains $2+2+6+2+3=15$ electrons. A neutral atom therefore has 15 protons and atomic number 15, identifying phosphorus.	2

Tier 2 · Standard

Mark scheme for 3.1.1.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Energy required to remove one electron from each atom in one mole of gaseous atoms $\mathrm{Mg(g)\rightarrow Mg^+(g)+e^-}$	The definition must specify one mole of gaseous atoms and removal of one electron from every atom. The equation therefore starts with one gaseous Mg atom and forms a gaseous $1+$ ion plus one electron.	3
02.1	The ion has charge $2+$ The two $4s$ electrons are removed $4s$ electrons are removed before $3d$ electrons when this ion forms	The ion configuration contains two fewer electrons than the atom. Both missing electrons were in the $4s$ sub-shell, so the species has lost two electrons and has charge $2+$ .	3
03.1	The species are $\mathrm{Ar}$ , $\mathrm{K^+}$ and $\mathrm{Ca^{2+}}$ $1s^2\,2s^2\,2p^6\,3s^2\,3p^6$ They have 18, 19 and 20 protons respectively Element identity is determined by proton number, not electron configuration	Neutral argon has 18 electrons, giving the full configuration $1s^2\,2s^2\,2p^6\,3s^2\,3p^6$ . Potassium must lose one electron and calcium must lose two to become isoelectronic with it. The shorthand $\mathrm{[Ar]}$ is acceptable for $\mathrm{K^+}$ and $\mathrm{Ca^{2+}}$ , but argon's own configuration must be written in full. Their nuclei still contain different proton numbers, so they remain three different elements despite sharing an electron configuration.	4

Tier 3 · Hard

Mark scheme for 3.1.1.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The large jump after the fourth ionisation shows that X has four outer-shell electrons X is in Group 14 and is silicon Outer electron configuration $3s^2\,3p^2$ $\mathrm{Si^{4+}(g)\rightarrow Si^{5+}(g)+e^-}$	The sharp rise from the fourth to the fifth value shows that four electrons are removed before an electron must be taken from a shell closer to the nucleus. X therefore has four outer electrons and is the Group 14 element in Period 3, silicon, with outer configuration $3s^2\,3p^2$ . The fifth ionisation removes one electron from each gaseous $\mathrm{Si^{4+}}$ ion: $\mathrm{Si^{4+}(g)\rightarrow Si^{5+}(g)+e^-}$ .	5
02.1	Chromium is $\mathrm{[Ar]3d^5\,4s^1}$ Chromium is an exception to the simple filling pattern $4s$ electrons are removed before $3d$ electrons $\mathrm{Cr^{3+}}$ is $\mathrm{[Ar]3d^3}$	The chromium atom has the exceptional half-filled $3d$ configuration $\mathrm{[Ar]3d^5\,4s^1}$ . Forming $\mathrm{Cr^{3+}}$ removes the one $4s$ electron first and then two $3d$ electrons, leaving $\mathrm{[Ar]3d^3}$ .	4
03.1	$\mathrm{Mg^{2+}}$ is $1s^2\,2s^2\,2p^6$ and its third ionisation removes a $2p$ electron $\mathrm{Al^{2+}}$ is $1s^2\,2s^2\,2p^6\,3s^1$ and its third ionisation removes the $3s$ electron The electron removed from $\mathrm{Mg^{2+}}$ is in a shell closer to the nucleus It is less shielded and more strongly attracted to the nucleus, so much more energy is required	Magnesium loses both $3s$ electrons before its third ionisation, leaving the neon configuration. Its next electron must come from the inner $2p$ sub-shell. Aluminium loses one $3p$ and one $3s$ electron first, so $\mathrm{Al^{2+}}$ still has one outer $3s$ electron. The inner $2p$ electron in $\mathrm{Mg^{2+}}$ is closer and less shielded, producing the much larger third ionisation energy.	5
04.1	$\mathrm{Cu(g)\rightarrow Cu^+(g)+e^-}$ $\mathrm{Cu^+}$ has configuration $\mathrm{[Ar]3d^{10}}$ $\mathrm{Cu^+(g)\rightarrow Cu^{2+}(g)+e^-}$ $\mathrm{Cu^{2+}}$ has configuration $\mathrm{[Ar]3d^9}$ The second ionisation removes a $3d$ electron from a positive ion rather than the outer $4s$ electron from a neutral atom The $3d$ electron is closer to the nucleus and experiences stronger attraction, so more energy is needed	Remove the $4s$ electron first, giving the filled $3d^{10}$ configuration. The next electron must then come from $3d$ . It is removed from an already positive species and from a sub-shell closer to the nucleus, so its electrostatic attraction is substantially stronger.	6
05.1	Each species has electron configuration $\mathrm{1s^2\,2s^2\,2p^6}$ The nuclear charges are $10+$ for Ne, $11+$ for $\mathrm{Na^+}$ and $12+$ for $\mathrm{Mg^{2+}}$ The electron configuration and shielding are the same in all three species Increasing nuclear charge therefore increases the attraction between the nucleus and the electron removed $\mathrm{IE_1(Ne)<IE_2(Na)<IE_3(Mg)}$	Ne, $\mathrm{Na^+}$ and $\mathrm{Mg^{2+}}$ are isoelectronic. The electron removed is therefore from the same $2p$ sub-shell with the same inner-shell shielding in every case. Nuclear charge rises from 10 to 12, so nuclear attraction and the energy required for removal rise in the order $\mathrm{IE_1(Ne)<IE_2(Na)<IE_3(Mg)}$ .	5

3.1.2.1 · Relative atomic mass and relative molecular mass

Tier 1 · Easy

Mark scheme for 3.1.2.1 Tier 1 · Easy
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01.1	$M_r=60.0$	Urea contains one C, one O, two N and four H atoms. Therefore $M_r=12.0+16.0+2(14.0)+4(1.0)=60.0$ .	2
02.1	Neon: relative atomic mass, $A_r$ Carbon dioxide: relative molecular mass, $M_r$ Magnesium chloride: relative formula mass	Use relative atomic mass for an element's atoms, relative molecular mass for discrete molecules, and relative formula mass for an ionic formula unit because an ionic solid does not contain separate molecules.	3

Tier 2 · Standard

Mark scheme for 3.1.2.1 Tier 2 · Standard
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01.1	Relative formula mass $=246.4$	The anhydrous part contributes $24.3+32.1+4(16.0)=120.4$ . Seven waters contribute $7[2(1.0)+16.0]=126.0$ . The total is $120.4+126.0=246.4$ .	3
02.1	$M_r$ is a ratio and has no units, so $M_r(\mathrm{H_2O})=18.0$ Molar mass has units, so the molar mass is $18.0\,\mathrm{g\,mol^{-1}}$	Relative molecular mass compares masses on a reference scale and is dimensionless. Molar mass is the mass per mole, so it carries units of $\mathrm{g\,mol^{-1}}$ .	3
03.1	One twelfth of the carbon-12 atom mass is $1.6605\times10^{-27}\,\mathrm{kg}$ $M_r=(7.9704\times10^{-26})/(1.6605\times10^{-27})=48.0$ $M_r$ has no units	Relative molecular mass compares the mean mass of a molecule with one twelfth of the mass of a carbon-12 atom. The reference mass is $(1.9926\times10^{-26})/12=1.6605\times10^{-27}\,\mathrm{kg}$ . Dividing the molecular mass by this reference gives $48.0$ ; the units cancel because it is a ratio.	3

Tier 3 · Hard

Mark scheme for 3.1.2.1 Tier 3 · Hard
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01.1	$A_r(\mathrm{X})=48.05$ Abundance of $^{47}\mathrm{X}=47.5\%$	The three oxygen atoms contribute $3\times16.0=48.0$ , so $2A_r(\mathrm{X})=144.1-48.0=96.1$ and $A_r(\mathrm{X})=48.05$ . Let the fractional abundance of $^{47}\mathrm{X}$ be $x$ . Then $47x+49(1-x)=48.05$ , so $x=0.475$ . The percentage abundance is therefore $47.5\%$ .	4
02.1	$x=10$	The anhydrous relative formula mass is $2(23.0)+12.0+3(16.0)=106.0$ . The water contributes $286.0-106.0=180.0$ . Since $M_r(\mathrm{H_2O})=18.0$ , $x=180.0/18.0=10$ .	3
03.1	$M_r(\mathrm{X_2O_3})=152.0$ and oxygen percentage $=(48.0/152.0)\times100=31.6\%$ $M_r(\mathrm{XO_2})=84.0$ and oxygen percentage $=(32.0/84.0)\times100=38.1\%$ Mass of $\mathrm{X_2O_3}=8.00\times152.0/48.0=25.3\,\mathrm{g}$ Mass of $\mathrm{XO_2}=8.00\times84.0/32.0=21.0\,\mathrm{g}$ $\mathrm{XO_2}$ uses less material	First calculate each relative formula mass and the oxygen contribution to it. The oxygen fractions are $48.0/152.0$ and $32.0/84.0$ . To reverse the calculation, divide the required oxygen mass by the relevant oxygen fraction. This gives $25.3\,\mathrm{g}$ of $\mathrm{X_2O_3}$ but only $21.0\,\mathrm{g}$ of $\mathrm{XO_2}$ .	5
04.1	$M_r(\mathrm{NH_4NO_3})=14.0+4(1.0)+14.0+3(16.0)=80.0$ $M_r(\mathrm{(NH_4)_2SO_4})=2(14.0+4(1.0))+32.1+4(16.0)=132.1$ Four moles of formula units have total relative mass $3(80.0)+132.1=372.1$ Mean relative formula mass $=372.1/4=93.0$	Calculate the relative formula mass of each component, then apply the stated mole ratio as a weighted mean. Three formula units have relative formula mass 80.0 for every one of relative formula mass 132.1, so the mean is $[3(80.0)+132.1]/4=93.025$ , or $93.0$ to three significant figures.	4
05.1	The anhydrous relative formula mass is $2(23.0)+2(32.1)+3(16.0)=158.2$ The water fraction gives $18.0x/(158.2+18.0x)=0.363$ $18.0x=57.4266+6.534x$ $x=5.01$ , so the whole-number value is $5$ The hydrate has relative formula mass $158.2+5(18.0)=248.2$	Express the water contribution as a fraction of the total relative formula mass. Solving the resulting equation gives a value within rounding distance of five waters per formula unit. Substitute that integer into the full formula-mass sum to obtain 248.2.	5

3.1.2.2 · The mole and the Avogadro constant

Tier 1 · Easy

Mark scheme for 3.1.2.2 Tier 1 · Easy
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01.1	$0.0500\,\mathrm{mol}$	Use $n=m/M$ : $n=5.30/106.0=0.0500\,\mathrm{mol}$ .	2
02.1	$0.0500\,\mathrm{mol}$	Rearrange $N=nN_A$ to give $n=N/N_A$ . Therefore $n=(3.01\times10^{22})/(6.02\times10^{23})=0.0500\,\mathrm{mol}$ .	2

Tier 2 · Standard

Mark scheme for 3.1.2.2 Tier 2 · Standard
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01.1	$7.53\times10^{21}$ molecules	Multiply moles by the Avogadro constant: $N=0.0125\times6.02\times10^{23}=7.525\times10^{21}$ , which is $7.53\times10^{21}$ molecules to three significant figures.	2
02.1	Mass $=2.12\,\mathrm{g}$ A $250.0\,\mathrm{cm^3}$ volumetric flask	Convert the volume to $0.2500\,\mathrm{dm^3}$ . The amount required is $n=cV=0.0800\times0.2500=0.0200\,\mathrm{mol}$ , so $m=nM=0.0200\times106.0=2.12\,\mathrm{g}$ . A volumetric flask sets the final volume accurately.	3
03.1	Molar mass $=(7.31\times10^{-23})(6.02\times10^{23})=44.0\,\mathrm{g\,mol^{-1}}$ Amount in $0.220\,\mathrm{g}=0.220/44.0=5.00\times10^{-3}\,\mathrm{mol}$ Number of molecules $=3.01\times10^{21}$	Multiplying the mass of one molecule by the number of molecules per mole gives $44.0\,\mathrm{g\,mol^{-1}}$ . The sample therefore contains $0.220/44.0=0.00500\,\mathrm{mol}$ . Multiplying this amount by $N_A$ gives $0.00500\times6.02\times10^{23}=3.01\times10^{21}$ molecules.	4

Tier 3 · Hard

Mark scheme for 3.1.2.2 Tier 3 · Hard
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01.1	$8.13\times10^{21}$ sulfate ions	Convert the volume: $18.0\,\mathrm{cm^3}=0.0180\,\mathrm{dm^3}$ . Formula units amount to $n=cV=0.250\times0.0180=0.00450\,\mathrm{mol}$ . Each formula unit gives three sulfate ions, so their amount is $0.0135\,\mathrm{mol}$ . Hence $N=0.0135\times6.02\times10^{23}=8.13\times10^{21}$ sulfate ions.	4
02.1	$2.41\times10^{21}$ chloride ions remain	$n(\mathrm{AlCl_3})=0.200\times0.0150=0.00300\,\mathrm{mol}$ , so initially $n(\mathrm{Cl^-})=3(0.00300)=0.00900\,\mathrm{mol}$ . The silver-ion amount is $0.250\times0.0200=0.00500\,\mathrm{mol}$ , so it removes $0.00500\,\mathrm{mol}$ chloride ions. The amount remaining is $0.00400\,\mathrm{mol}$ , containing $0.00400\times6.02\times10^{23}=2.408\times10^{21}$ ions, or $2.41\times10^{21}$ to three significant figures.	5
03.1	$n(\mathrm{NaCl})=0.585/58.5=0.0100\,\mathrm{mol}$ Each $\mathrm{Na^+}$ ion has 10 electrons and each $\mathrm{Cl^-}$ ion has 18 electrons Each formula unit therefore contains $28$ electrons in its ions Amount of electrons $=0.280\,\mathrm{mol}$ Number of electrons $=1.69\times10^{23}$	The sample contains $0.0100\,\mathrm{mol}$ of sodium chloride formula units. Sodium loses one of its 11 electrons to form $\mathrm{Na^+}$ , while chlorine gains one to give 18 in $\mathrm{Cl^-}$ , so each ion pair contains 28 electrons. Thus there are $28\times0.0100=0.280\,\mathrm{mol}$ electrons, or $0.280\times6.02\times10^{23}=1.69\times10^{23}$ electrons.	5
04.1	Amount of hydrate $=0.5619/561.9=1.000\times10^{-3}\,\mathrm{mol}$ Each formula unit contains $12+9=21$ oxygen atoms Amount of oxygen atoms $=21(1.000\times10^{-3})=0.02100\,\mathrm{mol}$ Number of oxygen atoms $=0.02100(6.022\times10^{23})=1.265\times10^{22}$	Convert the sample mass to moles of complete hydrated formula units. Three sulfate ions supply 12 oxygen atoms and nine waters supply another nine. Multiply the formula-unit amount by 21, then use the Avogadro constant.	4
05.1	Amount of X in the sample $=0.0264/63.5=4.157\times10^{-4}\,\mathrm{mol}$ $N_A=N/n$ $N_A=(2.50\times10^{20})/(4.157\times10^{-4})=6.013\times10^{23}\,\mathrm{mol^{-1}}$ Using $6.022\times10^{23}-6.01325\times10^{23}$ , the absolute difference is $8.75\times10^{20}\,\mathrm{mol^{-1}}$ Percentage difference $=(8.75\times10^{20}/6.022\times10^{23})\times100=0.145\%$	Use the measured mass and molar mass to find how many moles contain the stated atom count, then rearrange $N=nN_A$ . Keeping unrounded values gives $N_A=6.01326\times10^{23}\,\mathrm{mol^{-1}}$ and a percentage difference of $0.145\%$ .	5

3.1.2.3 · The ideal gas equation

Tier 1 · Easy

Mark scheme for 3.1.2.3 Tier 1 · Easy
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01.1	$1.00\times10^5\,\mathrm{Pa}$	Rearrange to $p=nRT/V$ . Then $p=[0.100\times8.31\times300]/(2.49\times10^{-3})=1.00\times10^5\,\mathrm{Pa}$ .	3
02.1	$p=9.60\times10^4\,\mathrm{Pa}$ $V=2.50\times10^{-4}\,\mathrm{m^3}$ $T=295\,\mathrm{K}$	The gas constant $R$ has the value $8.31$ in units of $\mathrm{J\,K^{-1}\,mol^{-1}}$ , so pressure must be in pascals, volume in cubic metres and temperature in kelvin. Multiply kilopascals by $10^3$ , multiply cubic centimetres by $10^{-6}$ and add 273 to the Celsius temperature.	3

Tier 2 · Standard

Mark scheme for 3.1.2.3 Tier 2 · Standard
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01.1	$0.873\,\mathrm{dm^3}$	Convert pressure to $105000\,\mathrm{Pa}$ . Then $V=nRT/p=(0.0350\times8.31\times315)/105000=8.73\times10^{-4}\,\mathrm{m^3}$ . Multiplying by $1000$ gives $0.873\,\mathrm{dm^3}$ .	4
02.1	$296\,\mathrm{K}$	Use $p=120000\,\mathrm{Pa}$ and $V=410\times10^{-6}\,\mathrm{m^3}$ . Rearranging gives $T=pV/(nR)=(120000\times410\times10^{-6})/(0.0200\times8.31)=296.03\,\mathrm{K}$ , which is $296\,\mathrm{K}$ to three significant figures.	4
03.1	$p=9.85\times10^4\,\mathrm{Pa}$ and $V=2.45\times10^{-4}\,\mathrm{m^3}$ $R=pV/(nT)$ $R=8.29\,\mathrm{J\,K^{-1}\,mol^{-1}}$ to three significant figures	Convert the measured pressure and volume to SI units, then rearrange $pV=nRT$ to $R=pV/(nT)$ . Substitution gives $R=[(9.85\times10^4)(2.45\times10^{-4})]/(0.0100\times291)=8.29\,\mathrm{J\,K^{-1}\,mol^{-1}}$ .	4

Tier 3 · Hard

Mark scheme for 3.1.2.3 Tier 3 · Hard
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01.1	$118\,\mathrm{g\,mol^{-1}}$	Use SI units: $p=98000\,\mathrm{Pa}$ and $V=85.0\times10^{-6}\,\mathrm{m^3}$ . The amount is $n=pV/(RT)=(98000\times85.0\times10^{-6})/(8.31\times373)=2.687\times10^{-3}\,\mathrm{mol}$ . Therefore $M=m/n=0.318/(2.687\times10^{-3})=118\,\mathrm{g\,mol^{-1}}$ .	5
02.1	Hydrochloric acid is limiting Hydrogen volume $=199\,\mathrm{cm^3}$	$n(\mathrm{Mg})=0.240/24.3=0.0098765\,\mathrm{mol}$ and $n(\mathrm{HCl})=0.800\times0.0200=0.0160\,\mathrm{mol}$ . The magnesium would require $0.019753\,\mathrm{mol}$ HCl, so HCl limits and forms $0.0160/2=0.00800\,\mathrm{mol}$ hydrogen. Then $V=nRT/p=(0.00800\times8.31\times305)/102000=1.9879\times10^{-4}\,\mathrm{m^3}=199\,\mathrm{cm^3}$ to three significant figures.	6
03.1	$p_2=112\times(1.80/1.50)\times(300/320)=126.0\,\mathrm{kPa}$ At the same final volume and temperature, $n_{\mathrm{remaining}}/n_{\mathrm{original}}=92.0/126.0=0.73016$ Fraction escaped $=1-0.73016=0.26984$ Percentage escaped $=27.0\%$	For a fixed amount, $pV/T$ is constant, so the leak-free final pressure is $112\times1.80\times300/(320\times1.50)=126.0\,\mathrm{kPa}$ . At the same final $V$ and $T$ , pressure is proportional to amount. The measured-to-expected pressure ratio is therefore the fraction remaining: $92.0/126.0=0.73016$ . Hence $(1-0.73016)\times100=27.0\%$ escaped. Using the stated $126\,\mathrm{kPa}$ intermediate gives the same final percentage.	5
04.1	$p=1.00\times10^5\,\mathrm{Pa}$ and $V=5.25\times10^{-5}\,\mathrm{m^3}$ $n(\mathrm{CO_2})=pV/(RT)=2.120\times10^{-3}\,\mathrm{mol}$ The $1:1$ reaction ratio gives $n(\mathrm{CaCO_3})=2.120\times10^{-3}\,\mathrm{mol}$ Mass of calcium carbonate $=(2.120\times10^{-3})(100.1)=0.2122\,\mathrm{g}$ Percentage purity $=(0.2122/0.250)\times100=84.9\%$	Convert the pressure and volume to SI units before using the gas equation. Carbonate and carbon dioxide are in a $1:1$ mole ratio, so the gas amount fixes the mass of pure calcium carbonate. Compare this mass with the total sample mass, retaining unrounded intermediates.	5
05.1	One cubic decimetre has volume $1.000\times10^{-3}\,\mathrm{m^3}$ and mass $1.742\,\mathrm{g}$ Amount in this volume $=pV/(RT)=(102000\times10^{-3})/(8.31\times310)=0.0395947\,\mathrm{mol}$ Molar mass $=1.742/0.0395947=43.996\,\mathrm{g\,mol^{-1}}=44.0\,\mathrm{g\,mol^{-1}}$ The gas is $\mathrm{CO_2}$ Because $M\propto1/V$ , the calculated molar mass is about $2\%$ too large	Choose a convenient sample volume of $1.000\,\mathrm{dm^3}$ so the density gives its mass directly. Use the ideal gas equation to find the amount in that volume, then divide mass by amount. The rearranged expression shows the inverse dependence of calculated molar mass on measured volume.	5

3.1.2.4 · Empirical and molecular formula

Tier 1 · Easy

Mark scheme for 3.1.2.4 Tier 1 · Easy
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01.1	$\mathrm{Al_2O_3}$	Moles are Al: $2.70/27.0=0.100$ and O: $2.40/16.0=0.150$ . Dividing by $0.100$ gives $1:1.5$ ; multiplying both by $2$ gives $2:3$ , so the empirical formula is $\mathrm{Al_2O_3}$ .	3
02.1	Empirical formula $\mathrm{CH_2O}$ The subscripts in $\mathrm{C_2H_4O_2}$ share a factor of two	An empirical formula is the simplest whole-number ratio. Dividing all the subscripts $2:4:2$ by two gives the ratio $1:2:1$ and formula $\mathrm{CH_2O}$ .	2

Tier 2 · Standard

Mark scheme for 3.1.2.4 Tier 2 · Standard
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01.1	Empirical formula $\mathrm{C_2H_4O}$ Molecular formula $\mathrm{C_4H_8O_2}$	For a $100\,\mathrm{g}$ sample, moles are C: $54.5/12.0=4.542$ , H: $9.1/1.0=9.1$ , O: $36.4/16.0=2.275$ . Dividing by $2.275$ gives approximately $2:4:1$ , so the empirical formula is $\mathrm{C_2H_4O}$ . Its formula mass is $44$ , and $88/44=2$ , giving $\mathrm{C_4H_8O_2}$ .	5
02.1	The value 1.5 must not be rounded to 2 Multiply both parts of $1:1.5$ by two The empirical formula is $\mathrm{X_2Y_3}$	Dividing both amounts by $0.0200$ gives the exact ratio $1:1.5$ . A common whole-number multiplier must be applied to every term, so doubling gives $2:3$ and formula $\mathrm{X_2Y_3}$ .	3
03.1	Mass of oxygen gained $=4.00-2.40=1.60\,\mathrm{g}$ $n(\mathrm{X})=2.40/48.0=0.0500\,\mathrm{mol}$ and $n(\mathrm{O})=1.60/16.0=0.100\,\mathrm{mol}$ Mole ratio X:O $=1:2$ Empirical formula $\mathrm{XO_2}$	The increase in mass is the oxygen that combined with X: $4.00-2.40=1.60\,\mathrm{g}$ . The amounts are $2.40/48.0=0.0500\,\mathrm{mol}$ of X and $1.60/16.0=0.100\,\mathrm{mol}$ of O. Dividing both by $0.0500$ gives the simplest whole-number ratio $1:2$ , so the empirical formula is $\mathrm{XO_2}$ .	4

Tier 3 · Hard

Mark scheme for 3.1.2.4 Tier 3 · Hard
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01.1	Empirical formula $\mathrm{CH_2O}$ Molecular formula $\mathrm{C_5H_{10}O_5}$	Moles of $\mathrm{CO_2}=1.320/44.0=0.0300$ , so there are $0.0300\,\mathrm{mol}$ C atoms with mass $0.360\,\mathrm{g}$ . Moles of $\mathrm{H_2O}=0.540/18.0=0.0300$ , so there are $0.0600\,\mathrm{mol}$ H atoms with mass $0.0600\,\mathrm{g}$ . Oxygen mass is $0.900-0.360-0.060=0.480\,\mathrm{g}$ , or $0.0300\,\mathrm{mol}$ . The ratio C:H:O is $1:2:1$ , giving $\mathrm{CH_2O}$ . Its mass is $30$ , and $150/30=5$ , so the molecular formula is $\mathrm{C_5H_{10}O_5}$ .	7
02.1	Empirical formula $\mathrm{CH_2}$ Molecular formula $\mathrm{C_4H_8}$	For $100\,\mathrm{g}$ , the relative amounts are C: $85.7/12.0=7.1417$ and H: $14.3/1.0=14.300$ . Dividing by $7.1417$ gives approximately $1:2$ , so the empirical formula is $\mathrm{CH_2}$ with mass 14.0. Gas amount is $n=pV/(RT)=(100000\times124.7\times10^{-6})/(8.31\times300)=0.00500201\,\mathrm{mol}$ . The molar mass is $0.280/0.00500201=55.98\,\mathrm{g\,mol^{-1}}$ . The multiplier is $55.98/14.0\approx4$ , giving $\mathrm{C_4H_8}$ .	7
03.1	$M_r(\mathrm{C_3H_7NO_2})=89.0$ ; nitrogen $=(14.0/89.0)\times100=15.7\%$ and oxygen $=(32.0/89.0)\times100=35.955\%$ , which is $36.0\%$ to 3 significant figures $M_r(\mathrm{C_2H_5NO})=59.0$ ; nitrogen $=(14.0/59.0)\times100=23.7\%$ and oxygen $=(16.0/59.0)\times100=27.1\%$ $\mathrm{C_3H_7NO_2}$ is consistent with both measured percentages $\mathrm{C_2H_5NO}$ is inconsistent with both measured percentages	Test each proposed empirical formula rather than deriving a ratio from the measurements. For $\mathrm{C_3H_7NO_2}$ , nitrogen and oxygen contribute 14 and 32 of 89 mass units, giving $15.7\%$ and $35.955\%$ — the latter is $36.0\%$ to 3 significant figures, so both match the analysis. For $\mathrm{C_2H_5NO}$ they contribute 14 and 16 of 59, giving values far from both measurements. The first candidate is therefore supported.	5
04.1	Carbon dioxide volume $=40.0-10.0=30.0\,\mathrm{cm^3}$ The hydrocarbon: $\mathrm{CO_2}$ volume ratio is $1:3$ , so each molecule contains three carbon atoms Oxygen consumed $=50.0-10.0=40.0\,\mathrm{cm^3}$ , giving a hydrocarbon: $\mathrm{O_2}$ ratio of $1:4$ For $\mathrm{C_xH_y}$ , complete combustion requires $x+y/4$ oxygen molecules, so $3+y/4=4$ and $y=4$ The molecular formula is $\mathrm{C_3H_4}$	At equal temperature and pressure, gas-volume ratios equal mole ratios. Alkali removes $30.0\,\mathrm{cm^3}$ carbon dioxide, fixing three carbon atoms per hydrocarbon molecule. The reaction consumes $40.0\,\mathrm{cm^3}$ oxygen per $10.0\,\mathrm{cm^3}$ hydrocarbon. Substitution into $x+y/4=4$ gives four hydrogen atoms.	5
05.1	Two nitrogen atoms contribute relative mass $2(14.0)=28.0$ $28.0/M_r=0.304$ $M_r=92.1$ , consistent with $92$ after rounding of the percentage The oxygen contribution is $92-28=64$ , corresponding to $64/16.0=4$ oxygen atoms The molecular formula is $\mathrm{N_2O_4}$ and $M_r=92.0$	The stated nitrogen atom count fixes the nitrogen mass contribution. Divide that contribution by the nitrogen mass fraction to infer the molecular mass; the small non-integer result reflects the rounded percentage. Subtract the nitrogen contribution and divide the remainder by 16.0 to obtain four oxygen atoms.	5

3.1.2.5 · Balanced equations and associated calculations

Tier 1 · Easy

Mark scheme for 3.1.2.5 Tier 1 · Easy
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01.1	$\mathrm{C_3H_8+5O_2\rightarrow3CO_2+4H_2O}$	Balance C first to give $3\mathrm{CO_2}$ , then H to give $4\mathrm{H_2O}$ . The products contain $10$ O atoms in total, requiring $5\mathrm{O_2}$ .	1
02.1	The student's equation does not conserve oxygen atoms $\mathrm{Fe_2O_3+3CO\rightarrow2Fe+3CO_2}$	Two iron atoms require coefficient 2 before Fe. Using three CO molecules then gives six oxygen atoms on each side and three carbon atoms on each side, producing the balanced $1:3:2:3$ ratio.	2

Tier 2 · Standard

Mark scheme for 3.1.2.5 Tier 2 · Standard
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01.1	$0.188\,\mathrm{g}$	Moles of HCl are $0.150\times0.0250=0.00375\,\mathrm{mol}$ . The $1:2$ ratio gives $0.00375/2=0.001875\,\mathrm{mol}$ calcium carbonate. Its mass is $0.001875\times100.1=0.1877\,\mathrm{g}$ , or $0.188\,\mathrm{g}$ to three significant figures.	4
02.1	Mean concordant titre $=18.40\,\mathrm{cm^3}$ $c(\mathrm{NaOH})=0.147\,\mathrm{mol\,dm^{-3}}$	The concordant titres are $18.35$ and $18.45\,\mathrm{cm^3}$ , with mean $18.40\,\mathrm{cm^3}$ . Acid amount is $0.1000\times0.01840=0.001840\,\mathrm{mol}$ . The $1:2$ ratio gives $0.003680\,\mathrm{mol}$ NaOH, so $c=0.003680/0.02500=0.1472\,\mathrm{mol\,dm^{-3}}$ , or $0.147\,\mathrm{mol\,dm^{-3}}$ to three significant figures.	5
03.1	$\mathrm{Ca(NO_3)_2(aq)+Na_2CO_3(aq)\rightarrow CaCO_3(s)+2NaNO_3(aq)}$ $\mathrm{Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)}$ $\mathrm{Na^+}$ and $\mathrm{NO_3^-}$ are spectator ions	Balance the soluble salts first. Split aqueous ionic compounds into ions, but do not split the solid precipitate. Sodium and nitrate ions occur unchanged on both sides and cancel, leaving the charge-balanced $1:1$ precipitation equation for calcium and carbonate ions.	4

Tier 3 · Hard

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01.1	Ammonia is the limiting reactant Maximum urea mass $=150\,\mathrm{kg}$ Atom economy $=76.9\%$ Percentage yield $=84.0\%$	Amounts are $85.0/17.0=5.00\,\mathrm{kmol}$ ammonia and $132/44.0=3.00\,\mathrm{kmol}$ carbon dioxide. Five kmol ammonia requires only $2.50\,\mathrm{kmol}$ carbon dioxide, so ammonia limits the reaction and forms $2.50\,\mathrm{kmol}$ urea. The maximum mass is $2.50\times60.0=150\,\mathrm{kg}$ . Atom economy is $[60.0/(2\times17.0+44.0)]\times100=76.9\%$ . Percentage yield is $(126/150)\times100=84.0\%$ .	7
02.1	The metal ion has charge $3+$ $\mathrm{2M(s)+6H^+(aq)\rightarrow2M^{3+}(aq)+3H_2(g)}$	$n(\mathrm{M})=0.270/27.0=0.0100\,\mathrm{mol}$ . The hydrogen amount is $n=pV/(RT)=(100000\times371.5\times10^{-6})/(8.31\times298)=0.015002\,\mathrm{mol}$ . The ratio is therefore $1.00$ mol M to $1.50$ mol $\mathrm{H_2}$ . Each mole of hydrogen requires two moles of electrons, so each mole of M loses three moles of electrons and forms $\mathrm{M^{3+}}$ . Whole-number balancing gives $\mathrm{2M+6H^+\rightarrow2M^{3+}+3H_2}$ .	6
03.1	Mass lost as $\mathrm{CO_2}=4.00-3.12=0.88\,\mathrm{g}$ Amount decomposed $=n(\mathrm{CO_2})=0.88/44.0=0.0200\,\mathrm{mol}$ Initial amount of $\mathrm{MCO_3}=4.00/100.0=0.0400\,\mathrm{mol}$ , so percentage decomposed $=(0.0200/0.0400)\times100=50.0\%$ Undecomposed $\mathrm{MCO_3}=0.0200\,\mathrm{mol}$ , with mass $0.0200\times100.0=2.00\,\mathrm{g}$ $M_r(\mathrm{MO})=100.0-44.0=56.0$ , so its mass is $0.0200\times56.0=1.12\,\mathrm{g}$	The only gaseous product is $\mathrm{CO_2}$ , so the decrease in solid mass gives its mass directly. Convert $0.88\,\mathrm{g}$ of $\mathrm{CO_2}$ to $0.0200\,\mathrm{mol}$ ; the $1:1$ equation makes this the amount of carbonate decomposed and oxide formed. Compare it with the initial $0.0400\,\mathrm{mol}$ to obtain $50.0\%$ conversion. The remaining carbonate is $0.0200\,\mathrm{mol}$ . Removing one $\mathrm{CO_2}$ unit from the formula mass gives $M_r(\mathrm{MO})=56.0$ ; the two solid masses, $2.00+1.12=3.12\,\mathrm{g}$ , check against the stated residue.	5
04.1	Required amount of C $=12.0/120=0.100\,\mathrm{mol}$ Theoretical C needed before the second-stage loss $=0.100/0.750=0.1333\,\mathrm{mol}$ The $2:1$ ratio requires $2(0.1333)=0.2667\,\mathrm{mol}$ B Theoretical B needed before the first-stage loss $=0.2667/0.800=0.3333\,\mathrm{mol}$ Minimum mass of A $=0.3333(50.0)=16.7\,\mathrm{g}$	Work backwards from the required product. Divide by the second-stage fractional yield to recover the theoretical C amount, apply the $2:1$ B:C ratio, then divide by the first-stage yield. The $1:1$ A:B ratio makes the resulting amount equal to the A amount, which converts to $16.7\,\mathrm{g}$ .	5
05.1	Route A atom economy $=(120/190)\times100=63.2\%$ Route A yield-adjusted product fraction $=0.632(0.920)\times100=58.1\%$ Route B atom economy $=(120/150)\times100=80.0\%$ Route B yield-adjusted product fraction $=0.800(0.720)\times100=57.6\%$ The lower operating temperature means lower energy cost; Route B's toxic gas requires containment and treatment. A justified conclusion using the calculated efficiencies and process evidence; for example, Route A has slightly greater actual product per reactant input, needs less heating and avoids toxic gas, while Route B may be preferred if theoretical atom economy is given priority	For each route, divide theoretical desired-product mass by the total stoichiometric reactant mass. Multiplying atom economy by fractional yield compares actual desired product with reactant input. The adjusted values are close, so a conclusion must state how it weighs operating temperature and by-product hazard against Route B's higher theoretical atom economy.	6

3.1.3.1 · Ionic bonding

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01.1	$\mathrm{Al_2(SO_4)_3}$	The lowest common total charge is 6: two $\mathrm{Al^{3+}}$ ions give $+6$ and three sulfate ions give $-6$ . Therefore the neutral formula is $\mathrm{Al_2(SO_4)_3}$ .	1
02.1	Electron transfer forms the ions but is not the bond An ionic bond is the electrostatic attraction between oppositely charged ions	The transfer explains how charged particles are produced. The bond is the strong electrostatic attraction that then acts between positive and negative ions throughout the lattice.	2

Tier 2 · Standard

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01.1	A giant lattice contains $\mathrm{Mg^{2+}}$ and $\mathrm{O^{2-}}$ ions There is strong electrostatic attraction between oppositely charged ions The attraction acts throughout the lattice	Name the charged particles and the force between them: magnesium oxide consists of $\mathrm{Mg^{2+}}$ and $\mathrm{O^{2-}}$ ions held by strong electrostatic attractions in every direction through a giant lattice.	3
02.1	Three bracketed $\mathrm{Li^+}$ ions, each with no outer-shell electron shown One bracketed $\mathrm{N^{3-}}$ ion with eight outer-shell electrons Five nitrogen electrons and three transferred lithium electrons are distinguished around N All ion charges are shown	Each lithium atom loses one electron and nitrogen gains three, forming three $\mathrm{Li^+}$ ions and one $\mathrm{N^{3-}}$ ion. The nitride ion has an octet made from its five original electrons and three transferred electrons; every ion must be bracketed and charged.	4
03.1	M forms $\mathrm{M^{2+}}$ by losing its two outer electrons X forms $\mathrm{X^{2-}}$ by gaining two electrons The simple ions combine in a $1:1$ ratio, so the formula is $\mathrm{MX}$ Two ammonium ions balance one $\mathrm{X^{2-}}$ ion, so the formula is $\mathrm{(NH_4)_2X}$	A Group 2 atom loses two electrons to form a $2+$ ion, whereas a Group 6 atom gains two electrons to form a $2-$ ion. Equal and opposite charges give the formula $\mathrm{MX}$ . Ammonium has charge $1+$ , so two ammonium ions are required to balance one $\mathrm{X^{2-}}$ ion; brackets preserve the polyatomic ion in $\mathrm{(NH_4)_2X}$ .	4

Tier 3 · Hard

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01.1	$\mathrm{X^{3+}}$ $\mathrm{X(NO_3)_3}$ $\mathrm{X(OH)_3}$	Three carbonate ions contribute total charge $3\times(-2)=-6$ , so two X ions must contribute $+6$ and each is $\mathrm{X^{3+}}$ . Three singly charged nitrate or hydroxide ions are then needed per X ion, giving $\mathrm{X(NO_3)_3}$ and $\mathrm{X(OH)_3}$ .	4
02.1	$\mathrm{MgO}$ contains $\mathrm{Mg^{2+}}$ and $\mathrm{O^{2-}}$ , whereas $\mathrm{NaF}$ contains singly charged ions The products of the ionic charges are greater in magnesium oxide $\mathrm{Mg^{2+}}$ is smaller than $\mathrm{Na^+}$ The stronger attractions in magnesium oxide require more energy to overcome	Electrostatic attraction strengthens with higher ionic charge and shorter ion separation. The doubly charged ions in magnesium oxide, together with the smaller magnesium ion compared with sodium, produce much stronger lattice attractions than the singly charged ions in sodium fluoride.	4
03.1	Let the numbers of $\mathrm{M^{3+}}$ and $\mathrm{X^{2-}}$ ions be $m$ and $x$ , so $m+x=20$ Charge neutrality requires $3m=2x$ Solving gives $m=8$ and $x=12$ The ratio $\mathrm{M:X}=8:12=2:3$ The formula is $\mathrm{M_2X_3}$	Use both constraints: the ion counts total 20 and the total positive and negative charges must be equal. Substituting $x=20-m$ into $3m=2x$ gives $3m=40-2m$ , so $m=8$ and $x=12$ . Simplifying the $8:12$ ratio gives the neutral ionic formula $\mathrm{M_2X_3}$ .	5
04.1	Three $\mathrm{Na^+}$ ions provide charge $3+$ in $\mathrm{Na_3PO_4}$ , so phosphate is $\mathrm{PO_4^{3-}}$ Three $\mathrm{Mg^{2+}}$ ions provide charge $6+$ in $\mathrm{Mg_3(PO_4)_2}$ , so two phosphate ions provide $6-$ and each is again $3-$ One $\mathrm{Al^{3+}}$ ion balances one $\mathrm{PO_4^{3-}}$ ion The aluminium salt is $\mathrm{AlPO_4}$	Apply charge neutrality to each known formula. Both sodium phosphate and magnesium phosphate require phosphate to carry charge $3-$ . Aluminium carries charge $3+$ , so the aluminium-to-phosphate ratio is $1:1$ and the formula is $\mathrm{AlPO_4}$ .	4
05.1	One $\mathrm{OH^-}$ and one $\mathrm{Cl^-}$ give a combined charge of $2-$ , balanced by one $\mathrm{Mg^{2+}}$ The ratio $\mathrm{Mg^{2+}:OH^-:Cl^-}=1:1:1$ gives the simplest neutral formula $\mathrm{Mg(OH)Cl}$ With twice as many chloride ions, $2\,\mathrm{OH^-}+4\,\mathrm{Cl^-}$ carry charge $6-$ , balanced by three $\mathrm{Mg^{2+}}$ ( $6+$ ) That compound's formula is $\mathrm{Mg_3(OH)_2Cl_4}$	Use one of each anion to satisfy the equal-number condition. Their combined charge is $1-+1-=2-$ , which is balanced by one magnesium ion of charge $2+$ , giving $\mathrm{Mg(OH)Cl}$ . For the second compound, take two hydroxide and four chloride ions: total anion charge $6-$ requires three $\mathrm{Mg^{2+}}$ , giving $\mathrm{Mg_3(OH)_2Cl_4}$ .	4

3.1.3.2 · Nature of covalent and dative covalent bonds

Tier 1 · Easy

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01.1	A shared pair of electrons	A single covalent bond consists of one electron pair shared between two atoms.	1
02.1	A double covalent bond Two shared pairs of electrons	Four shared electrons form two electron pairs. Two shared pairs between the same atoms constitute a double covalent bond.	2

Tier 2 · Standard

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01.1	$\mathrm{Cl_3B\leftarrow NH_3}$ The nitrogen atom donates the electron pair	Nitrogen has the lone pair and boron accepts it. The arrow must therefore start at N and point towards B, represented as $\mathrm{Cl_3B\leftarrow NH_3}$ .	3
02.1	$\mathrm{Cl^-}$ is the electron-pair donor and Al is the acceptor The arrow must start at a lone pair on Cl The arrow must point from Cl towards Al	A displayed dative-bond arrow follows the donated electron pair. Chloride supplies a lone pair and the electron-deficient aluminium atom accepts it, so the arrow tail is on the chloride lone pair and the head is at Al.	3
03.1	Three shared pairs of electrons form an N $\equiv$ N triple bond Each shared pair contains one electron from each nitrogen atom One lone pair remains on each nitrogen atom There are three shared pairs and two lone pairs in total	Each nitrogen atom has five outer electrons and needs three more for an octet. The atoms share three pairs, with one electron from each atom in every pair, forming a triple bond. Two electrons remain as a lone pair on each nitrogen.	4

Tier 3 · Hard

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01.1	$\mathrm{NH_3+H^+\rightarrow NH_4^+}$ The arrow goes from the nitrogen lone pair to $\mathrm{H^+}$ Once formed, the dative bond is an ordinary covalent bond and all four N-H bonds are equivalent	Nitrogen donates its lone pair to the electron-pair acceptor $\mathrm{H^+}$ , so the arrow runs from N to H. This produces $\mathrm{NH_4^+}$ . The origin of the pair no longer makes that bond different, so the four N-H bonds are equivalent.	4
02.1	$\mathrm{Ag^++2NH_3\rightarrow[Ag(NH_3)_2]^+}$ Each arrow starts at a nitrogen lone pair and points towards Ag Each ammonia donates two electrons Four electrons are donated in total	Each ammonia ligand donates one lone pair to the silver ion, so both arrows run $\mathrm{N\rightarrow Ag}$ . One lone pair contains two electrons; two ligands therefore donate four electrons altogether while the complex retains overall charge $1+$ .	4
03.1	Two single bonds use four electrons; completing both oxygen octets uses the other twelve, leaving carbon with only four electrons around it The single-bond model therefore fails to give carbon an octet The correct line formula is $\mathrm{O=C=O}$ The two double bonds contain four shared pairs in total Each oxygen has two lone pairs, giving four lone pairs in total	Start with the fixed total of 16 outer electrons. In the proposed structure, two single bonds use four electrons and the two oxygen atoms need the remaining twelve as lone pairs, so carbon has only two shared pairs and no octet. Converting one lone pair from each oxygen into an additional shared pair produces two C=O double bonds. Carbon and both oxygen atoms then have octets.	5
04.1	The diagram has a C $\equiv$ O triple bond containing three shared pairs One lone pair is shown on carbon One lone pair is shown on oxygen Both atoms have eight electrons in their outer shells Oxygen supplies both electrons in one shared pair, so the dative component is represented $\mathrm{C\leftarrow O}$	Ten electrons are sufficient for three bonding pairs and two lone pairs. Place the three shared pairs between C and O and one lone pair on each atom; each atom then counts six bonding electrons plus its lone pair. In a dot-and-cross representation, one bonding pair contains two oxygen electrons, so its arrow points from O to C.	5
05.1	The diagram shows two Al atoms joined through two Al-Cl-Al bridges, with two terminal Cl atoms on each Al A lone pair is shown on each bridging chlorine donor Each dative arrow starts at a bridging chlorine lone pair and points to the electron-deficient Al atom in the other unit Each dative bond contains one donated electron pair Two dative bonds transfer four electrons into shared pairs in total	Start from two trigonal $\mathrm{AlCl_3}$ units. Use one chlorine from each unit as a bridge: its existing ordinary bond remains to its original aluminium and a lone pair forms a dative bond to the other aluminium. The two lone-pair donations therefore create two coordinate bonds and involve four electrons.	5

3.1.3.3 · Metallic bonding

Tier 1 · Easy

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01.1	Electrostatic attraction between positive ions and delocalised electrons	Identify both components of the lattice and the force between them: positive metal ions are held by electrostatic attraction to delocalised electrons.	2
02.1	Ion charge $=3+$ Three delocalised electrons per atom	Each aluminium atom contributes its three outer electrons to the delocalised electron system. Losing three electrons leaves a positive metal ion with charge $3+$ .	2

Tier 2 · Standard

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01.1	The metal contains delocalised electrons These electrons are mobile and carry charge through the structure They remain delocalised when the lattice melts	Electrical conduction requires mobile charged particles. The delocalised electrons can move through a solid metal and are still present and mobile after the regular ion lattice breaks down on melting.	3
02.1	Layers of positive metal ions can slide past one another The delocalised electrons move with or between the layers Non-directional electrostatic attraction between the ions and electrons is maintained	Metallic attraction is not confined to fixed pairs of neighbouring ions. When layers shift, the mobile electron density continues to attract the positive ions, so the structure can change shape without the bonding failing all at once.	3
03.1	Every Group 1 metal forms $1+$ ions and supplies one delocalised electron per atom, so ion charge and electron count are the same throughout the group The metal ion radius increases down the group from $\mathrm{Li^+}$ to $\mathrm{Cs^+}$ The larger ions place the delocalised electrons further from the positive charge, so the electrostatic attraction between the ions and the delocalised electrons weakens Weaker metallic bonding means less energy is needed to break down the lattice, so the melting point falls	Isolate the one variable that changes: down Group 1 the ion charge ( $1+$ ) and delocalised-electron count (one per atom) are constant, but the ion radius grows. Metallic bonding strength depends on the attraction between the positive ions and the delocalised electron system, which weakens as the ions become larger, so successively less energy is needed to melt each metal.	4

Tier 3 · Hard

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01.1	Magnesium forms $\mathrm{Mg^{2+}}$ ions and supplies two delocalised electrons per atom Sodium forms $\mathrm{Na^+}$ ions and supplies one delocalised electron per atom $\mathrm{Mg^{2+}}$ has greater charge and is smaller than $\mathrm{Na^+}$ The attraction to the delocalised electrons is stronger in magnesium, so more energy is needed to melt it	Magnesium contributes two electrons and leaves a smaller, doubly charged ion. This creates stronger electrostatic attraction between the ion lattice and the denser sea of delocalised electrons than in sodium, so more energy is required to overcome the bonding.	4
02.1	The lattice particles are positive metal ions, not neutral atoms The outer electrons are delocalised rather than fixed between neighbours Mobile delocalised electrons carry charge through the solid The electrostatic attraction is non-directional Attraction remains when ion layers slide, allowing deformation	Localised pairs could not account for mobile charge carriers or continued bonding after layers move. A metal instead contains positive ions attracted to a mobile, delocalised electron system throughout the lattice, explaining both conduction and malleability.	5
03.1	Let the P and Q counts be $p$ and $q$ : $p+q=6$ Electron contribution gives $p+2q=9$ Solving gives three P cores and three Q cores Their total positive charge is $3(1)+3(2)=9+$ , balanced by nine delocalised electrons Electrostatic attraction between all the positive cores and the delocalised electron system holds the mixed lattice together	Use the ion-core count and electron count as simultaneous constraints. Subtracting $p+q=6$ from $p+2q=9$ gives $q=3$ , so $p=3$ . The nine electrons carry charge $9-$ and exactly balance the $9+$ from the three singly and three doubly charged cores. Because the electrons are delocalised, they attract the positive cores throughout the whole region rather than forming six separate local bonds.	5
04.1	$n(\text{Al atoms})=2.70/27.0=0.100\,\mathrm{mol\,cm^{-3}}$ $n(\text{delocalised electrons in Al})=(2.70/27.0)\times3=0.300\,\mathrm{mol\,cm^{-3}}$ $n(\text{Mg atoms})=1.74/24.3=0.0716\,\mathrm{mol\,cm^{-3}}$ $n(\text{delocalised electrons in Mg})=(1.74/24.3)\times2=0.143\,\mathrm{mol\,cm^{-3}}$ Aluminium has the greater delocalised-electron density and $3+$ ion cores rather than $2+$ ion cores, so the electrostatic attraction is stronger and its metallic bonds are stronger	In each $1.00\,\mathrm{cm^3}$ sample, divide mass by molar mass to find moles of atoms and multiply by the number of delocalised electrons per atom. This gives $0.300\,\mathrm{mol\,cm^{-3}}$ for aluminium and $0.143\,\mathrm{mol\,cm^{-3}}$ for magnesium. The greater electron density and higher ion-core charge in aluminium produce stronger electrostatic attraction between the positive cores and the delocalised electrons.	5
05.1	Sodium has a metallic lattice with electrostatic attraction between positive sodium ion cores and delocalised electrons Sodium chloride has an ionic lattice with electrostatic attraction between oppositely charged $\mathrm{Na^+}$ and $\mathrm{Cl^-}$ ions The ion-ion attractions in sodium chloride are much stronger than the ion-delocalised-electron attractions in sodium Much more energy is therefore needed to overcome the attractions in sodium chloride, giving it the much higher melting point	Compare the particles attracting one another rather than merely labelling both substances as lattices. Sodium is held by metallic ion-delocalised-electron attraction; sodium chloride is held by stronger ionic ion-ion attraction throughout its lattice. The stronger attraction requires more energy to overcome on melting.	4

3.1.3.4 · Bonding and physical properties

Tier 1 · Easy

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01.1	Molecular crystal Intermolecular forces between $\mathrm{I_2}$ molecules are overcome	Discrete molecules form a molecular crystal. Melting separates the molecules by overcoming intermolecular forces; it does not break the covalent I-I bonds inside them.	2
02.1	The ions are fixed in the solid lattice and cannot carry charge through it Magnesium chloride conducts when molten or dissolved because its ions can move	Containing charged particles is not sufficient for conduction; those particles must be mobile. Melting or dissolving releases the ions from fixed lattice positions so they can carry charge.	2

Tier 2 · Standard

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01.1	Graphite has a macromolecular or giant covalent structure Many strong covalent bonds require much energy to break Each carbon contributes a delocalised electron The delocalised electrons move along the layers and carry charge	Graphite consists of extended covalently bonded carbon layers, so melting requires many strong bonds to be overcome. One electron per carbon is delocalised and mobile within a layer, allowing electrical conduction along the layers.	4
02.1	Make equal-concentration aqueous solutions using deionised water and test their electrical conductivity The ionic solution conducts because it contains mobile ions The molecular solution does not conduct, or conducts negligibly, because it has no mobile charged particles Allow: melt each solid and test the melt — the ionic melt conducts, the molecular melt does not	Testing the solids alone would not distinguish them because an ionic solid has fixed ions. Dissolving equal amounts to comparable concentrations makes the ionic charge carriers mobile, while neutral dissolved molecules do not provide charge carriers.	3
03.1	In graphite, each carbon atom is covalently bonded to three others in layers Only weak forces act between the graphite layers, so the layers can slide over one another and graphite is soft and slippery In diamond, each carbon atom is covalently bonded to four others in a rigid three-dimensional network Strong covalent bonds extend throughout diamond with no layers able to slide, so diamond is hard	Graphite contains strong covalent bonds within each sheet but only weak forces between its sheets, allowing the layers to slide. Diamond has a rigid three-dimensional arrangement in which every carbon is joined to four others by strong covalent bonds, so deformation would require bonds in the network to be broken.	4

Tier 3 · Hard

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01.1	A is ionic: its ions are fixed when solid but mobile when molten B is molecular: weak intermolecular forces give a low melting point and there are no mobile charged particles C is metallic: delocalised electrons conduct and layers of ions can slide	Match each pair of observations to its charge carriers and forces. A needs ions that become mobile only on melting, so it is ionic. B has weak attractions and no charge carriers, so it is molecular. C has mobile electrons and non-directional bonding that permits reshaping, so it is metallic.	6
02.1	Silicon dioxide is macromolecular or giant covalent Many strong covalent bonds must be broken to melt silicon dioxide Carbon dioxide is simple molecular Only weak intermolecular forces between carbon dioxide molecules are overcome during sublimation	The phrase covalent bond does not by itself determine the physical change. Silicon dioxide is a continuous covalent network, whereas carbon dioxide consists of discrete molecules; separating those molecules overcomes intermolecular attractions rather than the strong C=O bonds.	4
03.1	Exactly four bent $\mathrm{H_2O}$ molecules form one connected lattice fragment, each molecule having two solid O-H covalent bonds Each O is labelled $\delta-$ and each H is labelled $\delta+$ Every dotted hydrogen bond runs from a $\delta+$ H towards a shown lone pair on a $\delta-$ O of another molecule, with O-H $\cdots$ O approximately linear Ice is a molecular crystal: the molecules remain discrete Melting is endothermic because energy is required to overcome hydrogen bonds between molecules The covalent O-H bonds within the water molecules remain intact	Arrange four bent H-O-H units in one continuous fragment, retaining solid lines for every covalent O-H bond. For each intermolecular link, align the donor O-H bond approximately with a lone pair drawn on the acceptor oxygen, then use a dotted line from H to that lone pair. On melting, energy is absorbed to weaken and overcome these intermolecular hydrogen bonds; the individual water molecules are not decomposed and their O-H bonds remain intact.	6
04.1	Melting iodine overcomes van der Waals' attractions between $\mathrm{I_2}$ molecules The covalent I-I bonds within the molecules remain intact $15.5\,\mathrm{kJ\,mol^{-1}}$ is much smaller because intermolecular van der Waals' attractions are weak $151\,\mathrm{kJ\,mol^{-1}}$ is the much greater energy needed to break strong covalent I-I bonds within the molecules	Iodine is a molecular crystal. Fusion separates its $\mathrm{I_2}$ molecules by overcoming weak intermolecular van der Waals' attractions, so it requires only $15.5\,\mathrm{kJ\,mol^{-1}}$ . Breaking the covalent bond inside each molecule is a different process and requires $151\,\mathrm{kJ\,mol^{-1}}$ , about ten times as much energy.	4
05.1	The solid diagram shows four $\mathrm{M^+}$ and two $\mathrm{X^{2-}}$ ions in a regular alternating lattice The molten diagram shows the same six charged ions in a disordered arrangement where they are no longer fixed in lattice positions The $2:1$ cation-to-anion ratio and overall charge neutrality are preserved in both diagrams Fixed ions cannot transport charge through the solid Both ion types can move through the liquid and carry charge, while the attraction remains ionic	Preserve particle identity, charge and stoichiometric ratio across the state change. Only the arrangement and mobility change: the ordered lattice becomes disordered, allowing ions to migrate through the liquid under a potential difference.	5

3.1.3.5 · Shapes of simple molecules and ions

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01.1	Trigonal planar $120^{\circ}$	Boron has three bonding pairs and no lone pairs in $\mathrm{BF_3}$ . Three charge clouds spread equally in one plane, giving a trigonal planar shape with $120^{\circ}$ angles.	2
02.1	Each multiple bond counts as one electron charge cloud There are two charge clouds around carbon, so the molecule is linear The O-C-O bond angle is $180^{\circ}$	VSEPR counts regions of electron density, not the number of shared pairs within a multiple bond. The two C=O regions repel to opposite sides of carbon, giving a linear arrangement.	3

Tier 2 · Standard

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01.1	There are three bonding pairs and one lone pair around N The electron pairs adopt a tetrahedral arrangement The molecular shape is trigonal pyramidal Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, compressing the angle from $109.5^{\circ}$ to $107^{\circ}$	Four charge clouds arrange tetrahedrally. Because one is a lone pair, the positions of the atoms form a trigonal pyramid. The lone pair repels bonding pairs more strongly, reducing the ideal tetrahedral angle from $109.5^{\circ}$ to $107^{\circ}$ .	4
02.1	Three bonding pairs and one lone pair A tetrahedral electron-pair arrangement Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, so the angle is compressed	A trigonal pyramid arises from four electron regions when one region is a lone pair. The regions adopt a tetrahedral arrangement, but stronger repulsion from the lone pair pushes the three bonds closer together.	3
03.1	Five bonding pairs and no lone pairs around phosphorus Trigonal bipyramidal shape Three equatorial bonds are separated by $120^{\circ}$ The two axial bonds are $90^{\circ}$ to the equatorial plane and $180^{\circ}$ from each other	Five electron charge clouds maximise their separation in a trigonal bipyramid. Three bonds occupy one equatorial plane at $120^{\circ}$ . The remaining two are axial, perpendicular to that plane, giving axial-equatorial angles of $90^{\circ}$ and an axial-axial angle of $180^{\circ}$ .	4

Tier 3 · Hard

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01.1	Two lone pairs on Br Six electron pairs in an octahedral arrangement Square planar shape Bond angles $90^{\circ}$ and $180^{\circ}$	Bromine has seven outer electrons and the negative charge adds one. Four electrons are used by Br in the four bonds, leaving four electrons as two lone pairs. Six charge clouds arrange octahedrally; the lone pairs occupy opposite positions to minimise repulsion, leaving four F atoms in a square plane with $90^{\circ}$ and $180^{\circ}$ angles.	5
02.1	$\mathrm{SF_4}$ has one equatorial lone pair and is see-saw-shaped; allow distorted tetrahedral Its equatorial-equatorial F-S-F angle is slightly less than $120^{\circ}$ , its axial-equatorial angles are slightly less than $90^{\circ}$ , and its axial F-S-F angle is slightly less than $180^{\circ}$ $\mathrm{XeF_2}$ has three equatorial lone pairs and is linear, with an F-Xe-F angle of $180^{\circ}$ Lone pairs prefer equatorial positions because these have fewer $90^{\circ}$ interactions Lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion, so the $\mathrm{SF_4}$ bond angles are compressed	Five regions give a trigonal-bipyramidal electron-pair arrangement. An equatorial site has only two $90^{\circ}$ interactions, so lone pairs occupy equatorial sites first. One equatorial lone pair leaves the four S-F bonds in a see-saw arrangement (also described as distorted tetrahedral). Because lone pair-bond pair repulsion exceeds bond pair-bond pair repulsion, the ideal $90^{\circ}$ , $120^{\circ}$ and axial $180^{\circ}$ angles are all compressed slightly. Three equatorial lone pairs in $\mathrm{XeF_2}$ leave the two Xe-F bonds opposite on the axial line, giving a linear molecule.	6
03.1	There are six bonding pairs and no lone pairs around sulfur The six electron charge clouds repel to an octahedral arrangement The molecular shape is octahedral Adjacent F-S-F angles are $90^{\circ}$ and opposite F-S-F angles are $180^{\circ}$ A correct drawing has four F atoms in a square plane around S, with one F above and one below that plane	Six bonding regions maximise their separation by pointing to the vertices of an octahedron, not to six positions in one plane. Place four bonds in a square plane through sulfur and the other two perpendicular to that plane. This gives $90^{\circ}$ between neighbouring bonds and $180^{\circ}$ between opposite bonds.	5
04.1	Iodine carries one lone pair ( $6-5=1$ ) The molecular shape is square pyramidal The F-I-F angles are approximately $90^{\circ}$ , with the axial-basal angles slightly less than $90^{\circ}$ Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion The lone pair occupies the sixth octahedral position, so it repels the four basal bonding pairs away from itself, closing the axial-basal angles to slightly below $90^{\circ}$	Six charge clouds adopt an octahedral parent arrangement. Five bonding pairs and one lone pair remove one vertex, leaving a square pyramid. The ideal inter-vertex angle is $90^{\circ}$ , but the lone pair repels bonding pairs more strongly than they repel each other, compressing the angles between the axial fluorine and the four basal fluorines to slightly below $90^{\circ}$ .	5
05.1	Three charge clouds around sulfur adopt a trigonal-planar electron-pair arrangement With one lone pair, the atom positions give a bent or V-shaped molecule The O-S-O angle is slightly less than $120^{\circ}$ Carbon dioxide has two bonding regions and no lone pairs on carbon, so it is linear with a $180^{\circ}$ angle The sulfur lone pair repels bonding regions more strongly than they repel each other, compressing the sulfur dioxide angle	Multiple bonds count as single regions in electron-pair repulsion theory. Three regions around sulfur form a trigonal-planar arrangement, but the lone pair is omitted from the shape name and compresses the remaining bond angle. Carbon dioxide has only two regions, so they point in opposite directions.	5

3.1.3.6 · Bond polarity

Tier 1 · Easy

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01.1	$\mathrm{H^{\delta+}-Cl^{\delta-}}$	Chlorine has the greater electronegativity, so it attracts the bonding pair more strongly. Chlorine is therefore $\delta^-$ and hydrogen is $\delta^+$ .	1
02.1	The differences are $0.3$ , $0.8$ and $1.3$ , respectively; $\mathrm{C-H<N-H<O-H}$ .	Compare the electronegativity differences: $0.3$ for $\mathrm{C-H}$ , $0.8$ for $\mathrm{N-H}$ and $1.3$ for $\mathrm{O-H}$ . A larger difference produces a more polar bond.	2

Tier 2 · Standard

Mark scheme for 3.1.3.6 Tier 2 · Standard
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01.1	$\mathrm{CO_2}$ is linear and its equal bond dipoles oppose and cancel; $\mathrm{SO_2}$ is bent, so its bond dipoles do not cancel and it has a permanent dipole.	Treat the bond dipoles as vectors. In linear $\mathrm{CO_2}$ , the two equal $\mathrm{C=O}$ dipoles act in opposite directions, giving zero resultant. The bent geometry of $\mathrm{SO_2}$ prevents its $\mathrm{S-O}$ dipoles from acting directly opposite each other, so their resultant is non-zero.	4
02.1	Boron is $\delta^+$ and each fluorine is $\delta^-$ ; the three equal bond dipoles cancel in the trigonal planar molecule, so $\mathrm{BF_3}$ has no permanent dipole.	Fluorine attracts each bonding pair more strongly than boron, fixing the partial-charge directions. The three identical $\mathrm{B-F}$ dipoles are arranged symmetrically at $120^\circ$ , so their vector sum is zero.	4
03.1	The $\mathrm{H-O}$ bond dipole points from H towards O and the $\mathrm{O-F}$ bond dipole points from O towards F; the two dipole vectors do not cancel in the bent molecule, so their resultant is non-zero.	For each bond, point the dipole towards the atom with the larger electronegativity. This gives H towards O for $\mathrm{H-O}$ and O towards F for $\mathrm{O-F}$ . Bond dipoles are vectors, so combine them using the molecular shape. Because the molecule is bent, the two vectors are not arranged to cancel, giving a permanent molecular dipole.	4

Tier 3 · Hard

Mark scheme for 3.1.3.6 Tier 3 · Hard
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01.1	Both contain a polar $\mathrm{C-F}$ bond; the four identical dipoles in symmetrical $\mathrm{CF_4}$ cancel, whereas the unsymmetrical bond arrangement in $\mathrm{CH_3F}$ gives a non-zero resultant dipole.	First mark each $\mathrm{C-F}$ bond as polar towards fluorine. The four equal bond-dipole vectors in a regular tetrahedral $\mathrm{CF_4}$ molecule sum to zero. Replacing three fluorine atoms with hydrogen removes that symmetry, so the bond dipoles in $\mathrm{CH_3F}$ cannot cancel and the molecule has a permanent dipole.	5
02.1	In X the polar $\mathrm{C-Cl}$ bond dipoles reinforce to give a resultant, so X is Z; in Y the symmetric opposing bond dipoles cancel, so Y is E.	Chlorine is more electronegative than carbon, so both $\mathrm{C-Cl}$ dipoles point towards chlorine. With the chlorine atoms on the same side, their components do not cancel and the higher-priority chlorine groups define the Z isomer. On opposite sides the arrangement is symmetric, the dipoles cancel and the higher-priority groups define the E isomer.	5
03.1	M is linear, so its equal bond dipoles oppose and cancel. N is bent, so its bond dipoles do not cancel; the bent shape implies at least one lone pair on X that affects the bond angle.	Work backwards from the measured molecular dipoles. Two identical bond-dipole vectors can sum to zero only when they act in exactly opposite directions, so M must be linear. A non-zero resultant for the same two bond types requires a non-linear arrangement, so N is bent. For an $\mathrm{XCl_2}$ molecule, repulsion involving one or more lone pairs on X produces this bent arrangement.	5
04.1	In A-B, A is $\delta^-$ and B is $\delta^+$ . In B-C, C is $\delta^-$ and B is $\delta^+$ . The two bond dipoles act in opposite directions because A-B-C is linear. The resultant is $1.4-0.6=0.8$ arbitrary units, directed towards A.	For each bond, direct the dipole towards the more electronegative atom. The A-B dipole therefore points from B towards A, while the B-C dipole points from B towards C. These vectors are collinear and opposed. Subtract their magnitudes and retain the direction of the larger vector: $1.4-0.6=0.8$ units towards A.	4
05.1	The X-Y result shows that Y is more electronegative than X. The Y-Z result shows that Z is more electronegative than Y. The electronegativity order is therefore $\mathrm{Z>Y>X}$ . XYZ has a permanent dipole because its unequal bond-dipole vectors cannot cancel in a bent arrangement.	A negative partial charge identifies the atom that attracts the bonding pair more strongly. Read each measured bond separately to obtain $Y>X$ and $Z>Y$ , then combine the inequalities. For the molecule, treat the two bond dipoles as vectors. They are neither equal nor directly opposed in a bent structure, so their vector sum is non-zero.	4

3.1.3.7 · Forces between molecules

Tier 1 · Easy

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01.1	Permanent dipole–dipole forces.	$\mathrm{CH_3Cl}$ is polar because its bond dipoles do not cancel. It has no hydrogen bonded to nitrogen, oxygen or fluorine, so it cannot form hydrogen bonds with itself.	1
02.1	$\mathrm{CH_3OH}$ ; it has hydrogen bonded to oxygen and a lone pair on oxygen.	Hydrogen bonding needs a hydrogen atom covalently bonded to N, O or F and a lone pair on N, O or F in a neighbouring molecule. Only methanol supplies both features between its own molecules.	3

Tier 2 · Standard

Mark scheme for 3.1.3.7 Tier 2 · Standard
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01.1	Ethane has more electrons and a larger electron cloud, so it is more polarisable and has stronger induced dipole–dipole forces; more energy is needed to separate its molecules.	Both molecules are non-polar, so compare their induced dipole–dipole forces. Ethane has more electrons, allowing larger temporary and induced dipoles. Its stronger attractions require a greater energy input during boiling.	3
02.1	Increasing branching makes a molecule more compact, reducing surface contact between molecules and weakening induced dipole–dipole attractions, so less energy is needed for boiling.	The isomers have the same electron number, so the main difference is molecular shape rather than electron-cloud size. Pentane has the greatest surface contact and strongest induced dipole–dipole attractions; the most highly branched isomer has the least contact and the lowest boiling point.	4
03.1	Show an $\mathrm{O^{\delta-}-H^{\delta+}}$ covalent bond in water and a lone pair on $\mathrm{N^{\delta-}}$ in ammonia, with a dotted line from the $\delta^+$ hydrogen to the nitrogen lone pair.	First show that oxygen attracts the bonding pair in $\mathrm{O-H}$ , making O $\delta^-$ and H $\delta^+$ . In ammonia, nitrogen is $\delta^-$ relative to hydrogen and carries a lone pair. The hydrogen bond is an intermolecular dotted line from the partially positive water H atom to that nitrogen lone pair; it is not the covalent $\mathrm{O-H}$ bond.	4

Tier 3 · Hard

Mark scheme for 3.1.3.7 Tier 3 · Hard
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01.1	Water molecules form hydrogen bonds, which are stronger than the intermolecular forces between $\mathrm{H_2S}$ molecules; in ice, hydrogen bonding produces an open lattice whose molecules are farther apart than in liquid water.	Oxygen is sufficiently electronegative and has lone pairs, so each water molecule can participate in hydrogen bonding. More energy is required to overcome these attractions than the permanent and induced dipole attractions in $\mathrm{H_2S}$ , raising water's boiling point. Freezing arranges water molecules into an open hydrogen-bonded lattice. On melting, some bonds are disrupted and molecules occupy gaps, so the liquid packs more closely and is denser.	5
02.1	The claim is incorrect: both substances have induced dipole–dipole and permanent dipole–dipole attractions, but propan-1-ol also forms hydrogen bonds, so its overall intermolecular attractions are stronger and its boiling point is higher.	Similar relative molecular masses suggest broadly similar induced dipole–dipole contributions, but mass alone does not identify every force. Both molecules are polar. The $\mathrm{O-H}$ group in propan-1-ol allows intermolecular hydrogen bonding, whereas propanone has no hydrogen bonded to oxygen, so more energy is required to separate propan-1-ol molecules.	5
03.1	Pure methoxymethane cannot hydrogen-bond to itself because none of its hydrogens is bonded to N, O or F. In water it can accept hydrogen bonds: an oxygen lone pair in methoxymethane attracts a $\delta^+$ hydrogen of an $\mathrm{O-H}$ bond in water, so the student's claim is too broad.	Test donor and acceptor roles separately. Methoxymethane has oxygen lone pairs and can therefore accept a hydrogen bond, but it has no hydrogen attached to O and cannot donate one. A pure sample lacks a suitable donor. Water supplies $\mathrm{O-H}$ donor groups, so hydrogen bonds can form between water hydrogens and ether oxygen lone pairs in the mixture.	5
04.1	The three molecules have the same electron count, so their induced dipole-dipole attractions are comparable. Propan-1-ol and propylamine can form hydrogen bonds, whereas butane cannot because it has no hydrogen bonded to N, O or F. Butane therefore has the weakest intermolecular forces and the lowest boiling point. O-H $\cdots$ O hydrogen bonds are stronger than N-H $\cdots$ N hydrogen bonds because oxygen is more electronegative than nitrogen. Propan-1-ol therefore has the strongest intermolecular forces and the highest boiling point, with propylamine between it and butane.	Begin by controlling for induced dipole-dipole attractions: equal electron counts make these broadly comparable. Then identify hydrogen-bond donors. Both the alcohol and amine hydrogen-bond, but butane cannot. Finally compare the two hydrogen bonds: the greater electronegativity of oxygen gives stronger O-H $\cdots$ O attractions than the N-H $\cdots$ N attractions in propylamine, producing the order 370 K, 321 K, 273 K.	5
05.1	HF molecules form hydrogen bonds. HCl, HBr and HI do not form hydrogen bonds because their hydrogen atoms are not bonded to N, O or F. The extra hydrogen bonding makes the intermolecular forces in HF much stronger, so HF has the much higher boiling point. From HCl to HI, the number of electrons and the polarisability of the electron cloud increase. Induced dipole-dipole attractions therefore become stronger down the group, so the boiling points rise from HCl to HI.	Treat HF separately from the down-group trend. Fluorine permits hydrogen bonding, which dominates the comparison with HCl. Among HCl, HBr and HI, hydrogen bonding is absent, so increasing electron count and electron-cloud polarisability strengthen induced dipole-dipole attractions and raise the boiling point.	5

3.1.4.1 · Enthalpy change

Tier 1 · Easy

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01.1	Exothermic; $\Delta H$ is negative.	Energy leaving the reacting chemicals warms the surroundings, so the reaction is exothermic. The products have lower enthalpy than the reactants, giving $\Delta H<0$ .	2
02.1	The sign is wrong; $\Delta H=-42\,\text{kJ mol}^{-1}$ .	Use $\Delta H=H(\text{products})-H(\text{reactants})$ . Products lying below reactants give a negative enthalpy change, so the reaction is exothermic.	2

Tier 2 · Standard

Mark scheme for 3.1.4.1 Tier 2 · Standard
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01.1	Formation: the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions. Combustion: the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions.	For each definition, state the one-mole basis and the chemical process. Standard conditions mean $100\,\text{kPa}$ and a stated temperature, with substances in their standard states.	4
02.1	A is neither as written because it forms two moles of sodium chloride; B is formation because one mole of sodium chloride forms from its elements in their standard states; C is combustion because one mole of carbon monoxide burns completely in oxygen.	A standard formation equation must form exactly one mole of compound from elements in their standard states. A standard combustion equation completely burns one mole of a substance in oxygen. Apply both the process and one-mole tests rather than identifying equations only from their products.	5
03.1	$\mathrm{M(s)+Q_2(g)\rightarrow MQ_2(l)}$ $\mathrm{MQ_2}$ is liquid at $298\,\text{K}$ because this temperature lies between its melting and boiling points.	A formation equation starts from the elements in their standard states and forms exactly one mole of compound. One mole of $\mathrm{MQ_2}$ needs one M atom and one $\mathrm{Q_2}$ molecule. At $298\,\text{K}$ the compound is above its melting point but below its boiling point, so its standard state here is liquid.	4

Tier 3 · Hard

Mark scheme for 3.1.4.1 Tier 3 · Hard
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01.1	$\Delta_\mathrm{c}H^\circ=-888\,\text{kJ mol}^{-1}$	The energy released per mole is $222/0.250=888\,\text{kJ mol}^{-1}$ . Combustion is exothermic, so attach a negative sign: $\Delta_\mathrm{c}H^\circ=-888\,\text{kJ mol}^{-1}$ .	3
02.1	Amount $n=2.20/55.0=0.0400\,\text{mol}$ Mass $=0.0400\times46.0=1.84\,\text{g}$	Use the magnitude because the negative sign records that energy is released: $n=q/\|\Delta H\|=2.20/55.0=0.0400\,\text{mol}$ . Then $m=nM_\mathrm{r}=0.0400\times46.0=1.84\,\text{g}$ , exact with no rounding boundary.	3
03.1	Amount of Y $=12.0/48.0=0.250\,\text{mol}$ . $41.0\,\text{kJ}$ is absorbed.	Reverse the given equation for the decomposition, which changes the enthalpy sign to $+164\,\text{kJ}$ per mole of Y decomposed. The sample contains $12.0/48.0=0.250\,\text{mol}$ of Y, so $q=0.250\times164=41.0\,\text{kJ}$ . The positive sign means the reacting chemicals absorb heat.	4
04.1	$0.500\,\text{mol}$ of A permits $0.250\,\text{mol}$ of reaction as written. $0.900\,\text{mol}$ of B permits $0.300\,\text{mol}$ of reaction as written, so A is limiting. The reaction releases $0.250\times240=60.0\,\text{kJ}$ of heat. $0.250\,\text{mol}$ of C forms. $0.900-3(0.250)=0.150\,\text{mol}$ of B remains.	Divide each starting amount by its equation coefficient. A gives an extent of $0.500/2=0.250\,\text{mol}$ , while B gives $0.900/3=0.300\,\text{mol}$ , so A fixes the extent. Multiply the thermochemical enthalpy and the coefficient of C by $0.250$ . The reaction consumes $3\times0.250=0.750\,\text{mol}$ of B, leaving $0.150\,\text{mol}$ .	5
05.1	$\mathrm{C_2H_6(g)+\tfrac{7}{2}O_2(g)\rightarrow2CO_2(g)+3H_2O(l)}$ The equation shows complete combustion in oxygen and uses the standard states of all substances. It represents the combustion of exactly one mole of ethane, as required by the definition of $\Delta_\mathrm{c}H^\circ$ . Clearing the fraction would double every coefficient and would describe two moles of ethane, so the enthalpy change would also be doubled rather than being $\Delta_\mathrm{c}H^\circ$ for one mole.	Balance carbon and hydrogen first to obtain two carbon dioxide molecules and three liquid water molecules from one ethane molecule. These products require seven oxygen atoms, hence $\tfrac{7}{2}\mathrm{O_2}$ . Keep the half-integer because a standard enthalpy of combustion is defined per one mole of the substance burned.	4

3.1.4.2 · Calorimetry

Tier 1 · Easy

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01.1	$q=2.7\times10^3\,\text{J}$ (or $2.7\,\text{kJ}$ )	Substitute into $q=mc\Delta T$ : $q=100\times4.18\times6.5=2717\,\text{J}$ . To two significant figures this is $2.7\times10^3\,\text{J}$ .	2
02.1	Dependent variable: the measured temperature change; controls include the volumes and concentrations of both solutions and their initial temperatures.	Only the calorimeter design should change. Measure the temperature rise, while keeping the reacting amounts and starting thermal conditions the same so differences can be attributed to heat transfer through the apparatus.	3

Tier 2 · Standard

Mark scheme for 3.1.4.2 Tier 2 · Standard
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01.1	$\Delta H=-56.8\,\text{kJ mol}^{-1}$	The solution mass is $100.0\,\text{g}$ , so $q=100.0\times4.18\times6.80=2842.4\,\text{J}=2.8424\,\text{kJ}$ . Each reagent supplies $1.00\times0.0500=0.0500\,\text{mol}$ , forming $0.0500\,\text{mol}$ of water. Thus $\Delta H=-2.8424/0.0500=-56.8\,\text{kJ mol}^{-1}$ .	5
02.1	$8.9\,\text{K}$ is anomalous; the mean of the concordant results is $6.3\,\text{K}$ ; repeats reveal anomalies and allow a mean that reduces random uncertainty.	The first, second and fourth values lie within $0.2\,\text{K}$ , whereas $8.9\,\text{K}$ is isolated. Excluding that anomaly gives $(6.3+6.4+6.2)/3=6.3\,\text{K}$ .	4
03.1	The mass changing temperature should be $50.0\,\text{g}$ , so using $45.0\,\text{g}$ underestimates the magnitude of $q$ . The calculated positive molar enthalpy of solution is therefore too small in magnitude.	After complete dissolution, both the original water and dissolved solid form the solution whose temperature falls. The appropriate mass is therefore $45.0+5.0=50.0\,\text{g}$ . Since $q$ is directly proportional to $m$ , the smaller mass gives too small a heat magnitude. Dividing that underestimated heat by the unchanged amount dissolved gives an endothermic $\Delta H$ that is insufficiently positive.	4

Tier 3 · Hard

Mark scheme for 3.1.4.2 Tier 3 · Hard
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01.1	$\Delta_\mathrm{c}H=-1.60\times10^3\,\text{kJ mol}^{-1}$ ; for example, heat is lost to the surroundings or used to warm the apparatus.	The water gains $q=250\times4.18\times18.4=19228\,\text{J}=19.228\,\text{kJ}$ . The amount burned is $0.720/60.0=0.0120\,\text{mol}$ . Therefore $\Delta_\mathrm{c}H=-19.228/0.0120=-1602\,\text{kJ mol}^{-1}$ , or $-1.60\times10^3\,\text{kJ mol}^{-1}$ . Heat loss means the measured water temperature rise accounts for less energy than the fuel actually released.	6
02.1	(a) The extrapolated temperatures are $20.00$ and $24.40\,^{\circ}\text{C}$ , so $\Delta T=4.40\,\text{K}$ and $\Delta H=-46.0\,\text{kJ mol}^{-1}$ (accept $-45.98\,\text{kJ mol}^{-1}$ ). (b) Extrapolation estimates the temperature rise at mixing before heat loss lowers the observed readings.	The pre-mixing trend falls by $0.1\,^{\circ}\text{C min}^{-1}$ , giving $20.0\,^{\circ}\text{C}$ at $3.0\,\text{min}$ . The cooling trend falls by $0.2\,^{\circ}\text{C min}^{-1}$ , giving $24.4\,^{\circ}\text{C}$ when extrapolated back to $3.0\,\text{min}$ . Thus $q=100.0\times4.18\times4.40=1839.2\,\text{J}=1.8392\,\text{kJ}$ and $\Delta H=-1.8392/0.0400=-45.98\,\text{kJ mol}^{-1}$ , reported as $-46.0\,\text{kJ mol}^{-1}$ .	7
03.1	The solution absorbs $1.87264\,\text{kJ}$ and the apparatus absorbs $0.1792\,\text{kJ}$ , so the reaction releases $2.05184\,\text{kJ}$ and $\Delta H=-51.3\,\text{kJ mol}^{-1}$ . Ignoring the apparatus gives $-46.8\,\text{kJ mol}^{-1}$ , which is too small in magnitude and therefore less exothermic than the corrected value.	The solution gains $q=mc\Delta T=80.0\times4.18\times5.60=1872.64\,\text{J}$ . The apparatus gains $q=C\Delta T=32.0\times5.60=179.2\,\text{J}$ . Retaining unrounded values, the reaction releases $2.05184\,\text{kJ}$ , giving $\Delta H=-2.05184/0.0400=-51.296\,\text{kJ mol}^{-1}$ , or $-51.3\,\text{kJ mol}^{-1}$ . Omitting the apparatus gives $-1.87264/0.0400=-46.816\,\text{kJ mol}^{-1}$ , or $-46.8\,\text{kJ mol}^{-1}$ , so some released energy is missed and the result is insufficiently exothermic.	6
04.1	The mass-weighted temperature before reaction is $[40.0(18.00)+60.0(22.00)]/100.0=20.40\,^{\circ}\text{C}$ . The temperature rise caused by reaction is $26.80-20.40=6.40\,\text{K}$ . The solutions gain $q=100.0\times4.18\times6.40=2675.2\,\text{J}=2.6752\,\text{kJ}$ . The molar enthalpy change is $\Delta H=-2.6752/0.0500=-53.5\,\text{kJ mol}^{-1}$ .	Before attributing a rise to reaction, calculate the mass-weighted initial temperature. Equal specific heat capacities give $20.40\,^{\circ}\text{C}$ , so the reaction accounts for a $6.40\,\text{K}$ rise. Calculate the heat gained by the full solution in joules, convert it to kilojoules, divide by the amount of product and attach a negative sign because the solution warmed.	4
05.1	The minimum permitted rise is $4.90\,\text{K}$ , giving $q=2.0482\,\text{kJ}$ . This limit gives $\Delta H=-2.0482/0.0400=-51.2\,\text{kJ mol}^{-1}$ . The maximum permitted rise is $5.10\,\text{K}$ , giving $q=2.1318\,\text{kJ}$ . This limit gives $\Delta H=-2.1318/0.0400=-53.3\,\text{kJ mol}^{-1}$ . The reference value lies outside $-53.3$ to $-51.2\,\text{kJ mol}^{-1}$ , so the stated random uncertainty cannot explain the result; a systematic over-reading of the temperature rise could make the result too exothermic.	Propagate the thermometer uncertainty by recalculating at both limiting temperature rises, keeping unrounded heat values. Because the experiment is exothermic, the larger rise gives the more negative limit. The entire allowed interval is more negative than $-50.0\,\text{kJ mol}^{-1}$ , so random temperature uncertainty of the stated size is insufficient. A calibration bias that exaggerates $\Delta T$ has the required direction.	5

3.1.4.3 · Applications of Hess's law

Tier 1 · Easy

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01.1	$-55\,\text{kJ mol}^{-1}$	Add the enthalpy changes for the two consecutive routes: $+25+(-80)=-55\,\text{kJ mol}^{-1}$ .	1
02.1	Reversing the equation reverses the sign, so $\Delta H=-48\,\text{kJ mol}^{-1}$ .	The reverse route undoes the original enthalpy change. Keep the magnitude but change the sign whenever the chemical equation is reversed.	2

Tier 2 · Standard

Mark scheme for 3.1.4.3 Tier 2 · Standard
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01.1	$\Delta_\mathrm{r}H^\circ=-99\,\text{kJ mol}^{-1}$	Apply products minus reactants. The standard formation enthalpy of $\mathrm{O_2(g)}$ is zero, so $\Delta_\mathrm{r}H^\circ=-396-[-297+\tfrac12(0)]=-99\,\text{kJ mol}^{-1}$ .	2
02.1	$\Delta H=-111\,\text{kJ mol}^{-1}$	Use the first equation in its written direction, then reverse the second: $\mathrm{CO_2(g)\rightarrow CO(g)+\tfrac12O_2(g)}$ , $\Delta H=+283\,\text{kJ mol}^{-1}$ . Adding and cancelling $\mathrm{CO_2}$ gives the target equation and $-394+283=-111\,\text{kJ mol}^{-1}$ .	4
03.1	$\Delta_\mathrm{r}H^\circ=-1124\,\text{kJ mol}^{-1}$ .	Use products minus reactants. The product total is $2(-286)+2(-297)=-1166\,\text{kJ mol}^{-1}$ . The reactant total is $2(-21)+3(0)=-42\,\text{kJ mol}^{-1}$ because the standard enthalpy of formation of $\mathrm{O_2(g)}$ is zero. Therefore $\Delta_\mathrm{r}H^\circ=-1166-(-42)=-1124\,\text{kJ mol}^{-1}$ .	4

Tier 3 · Hard

Mark scheme for 3.1.4.3 Tier 3 · Hard
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01.1	$\Delta H=-124\,\text{kJ mol}^{-1}$	All three substances can be combusted to the same final products. Therefore use combustion of the reactants minus combustion of the product: $\Delta H=[-2058+(-286)]-(-2220)=-2344+2220=-124\,\text{kJ mol}^{-1}$ .	3
02.1	$\Delta_\mathrm{f}H^\circ[\mathrm{CaCO_3(s)}]=-1.21\times10^3\,\text{kJ mol}^{-1}$ or $-1207\,\text{kJ mol}^{-1}$	Let the unknown formation enthalpy be $x$ . Products minus reactants gives $178=[-635+(-394)]-x=-1029-x$ . Therefore $x=-1207\,\text{kJ mol}^{-1}$ , reported to three significant figures as $-1.21\times10^3\,\text{kJ mol}^{-1}$ . The target value refers to forming one mole of solid calcium carbonate from its elements in their standard states.	4
03.1	$\Delta_\mathrm{c}H^\circ(\mathrm{X})=-1316\,\text{kJ mol}^{-1}$	Let the combustion enthalpy of X be $x$ . Combusting the reactants to the common products releases $x+(-286)$ , whereas combusting Y releases $-1510\,\text{kJ mol}^{-1}$ . Therefore $-92=[x-286]-(-1510)=x+1224$ . Solving gives $x=-1316\,\text{kJ mol}^{-1}$ .	4
04.1	Double the first equation to give $\mathrm{2A+2B\rightarrow2C}$ . Doubling its coefficients gives $\Delta H=2(-72)=-144\,\text{kJ mol}^{-1}$ . Adding $\mathrm{2C\rightarrow D}$ cancels the intermediate 2C and gives the target equation. $\Delta H=-144+40=-104\,\text{kJ mol}^{-1}$ .	Hess's law permits the target route to be assembled from the supplied routes. The target consumes two A and two B, so multiply the whole first equation, including its enthalpy, by two. Add the second equation in its stated direction. The 2C produced in the first step is consumed in the second and cancels, leaving the required overall equation and an enthalpy sum of $-104\,\text{kJ mol}^{-1}$ .	4
05.1	The route $\mathrm{A\rightarrow B\rightarrow C}$ has $\Delta H=35-82=-47\,\text{kJ mol}^{-1}$ . The alternative route must satisfy $-20+\Delta H(\mathrm{D\rightarrow C})=-47$ . $\Delta H(\mathrm{D\rightarrow C})=-27\,\text{kJ mol}^{-1}$ . Reversing $\mathrm{A\rightarrow C}$ changes the sign of its enthalpy. $\Delta H(\mathrm{C\rightarrow A})=+47\,\text{kJ mol}^{-1}$ .	First obtain the common A-to-C change from the complete route through B: $+35+(-82)=-47\,\text{kJ mol}^{-1}$ . Hess's law requires the route through D to have the same total, so solve $-20+x=-47$ to obtain $x=-27\,\text{kJ mol}^{-1}$ . The C-to-A change is simply the reverse of the established A-to-C route, so its magnitude is unchanged and its sign is positive.	5

3.1.4.4 · Bond enthalpies

Tier 1 · Easy

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01.1	The mean energy required to break one mole of a specified covalent bond in gaseous molecules.	Include energy required, one mole of the named bond, bond breaking and gaseous molecules. The word mean shows that the value is averaged across compounds.	2
02.1	Bond breaking is endothermic, so its contribution is $+413\,\text{kJ mol}^{-1}$ ; bond formation is exothermic, so its contribution is $-413\,\text{kJ mol}^{-1}$ .	Energy is required to separate bonded atoms and released when a bond forms. This is why bond-enthalpy estimates use bonds broken minus bonds formed.	2

Tier 2 · Standard

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01.1	$\Delta H=-183\,\text{kJ mol}^{-1}$	Break one $\mathrm{H-H}$ and one $\mathrm{Cl-Cl}$ bond, then form two $\mathrm{H-Cl}$ bonds. Thus $\Delta H=(436+243)-2(431)=679-862=-183\,\text{kJ mol}^{-1}$ .	3
02.1	$E(\mathrm{C-Cl})=338\,\text{kJ mol}^{-1}$	Only one $\mathrm{C-H}$ and one $\mathrm{Cl-Cl}$ bond are broken; one $\mathrm{C-Cl}$ and one $\mathrm{H-Cl}$ bond are formed. If $E(\mathrm{C-Cl})=x$ , then $-114=(413+243)-(x+432)=224-x$ , so $x=338\,\text{kJ mol}^{-1}$ .	4
03.1	Amount of $\mathrm{N-H}$ bonds $=3\times0.500=1.50\,\text{mol}$ ; energy required $=1.50\times388=582\,\text{kJ}$ .	Each ammonia molecule contains three $\mathrm{N-H}$ bonds, so $0.500\,\text{mol}$ of molecules contains $1.50\,\text{mol}$ of these bonds. Mean bond enthalpy is energy per mole of bonds broken in gaseous molecules. Thus the required energy is $1.50\times388=582\,\text{kJ}$ , with a positive sign because bond breaking is endothermic.	3

Tier 3 · Hard

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01.1	$\Delta H=-126\,\text{kJ mol}^{-1}$ ; mean bond enthalpies are averages from bonds in different molecular environments.	The unchanged four $\mathrm{C-H}$ bonds cancel. Break one $\mathrm{C{=}C}$ and one $\mathrm{H-H}$ bond; form one $\mathrm{C-C}$ and two additional $\mathrm{C-H}$ bonds. Hence $\Delta H=(612+436)-[348+2(413)]=1048-1174=-126\,\text{kJ mol}^{-1}$ . The estimate uses averaged bond data rather than molecule-specific enthalpies.	4
02.1	A new $\mathrm{C-H}$ bond is also formed; $\Delta H=(612+366)-(348+413+276)=-59\,\text{kJ mol}^{-1}$ .	Addition removes the carbon–carbon double bond and the hydrogen–bromine bond. The product contains a carbon–carbon single bond plus new $\mathrm{C-H}$ and $\mathrm{C-Br}$ bonds. Using all changed bonds gives $978-1037=-59\,\text{kJ mol}^{-1}$ .	5
03.1	Mean bond enthalpies predict $\Delta H=0\,\text{kJ mol}^{-1}$ because the same numbers and types of bonds are broken and formed. Accurate values can differ because mean bond enthalpies average a bond over different molecular environments and do not distinguish the environments in the two isomers.	Deduce the bond inventory from each displayed structure: both contain three $\mathrm{C-C}$ and ten $\mathrm{C-H}$ bonds. The mean-energy total for bonds broken therefore equals that for bonds formed, so the estimate is zero. This cancellation exposes the limitation of mean data: actual bond energies depend on their molecular environments, which differ between the straight-chain and branched isomers.	4
04.1	Ten C-H bonds occur on each side and cancel from the calculation. Breaking the three C-C bonds in butane requires $3(348)=1044\,\text{kJ mol}^{-1}$ . Forming one C-C bond and one C=C bond releases $348+612=960\,\text{kJ mol}^{-1}$ . $\Delta H=1044-960=+84\,\text{kJ mol}^{-1}$ . Mean bond enthalpies apply to gaseous species; using a liquid would also require its enthalpy of vaporisation.	Butane contains three C-C and ten C-H bonds. Ethane and ethene together contain one C-C, one C=C and ten C-H bonds, so the C-H terms cancel. Apply broken minus formed to the remaining bonds: $3(348)-(348+612)=+84\,\text{kJ mol}^{-1}$ . Because mean bond enthalpies are gas-phase quantities, a non-gaseous reactant or product needs the appropriate phase-change enthalpy as well.	5

3.1.5.1 · Collision theory

Tier 1 · Easy

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01.1	The minimum energy required for a reaction to occur following a collision between particles.	The definition must state both minimum energy and that this energy allows colliding particles to react.	2
02.1	Q and R have sufficient energy to react; P does not. Q ( $91$ ) and R ( $84$ ) have energy at or above $E_\mathrm{a}=80\,\text{kJ mol}^{-1}$ ; P has only $72\,\text{kJ mol}^{-1}$ , below the activation energy.	Compare each collision energy with $E_\mathrm{a}=80\,\text{kJ mol}^{-1}$ . The energies of Q and R exceed the activation energy, whereas P has only $72\,\text{kJ mol}^{-1}$ and therefore cannot cross the energy barrier.	3

Tier 2 · Standard

Mark scheme for 3.1.5.1 Tier 2 · Standard
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01.1	Most colliding particles have energy below the activation energy.	A collision is necessary but not sufficient. Compare the collision energy with $E_\mathrm{a}$ : only collisions with at least the activation energy can lead to reaction.	3
02.1	The gas particle concentration or pressure has increased; the number of successful collisions per second, and hence the rate, doubles.	An unchanged energetic fraction is consistent with constant temperature and unchanged activation energy. A higher particle number per unit volume raises collision frequency. Since $1.0\times10^9$ is twice $5.0\times10^8$ and the success fractions are fixed, the successful-collision frequency doubles.	4
03.1	Reaction P is faster because a greater proportion of its collisions have energy equal to or greater than its lower activation energy.	At the same temperature the two reactions have the same energy distribution. The lower threshold for P leaves a larger area of the distribution at or above $E_\mathrm{a}$ . Since the collision frequencies are equal, P has more successful collisions per second and therefore a higher rate.	3

Tier 3 · Hard

Mark scheme for 3.1.5.1 Tier 3 · Hard
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01.1	$8.0\times10^4$ successful collisions per second and approximately $2.0\times10^{10}$ unsuccessful collisions per second; the activation energy is the minimum collision energy needed for reaction.	Collisions with sufficient energy occur at $(2.0\times10^{10})(4.0\times10^{-6})=8.0\times10^4\,\text{s}^{-1}$ . The unsuccessful frequency is $(1-4.0\times10^{-6})(2.0\times10^{10})$ , which is $2.0\times10^{10}\,\text{s}^{-1}$ to two significant figures. Collisions below $E_\mathrm{a}$ cannot cross the reaction's energy barrier.	4
02.1	The energetic fraction should remain $0.015$ because temperature and activation energy are unchanged; the doubled collision frequency therefore doubles, rather than quadruples, the number of successful collisions per second.	Concentration changes particle density, not the Maxwell–Boltzmann energy distribution. The table incorrectly changes the energetic fraction. With the energetic fraction fixed, successful collisions scale directly with total collision frequency.	4
03.1	At $T_1$ , the successful fraction is $0.030$ , so $E_\mathrm{a}=80\,\text{kJ mol}^{-1}$ . At $T_2$ , the successful-collision frequency is $3.6\times10^8\,\text{s}^{-1}$ , an increase by a factor of $3.0$ .	The successful fraction at $T_1$ is $(1.20\times10^8)/(4.0\times10^9)=0.030$ , which matches the cumulative fraction at or above $80\,\text{kJ mol}^{-1}$ . At $T_2$ , $(5.0\times10^9)(0.072)=3.6\times10^8\,\text{s}^{-1}$ . The increase factor is $(3.6\times10^8)/(1.20\times10^8)=3.0$ .	5
04.1	R: $7.2\times10^7\,\text{s}^{-1}$ . S: $9.0\times10^7\,\text{s}^{-1}$ . S is faster. The ratio is $9.0/7.2=1.25$ .	Do not compare either total collision frequency or energetic fraction alone. For each reaction multiply the total frequency by the fraction at or above its activation energy. The smaller total for S is outweighed by its larger energetic fraction, giving the larger successful-collision frequency.	4

3.1.5.2 · Maxwell–Boltzmann distribution

Tier 1 · Easy

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01.1	Horizontal axis: molecular energy; vertical axis: number (or fraction) of molecules.	The distribution describes how the molecules in a sample are spread across energies, so energy is the independent horizontal quantity and molecular population is vertical.	2
02.1	The curve must start at the origin, and the high-energy tail must approach the energy axis without touching or crossing it.	The number of molecules at zero energy is represented by the origin. There is no maximum molecular energy, so the population becomes very small at high energy but the curve remains above the axis.	2

Tier 2 · Standard

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01.1	Draw a lower peak shifted to higher energy, make the curve broader with a longer high-energy tail, and keep its total area equal to that of the original curve.	Increasing temperature spreads molecular energies over a wider range and increases the most probable energy, so move and lower the peak. The sample size is unchanged, so preserve the area under the curve and retain a tail that approaches the axis.	4
02.1	The area represents the number of molecules and must remain unchanged; the higher-temperature curve should be adjusted to have the same total area while retaining its lower, right-shifted peak and broader shape.	Heating redistributes the fixed molecular population among energy values; it does not create molecules. Therefore the hotter curve can move and broaden, but its area must equal that of the original curve.	3
03.1	The curves have the same most-probable-energy position and the same distribution shape, but B has twice the total area and twice the vertical population at each energy.	Temperature determines how the molecular population is distributed across energies, so equal temperatures give the same relative shape and peak-energy position. Area represents molecule number. Doubling the number therefore scales the curve vertically and doubles its area without shifting the energy values.	3
04.1	The most probable energy is read from the energy (horizontal) axis at the value directly below the maximum, not from the peak height. The height of the peak gives the number of molecules that have the most probable energy. When the sample is heated, the most probable energy moves to a higher value.	Separate the two axes: the horizontal axis carries energy, so the most probable energy is the energy coordinate beneath the curve maximum; the vertical coordinate at that point is how many molecules possess it. Heating spreads the fixed molecular population over a wider energy range, so the maximum, and with it the most probable energy, moves to a higher energy.	3

Tier 3 · Hard

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01.1	The higher-temperature curve has a lower peak but a larger high-energy tail; equal total areas require the curves to cross, and the fraction above $E^$ is larger at $T_2$ when $E^$ lies in the high-energy region beyond the crossing.	The samples contain equal numbers of molecules, so the areas under their curves are equal. The $T_2$ curve lies below $T_1$ around the lower, sharper $T_1$ peak but above it in the high-energy tail; it must cross to redistribute the same area. For a threshold drawn in that tail, the area to the right—and hence the fraction above $E^*$ —is greater at $T_2$ .	5
02.1	B is hotter; it has three times the fraction at or above $E_\mathrm{a}$ and its full curve has a lower peak at higher energy, is broader and has the same total area as A.	Each sample contains $100\,000$ molecules, so their distribution areas are equal. The larger high-energy population identifies B as the higher-temperature sample. Heating spreads the distribution and shifts its most probable energy to the right while conserving area.	4
03.1	Route P has $9+5=14\%$ of molecules at or above its activation energy; route Q has $5\%$ . With the other factors identical, P has $14/5=2.8$ times the successful-collision frequency of Q.	For the lower threshold P, include both regions to its right: the between-threshold region and the region above Q, giving $14\%$ . Only the final $5\%$ clears the higher threshold Q. With the collision frequency identical, the successful frequencies scale with these energetic fractions, so the ratio is $14/5=2.8$ .	5
04.1	At $T_1$ , the fraction at or above $E_1$ is $18+10=28\%$ . At $T_2$ , the fraction at or above $E_1$ is $21+19=40\%$ . The population energetic enough to clear $E_1$ increases by a factor of $40/28=1.43$ . The fraction energetic enough to clear $E_2$ increases from $10\%$ to $19\%$ . The population energetic enough to clear $E_2$ increases by a factor of $19/10=1.90$ . The higher-threshold $E_2$ population is proportionally more temperature-sensitive because it has the larger factor increase.	For the lower threshold, include both energy regions to its right; for the higher threshold, include only the final region. The unchanged molecule number means the counts in each energetic population scale directly with the percentage ratios. Retain the unrounded $40/28$ value until reporting $1.43$ ; these data alone do not determine the separate change in total collision frequency.	6

3.1.5.3 · Effect of temperature on reaction rate

Tier 1 · Easy

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01.1	The rate increases.	At a higher temperature, collisions occur more frequently and a greater fraction of them have sufficient energy, so successful collisions occur more often.	1
02.1	Independent variable: temperature; measured dependent variable: time for the cross to disappear, from which $1/t$ can be used as a rate measure; controls include reactant volumes and concentrations.	Change only the temperature. Use the same total reacting amounts and concentrations, the same cross and viewing arrangement, and the same endpoint rule so that changes in time can be attributed to temperature.	4

Tier 2 · Standard

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01.1	The higher-temperature distribution has a lower, broader peak shifted right; the area beyond the unchanged activation energy increases substantially, so a larger fraction of molecules undergo successful collisions.	Draw or imagine a fixed vertical $E_\mathrm{a}$ line on both distributions. Although the curve changes modestly overall, the high-energy tail beyond the line can grow by a large proportion. This makes the frequency of collisions with $E\ge E_\mathrm{a}$ rise sharply.	4
02.1	$1/t=0.0125$ , $0.0200$ and $0.0400\,\text{s}^{-1}$ , respectively; the rate measure increases with temperature and the increase is not linear over these intervals.	Use the reciprocal time because each run reaches the same visible endpoint: $1/80.0=0.0125$ , $1/50.0=0.0200$ and $1/25.0=0.0400\,\text{s}^{-1}$ . Equal $10\,^{\circ}\text{C}$ rises do not produce equal increases in $1/t$ .	4
03.1	The higher-temperature curve has a steeper initial gradient, reaches its plateau sooner and has the same final plateau volume.	Higher temperature increases the initial rate, so product volume rises more rapidly and the initial tangent is steeper. The fixed limiting amount determines the total gas produced, not the temperature, so both curves finish at the same volume; the faster reaction reaches it earlier.	3

Tier 3 · Hard

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01.1	The rate increases by a factor of $1.7$ ; at the higher temperature a greater fraction of molecules has energy at least equal to $E_\mathrm{a}$ , so successful collisions are more frequent.	The two average rates are $8.4/10.0=0.84\,\text{cm}^3\text{s}^{-1}$ and $14.1/10.0=1.41\,\text{cm}^3\text{s}^{-1}$ . Their ratio is $1.41/0.84=1.678\ldots$ , reported as $1.7$ to two significant figures. Heating broadens and shifts the energy distribution, greatly increasing the area beyond the unchanged activation energy; collision frequency also rises slightly.	5
02.1	Expected hydrogen volume $=(0.450/24.3)\times24.0\times1000=444\,\text{cm}^3$ (to 3 significant figures). Expected mass loss $=(0.450/24.3)\times2.00=0.0370\,\text{g}$ (to 3 significant figures). The gas syringe gives a much larger fractional signal and measures hydrogen directly, so its early volume–time gradient is better resolved, but leaks, plunger friction and gas-temperature differences can bias it. Mass loss needs no sealed gas-collection system, but the entire hydrogen loss is only $0.0370\,\text{g}$ and evaporation or acid spray would also reduce the reading. A leak-tested, low-friction gas syringe is preferred; use the initial tangent, keep acid volume and concentration and magnesium mass and surface area fixed, pre-equilibrate the apparatus, and repeat each temperature.	The unrounded amount of magnesium is $0.450/24.3=0.0185185\ldots\,\text{mol}$ , predicting $444.444\ldots\,\text{cm}^3$ of hydrogen and a mass loss of $0.0370370\ldots\,\text{g}$ . Compare each total change with the instrument scale, then consider method-specific systematic errors. Initial rate is obtained from the tangent at $t=0$ , not from the time to completion.	6
03.1	The measured initial rate is too low for $50.0\,^{\circ}\text{C}$ because the early reaction occurs below the stated temperature, where fewer molecules have energy at least equal to $E_\mathrm{a}$ . Equilibrate the reactant containers separately in the bath, then mix promptly, start timing at mixing, keep the bath temperature constant and repeat.	Initial rate refers to the conditions at the start. Here the start is cooler than the quoted temperature, so the high-energy fraction and successful-collision frequency are smaller than they should be. Pre-equilibrating the separate reactants avoids allowing reaction before the target temperature is reached. Mixing only after both have equilibrated makes $t=0$ correspond to the intended temperature.	5
04.1	The mean at $30\,^{\circ}\text{C}$ is $0.0200\,\text{s}^{-1}$ . The mean at $35\,^{\circ}\text{C}$ is $0.0270\,\text{s}^{-1}$ . The mean rate measure increases by a factor of $0.0270/0.0200=1.35$ . The repeat ranges are $0.0196$ - $0.0204$ and $0.0268$ - $0.0272\,\text{s}^{-1}$ . The ranges do not overlap, so these repeats resolve an increase at the higher temperature.	Average each concordant triplet without rounding intermediate sums. Compare the means as a ratio because each run reaches the same endpoint. Then inspect the full spread of repeats at each temperature: the smallest high-temperature value still exceeds the largest low-temperature value, supporting a resolved increase within this dataset.	5
05.1	In the cooler run, the total volume is $62.0\,\text{cm}^3$ and the thiosulfate concentration is $0.0581\,\text{mol dm}^{-3}$ . The acid concentration in the cooler mixture is $0.155\,\text{mol dm}^{-3}$ . In the hotter run, the total volume is $50.0\,\text{cm}^3$ and the thiosulfate concentration is $0.0720\,\text{mol dm}^{-3}$ . The acid concentration in the hotter mixture is $0.192\,\text{mol dm}^{-3}$ . Both mixed concentrations are $62/50=1.24$ times larger in the hot run, so concentration and temperature effects are confounded. Use the same volumes in every run and bring the separate reactants to each target temperature in a thermostatically controlled bath before mixing.	For each reactant use $c_2=c_1V_1/V_{\text{total}}$ . The cooler mixture totals $62.0\,\text{cm}^3$ , whereas the hotter mixture totals only $50.0\,\text{cm}^3$ . Reducing water therefore raises both reacting concentrations by the same $62/50$ factor as well as changing temperature. A fair comparison keeps composition fixed and changes only the equilibrated reaction temperature.	6

3.1.5.4 · Effect of concentration and pressure

Tier 1 · Easy

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01.1	There are more reactant particles per unit volume, so collision frequency increases.	Concentration measures amount per volume. A larger particle population in the same space creates more encounters each second and therefore more opportunities for successful collisions.	2
02.1	Increasing the solution concentration and compressing the reacting gases.	Both selected changes increase particle number density and hence collision frequency. A catalyst changes the reaction route, while heating changes the molecular-energy distribution.	2

Tier 2 · Standard

Mark scheme for 3.1.5.4 Tier 2 · Standard
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01.1	The pressure and number of particles per unit volume increase, so particles are closer together and collide more frequently.	The same gas particles now occupy half the volume, doubling their number density. The shorter typical separation produces more collisions per second. Because temperature is constant, do not attribute the change to faster particles.	3
02.1	Each doubling of $[\mathrm{X}]$ doubles the measured rate; more X particles per unit volume cause more frequent collisions with the other reactant, while the energetic fraction is unchanged at constant temperature.	Read the empirical pattern directly from the table. Increasing $[\mathrm{X}]$ raises particle density and collision frequency. Since temperature is fixed, do not claim that particles move faster or that a larger fraction exceeds the activation energy.	4
03.1	The new concentration is $0.200\,\text{mol dm}^{-3}$ ; fewer reactant particles occupy each unit volume, so collisions with the second reactant occur less frequently and the rate decreases.	The amount is unchanged by dilution, so use $c_1V_1=c_2V_2$ : $c_2=(0.800\times25.0)/100.0=0.200\,\text{mol dm}^{-3}$ . Dilution spreads the same particles through four times the volume. At the same temperature their energy distribution is unchanged, but their collision frequency with the other reactant falls.	4

Tier 3 · Hard

Mark scheme for 3.1.5.4 Tier 3 · Hard
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01.1	Collision frequency and the number of successful collisions per second increase; the activation energy and particle energies remain unchanged because the reaction route and temperature are unchanged.	More particles of the chosen reactant occupy each unit volume, increasing its collision frequency with the other reactant. The temperature is fixed, so the particles have not gained energy. The reaction pathway is also unchanged, so $E_\mathrm{a}$ is constant. More collisions at the same energetic-success fraction give more successful collisions each second and therefore a higher rate.	5
02.1	Adding helium at constant volume leaves the reacting gases' concentrations and collision frequency with each other unchanged, so their reaction rate is unchanged; halving the volume doubles their concentrations, increases collision frequency and increases the rate.	Total pressure alone is not the cause of the kinetic change. In A, the same reactant particles still occupy the same volume, so their number densities and partial pressures are unchanged. In B, the same reactant particles occupy half the volume, so encounters between them become more frequent. Temperature is fixed in both experiments, so their energy distribution is unchanged.	5
03.1	The claim is incorrect: expansion increases the volume occupied by the reacting gases, lowering their concentrations and partial pressures. Their collision frequency and reaction rate decrease, although helium does not react and the molecular-energy distribution and activation energy remain unchanged.	The boundary condition matters. Keeping total pressure constant while adding helium requires the volume to increase. The original numbers of reactant molecules are then spread through a larger volume, so their number densities and partial pressures fall. Encounters between reactant molecules become less frequent. Inert helium does not change the pathway or particle energies, but its volume-expansion effect can still reduce the rate.	5
04.1	For a fixed amount, $[\mathrm{X}]$ is inversely proportional to volume. The concentration of X increases by a factor of $2.40/1.50=1.60$ . More X particles reach and collide with the fixed solid surface per second. The rate increases, while the energetic fraction and activation energy remain unchanged at constant temperature and unchanged route.	Compression changes the number density of the gaseous reactant by the inverse volume ratio. The solid is a separate phase, so do not invent a concentration change for it; its relevant quantity here is exposed surface area, which the prompt holds fixed. The higher gas number density increases surface-collision frequency without changing molecular energies.	4
05.1	In P, $[\mathrm{X}]=0.0400/0.500=0.0800\,\text{mol dm}^{-3}$ . In Q, $[\mathrm{X}]=0.0600/0.500=0.120\,\text{mol dm}^{-3}$ . In P, $[\mathrm{Y}]=0.0600/0.500=0.120\,\text{mol dm}^{-3}$ . In Q, $[\mathrm{Y}]=0.0300/0.500=0.0600\,\text{mol dm}^{-3}$ . Considered alone, raising $[\mathrm{X}]$ from $0.0800$ to $0.120\,\text{mol dm}^{-3}$ increases the frequency of X-Y collisions, whereas lowering $[\mathrm{Y}]$ from $0.120$ to $0.0600\,\text{mol dm}^{-3}$ decreases it, so the two changes act in opposite directions on the same X-Y collision frequency.	Divide each amount by the common $0.500\,\text{dm}^3$ volume. Reactive collisions are those between X and Y, so their frequency rises with either reactant's concentration. Taking each change alone: the increase in $[\mathrm{X}]$ makes X-Y collisions more frequent, while the decrease in $[\mathrm{Y}]$ makes them less frequent — opposite influences on the one X-Y collision frequency.	5

3.1.5.5 · Catalysts

Tier 1 · Easy

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01.1	A substance that increases reaction rate without being changed in chemical composition or amount by the overall reaction.	State both the kinetic effect and that the catalyst is regenerated overall; simply saying that it is not used up is incomplete if composition is ignored.	2
02.1	Y is the catalyst and XY is the intermediate.	Y is consumed in the first step and regenerated in the second, so it cancels from the overall equation. XY is formed in one step and consumed in the next, so it is an intermediate.	2

Tier 2 · Standard

Mark scheme for 3.1.5.5 Tier 2 · Standard
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01.1	The catalyst provides an alternative route with lower activation energy; the distribution curve is unchanged, but the area beyond the lower $E_\mathrm{a}$ is larger, so more collisions are successful.	Keep one molecular-energy curve because temperature is unchanged. Draw the catalysed activation-energy line to the left of the uncatalysed line. The additional area to the right of the lower threshold represents the increased fraction of molecules able to react.	4
02.1	Q is catalysed; $E_\mathrm{a}=95-40=55\,\text{kJ mol}^{-1}$ and $\Delta H=10-40=-30\,\text{kJ mol}^{-1}$ .	The catalysed route is the one with the lower barrier between the same reactant and product levels. Subtract the reactant level from the peak for $E_\mathrm{a}$ , and subtract reactant enthalpy from product enthalpy for $\Delta H$ .	4
03.1	The time halves for the same product amount, so the average rate doubles. The measured mass change is $0.001\pm0.002\,\text{g}$ , so it is consistent with zero; with unchanged composition, the evidence is consistent with a catalyst and the claim is not justified.	Compare rates through the time taken to reach the same endpoint: $240/120=2$ , so the additive increases rate. The apparent mass loss is $1.250-1.249=0.001\,\text{g}$ . A difference uses two readings, so their absolute uncertainties add to $\pm0.002\,\text{g}$ ; the interval for the mass change includes zero. Together with unchanged chemical composition, this supports the definition of a catalyst as unchanged in composition and amount by the overall reaction. A catalyst may participate in steps and be regenerated; an insignificant weighing difference does not make it a reactant.	4

Tier 3 · Hard

Mark scheme for 3.1.5.5 Tier 3 · Hard
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01.1	Uncatalysed reverse $E_\mathrm{a}=180\,\text{kJ mol}^{-1}$ ; catalysed reverse $E_\mathrm{a}=117\,\text{kJ mol}^{-1}$ ; the catalyst changes the route, not the reactant and product enthalpies.	The products lie $35\,\text{kJ mol}^{-1}$ below the reactants. Therefore the reverse barrier is $35\,\text{kJ mol}^{-1}$ larger than the corresponding forward barrier: $145+35=180$ and $82+35=117\,\text{kJ mol}^{-1}$ . The initial and final energy levels are unchanged, so their difference $\Delta H$ is unchanged.	5
02.1	The claim is incorrect: both mixtures reach the same equilibrium composition and therefore the same $K_\mathrm{c}$ ; the catalyst only allows equilibrium to be reached sooner by increasing both forward and reverse rates.	Compare the plateau concentrations rather than the times taken to reach them. Their equality shows that the equilibrium position is unchanged. At a fixed temperature $K_\mathrm{c}$ is unchanged because a catalyst changes activation energies, not the thermodynamic equilibrium ratio.	4
03.1	For Q, forward $E_\mathrm{a}=120-25=95\,\text{kJ mol}^{-1}$ and reverse $E_\mathrm{a}=120-70=50\,\text{kJ mol}^{-1}$ . Q can represent catalysis because it lowers the barriers while keeping $\Delta H=70-25=+45\,\text{kJ mol}^{-1}$ unchanged; R cannot because moving the product level changes $\Delta H$ .	Activation energy is the height from the relevant starting level to the pathway maximum. For Q these differences are $95$ and $50\,\text{kJ mol}^{-1}$ . A catalyst changes the route but not the enthalpies of reactants and products. Q obeys this condition. R would change the reaction enthalpy from $+45$ to $+15\,\text{kJ mol}^{-1}$ , so it is not merely a catalysed profile for the same reaction.	6
04.1	With a catalyst at constant temperature, the Maxwell–Boltzmann curve is unchanged. The catalyst provides a lower activation energy, so the $E_\mathrm{a}$ marker moves left and the area to its right increases. On heating, the curve becomes broader with a lower peak shifted to the right. For the heated uncatalysed reaction, the $E_\mathrm{a}$ marker stays fixed. Heating increases the area beyond the fixed $E_\mathrm{a}$ , while catalysis increases the area beyond a lowered $E_\mathrm{a}$ ; both therefore increase the fraction able to react.	Separate a pathway change from a temperature change. A catalyst lowers the threshold without redistributing molecular energies, so retain the same curve and move only the activation-energy line left. Heating redistributes the molecules to a broader, right-shifted curve but does not alter the activation energy of the unchanged route. In both diagrams compare the area to the right of the relevant threshold.	5

3.1.6.1 · Chemical equilibria and Le Chatelier's principle

Tier 1 · Easy

Mark scheme for 3.1.6.1 Tier 1 · Easy
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01.1	The forward and reverse reactions have equal rates; reactant and product concentrations remain constant.	Equilibrium is dynamic because both reactions continue. Equal rates create no net concentration change, but the concentrations need not be equal to each other.	2
02.1	Equilibrium is first shown at $60\,\text{s}$ ; the concentrations remain constant because the forward and reverse reactions continue at equal rates, not because both reactions stop.	The first repeated constant composition occurs from $60\,\text{s}$ onward. Constant macroscopic concentrations arise from equal opposing rates at dynamic equilibrium.	3

Tier 2 · Standard

Mark scheme for 3.1.6.1 Tier 2 · Standard
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01.1	Higher pressure increases the ammonia yield; higher temperature decreases it; a catalyst has no effect on the equilibrium yield.	There are four moles of gas on the left and two on the right, so higher pressure shifts equilibrium right. Heat behaves like a product for an exothermic forward reaction, so higher temperature shifts equilibrium left. A catalyst accelerates both directions and does not alter the equilibrium position.	3
02.1	A gives the higher ethanol equilibrium yield because lower temperature favours the exothermic direction and higher pressure favours the change from two gaseous moles to one, but it is slower and more expensive to compress. B is faster and cheaper to pressurise but gives lower equilibrium conversion. Either recommendation is acceptable if it balances yield, throughput, energy cost and safety using the data.	Separate equilibrium advantages from rate and engineering disadvantages. In the exothermic forward reaction, A's lower temperature raises equilibrium conversion but slows the reaction. Its higher pressure favours the one-mole gas side but increases compression and equipment costs. B sacrifices equilibrium yield for faster production and lower compression demand. A defensible decision must compare more than yield alone and may mention recycling unreacted gases.	6
03.1	Immediately after removal the forward rate is initially unchanged but the reverse rate is lower, so the forward rate is greater and the equilibrium shifts right to replace B. At the new equilibrium the forward and reverse rates are equal again.	Before the disturbance the two rates are equal. Removing B immediately lowers its concentration and therefore the reverse rate, while A has not yet changed so the forward rate initially retains its previous value. Net conversion of A to B follows until the opposing rates become equal at a new constant composition.	4

Tier 3 · Hard

Mark scheme for 3.1.6.1 Tier 3 · Hard
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01.1	Lower temperature and higher pressure favour $\mathrm{SO_3}$ , but low temperature gives a slow rate and very high pressure is costly; a moderate temperature and economically suitable pressure are compromises, while a catalyst raises rate without changing equilibrium yield.	Because the forward reaction is exothermic, lowering temperature shifts equilibrium right but also slows the reaction, so a moderate temperature balances yield and rate. Three gaseous moles form two, so pressure favours products, but progressively higher pressure brings engineering and energy costs. A catalyst lowers activation energies for both directions, reaching the same equilibrium faster.	6
02.1	$K_\mathrm{c}=[\mathrm{B}]/[\mathrm{A}]=0.50$ for both mixtures; adding B shifts the equilibrium position to the left so some B is converted into A, but $K_\mathrm{c}$ does not change because temperature is fixed.	Before addition, $K_\mathrm{c}=0.40/0.80=0.50$ ; after re-equilibration, $K_\mathrm{c}=0.50/1.00=0.50$ . The changed concentrations show a different equilibrium composition after the disturbance, while the unchanged ratio shows that the constant itself is fixed at this temperature.	5
03.1	Higher temperature favours the endothermic forward direction and shifts equilibrium right, whereas higher pressure favours the side with fewer gas moles and shifts it left. The net positional change cannot be deduced without quantitative information; temperature increases the equilibrium constant for this endothermic reaction, while pressure does not change the constant.	Analyse the changes separately. Heating favours the direction that absorbs heat, so it favours B. Compression favours one mole of gaseous A over two moles of gaseous B. Since these effects oppose each other, Le Chatelier's principle alone cannot say which dominates. Only the temperature change alters the equilibrium constant; pressure changes the composition needed to satisfy that temperature-dependent constant.	5
04.1	Compression favours the product side, so that side has fewer moles of gas. The pressure observation gives only the relative gaseous mole totals, not the exact equation coefficients. Heating lowers product yield, so the reverse direction is endothermic. The forward reaction is therefore exothermic. The catalyst increases both opposing rates and reaches the same equilibrium sooner without changing its position or constant.	Work backwards from each isolated disturbance. Higher pressure favours fewer gaseous particles, identifying the product side as the lower-mole side. Higher temperature favours the endothermic direction; because product yield falls, that direction is reverse, making the forward reaction exothermic. The unchanged final composition separates the catalyst's kinetic effect from equilibrium position.	5
05.1	Labelled AB forming from labelled $\mathrm{A_2}$ shows that the forward reaction continues. Labelled $\mathrm{A_2}$ forming from labelled AB shows that the reverse reaction continues. Both directions therefore occur at equilibrium, so the equilibrium is dynamic. Constant macroscopic concentrations show that the forward and reverse rates are equal. Equal opposing rates give no net composition change even though individual molecules continue to react.	Use each tracer direction separately: transfer of the label from a reactant into product requires forward reaction, while transfer from product back into reactant requires reverse reaction. The unchanging bulk concentrations rule out a continuing net reaction in either direction. The only consistent description is simultaneous opposing reactions at equal rates.	5

3.1.6.2 · Equilibrium constant Kc for homogeneous systems

Tier 1 · Easy

Mark scheme for 3.1.6.2 Tier 1 · Easy
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01.1	$K_\mathrm{c}=\dfrac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}$	Place the product concentration in the numerator and reactant concentrations in the denominator. Use each equation coefficient as the corresponding power, giving power $2$ for $[\mathrm{HI}]$ .	1
02.1	The coefficient 2 has been used as a multiplier instead of a power; $K_\mathrm{c}=\dfrac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2}$	Write every equilibrium concentration in square brackets. A stoichiometric coefficient becomes a power in the expression; it is not a multiplier placed before the concentration.	3

Tier 2 · Standard

Mark scheme for 3.1.6.2 Tier 2 · Standard
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01.1	$K_\mathrm{c}=24.0$ ; no units	Substitute into $K_\mathrm{c}=[\mathrm{HI}]^2/([\mathrm{H_2}][\mathrm{I_2}])$ : $K_\mathrm{c}=1.20^2/(0.200\times0.300)=1.44/0.0600=24.0$ . The total concentration power is two in both numerator and denominator, so the units cancel.	3
02.1	$[\mathrm{NO_2}]=0.300\,\text{mol dm}^{-3}$	$K_\mathrm{c}=[\mathrm{NO_2}]^2/[\mathrm{N_2O_4}]$ , so $[\mathrm{NO_2}]^2=(0.360)(0.250)=0.0900\,\text{mol}^2\text{dm}^{-6}$ . Taking the positive square root gives $0.300\,\text{mol dm}^{-3}$ .	3
03.1	$K_\mathrm{c}=\dfrac{[\mathrm{C}]}{[\mathrm{A}]^2[\mathrm{B}]}$ ; units $=\mathrm{mol^{-2}\,dm^6}$ .	Use equation coefficients as powers: one for C, two for A and one for B. The net concentration power is $1-(2+1)=-2$ . Therefore the units are $(\text{mol dm}^{-3})^{-2}=\text{mol}^{-2}\text{dm}^{6}$ .	3

Tier 3 · Hard

Mark scheme for 3.1.6.2 Tier 3 · Hard
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01.1	$K_\mathrm{c}=1.30$ with no units; increasing temperature decreases $K_\mathrm{c}$ .	Forming $0.500\,\text{mol}$ of C uses $0.250\,\text{mol}$ each of A and B. Equilibrium amounts are therefore $0.550$ , $0.350$ and $0.500\,\text{mol}$ , giving concentrations $0.275$ , $0.175$ and $0.250\,\text{mol dm}^{-3}$ . Hence $K_\mathrm{c}=0.250^2/(0.275\times0.175)=1.30$ ; the concentration powers cancel. Heating favours the endothermic reverse direction, so the product-to-reactant ratio and $K_\mathrm{c}$ decrease.	6
02.1	$[\mathrm{CO}]=[\mathrm{H_2O}]=0.200\,\text{mol dm}^{-3}$ and $[\mathrm{CO_2}]=[\mathrm{H_2}]=0.400\,\text{mol dm}^{-3}$ ; $K_\mathrm{c}$ has no units.	Let $x$ be the equilibrium concentration of each product. Then $K_\mathrm{c}=x^2/(0.600-x)^2=4.00$ . Taking the positive square root gives $x/(0.600-x)=2.00$ , so $x=0.400\,\text{mol dm}^{-3}$ and each reactant concentration is $0.200\,\text{mol dm}^{-3}$ . The total concentration power is two above and below the fraction, so the units cancel.	6
03.1	$K_\mathrm{c}=6.40\,\mathrm{mol^{-1}\,dm^3}$ .	Convert equilibrium amounts to concentrations: $[\mathrm{P}]=0.250/0.500=0.500\,\text{mol dm}^{-3}$ , $[\mathrm{Q}]=0.200/0.500=0.400\,\text{mol dm}^{-3}$ and $[\mathrm{R}]=0.400/0.500=0.800\,\text{mol dm}^{-3}$ . Then $K_\mathrm{c}=[\mathrm{R}]^2/([\mathrm{P}]^2[\mathrm{Q}])=0.800^2/(0.500^2\times0.400)=0.640/0.100=6.40$ . The net concentration power is $2-(2+1)=-1$ , so the units are $\mathrm{mol^{-1}\,dm^3}$ .	4
04.1	$K_\mathrm{c}=\dfrac{[\mathrm{C}]}{[\mathrm{A}][\mathrm{B}]}$ . For volume V, substitution gives $4.00=(0.120/V)/[(0.200/V)(0.300/V)]$ . This simplifies to $4.00=2.00V$ , so $V=2.00\,\text{dm}^3$ . $[\mathrm{A}]=0.100\,\text{mol dm}^{-3}$ and $[\mathrm{B}]=0.150\,\text{mol dm}^{-3}$ . $[\mathrm{C}]=0.0600\,\text{mol dm}^{-3}$ .	Express every equilibrium concentration as amount divided by the same unknown volume. Unlike an equation with equal total concentration powers, this expression retains one factor of volume. Simplify before solving: $K_\mathrm{c}=0.120V/(0.200\times0.300)=2.00V$ . Divide each amount by the resulting $2.00\,\text{dm}^3$ and verify that $0.0600/(0.100\times0.150)=4.00$ .	5
05.1	For $\mathrm{B(g)\rightleftharpoons2A(g)}$ , the constant is the reciprocal: $1/0.0400=25.0$ . The reverse constant has units $\text{mol dm}^{-3}$ . For $\mathrm{4A(g)\rightleftharpoons2B(g)}$ , the constant is $(0.0400)^2=0.00160$ . The doubled-equation constant has units $\text{mol}^{-2}\text{ dm}^6$ . Heating decreases the original constant because its forward reaction is exothermic. Heating increases the reverse constant because the reverse reaction is endothermic.	Write the concentration expressions to establish the transformations. Reversing swaps numerator and denominator, so both the numerical constant and its units are reciprocated. Doubling every coefficient doubles every power, so the complete original expression is squared. Temperature favours the endothermic direction; therefore heating changes the two reciprocal constants in opposite numerical directions.	6

3.1.7 · Oxidation, reduction and redox equations

Tier 1 · Easy

Mark scheme for 3.1.7 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$+7$	Let the oxidation state of manganese be $x$ . Oxygen is $-2$ , so $x+4(-2)=-1$ . Therefore $x=+7$ .	2
02.1	$\mathrm{Cl_2}$ is the oxidising agent. $\mathrm{Br^-}$ is the reducing agent.	$\mathrm{Cl_2}$ accepts electrons to form $\mathrm{Cl^-}$ , so it is reduced and is the oxidising agent. $\mathrm{Br^-}$ donates electrons to form $\mathrm{Br_2}$ , so it is oxidised and is the reducing agent.	2

Tier 2 · Standard

Mark scheme for 3.1.7 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$\mathrm{2Fe^{2+}+Cl_2\rightarrow2Fe^{3+}+2Cl^-}$	Multiply the iron half-equation by $2$ so that it releases $2e^-$ . Add the half-equations and cancel the electrons to obtain $\mathrm{2Fe^{2+}+Cl_2\rightarrow2Fe^{3+}+2Cl^-}$ .	3
02.1	$\mathrm{NO_3^-+4H^++3e^-\rightarrow NO+2H_2O}$	Add two water molecules to the right to balance oxygen, then four $\mathrm{H^+}$ ions to the left to balance hydrogen. The left side then has charge $+3$ , so add three electrons to the left to make the total charge zero on both sides.	3
03.1	$\mathrm{2Fe^{3+}+2e^-\rightarrow2Fe^{2+}}$ ; also accept $\mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}}$ . $\mathrm{Fe^{3+}}$ is the oxidising agent.	Remove the stated oxidation half-equation from the overall equation. The remaining iron species require $\mathrm{2Fe^{3+}+2e^-\rightarrow2Fe^{2+}}$ . Each $\mathrm{Fe^{3+}}$ ion accepts an electron, so it is reduced; the electron acceptor is the oxidising agent.	3

Tier 3 · Hard

Mark scheme for 3.1.7 Tier 3 · Hard
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01.1	$\mathrm{Cr_2O_7^{2-}+14H^++3Sn^{2+}\rightarrow2Cr^{3+}+7H_2O+3Sn^{4+}}$	The reduction half-equation is $\mathrm{Cr_2O_7^{2-}+14H^++6e^-\rightarrow2Cr^{3+}+7H_2O}$ . The oxidation half-equation is $\mathrm{Sn^{2+}\rightarrow Sn^{4+}+2e^-}$ . Multiply the tin half-equation by $3$ , add the equations and cancel $6e^-$ to give the stated overall equation.	5
02.1	The proposed left-hand charge is $+14$ , whereas the right-hand charge is $+4$ . The hydrogen atoms also fail to balance: 26 on the left, 16 on the right. $\mathrm{2MnO_4^-+5H_2O_2+6H^+\rightarrow2Mn^{2+}+5O_2+8H_2O}$	The proposed equation has total charge $+14$ on the left and $+4$ on the right; it also has 26 hydrogen atoms on the left but only 16 on the right. The reduction half-equation is $\mathrm{MnO_4^-+8H^++5e^-\rightarrow Mn^{2+}+4H_2O}$ and the oxidation half-equation is $\mathrm{H_2O_2\rightarrow O_2+2H^++2e^-}$ . Multiply these by $2$ and $5$ respectively, add them, then cancel ten electrons and ten of the sixteen protons. The corrected equation has total charge $+4$ on each side and balances every atom.	5
03.1	Iodine changes from $+5$ in $\mathrm{IO_3^-}$ to $0$ , so one iodate ion gains five electrons. Five $\mathrm{I^-}$ ions each change from $-1$ to $0$ and lose one electron, giving a $1{:}5$ iodate-to-iodide ratio. $\mathrm{IO_3^-+5I^-+6H^+\rightarrow3I_2+3H_2O}$	The iodine atom in $\mathrm{IO_3^-}$ has oxidation state $+5$ and is reduced to $0$ , a gain of five electrons. Each iodide iodine is oxidised from $-1$ to $0$ , so five iodide ions supply those five electrons. The six iodine atoms form three $\mathrm{I_2}$ molecules. Add three water molecules to balance oxygen and six protons to balance hydrogen. The left-hand charge is $-1-5+6=0$ , matching the neutral products.	5
04.1	The three M atoms have a total oxidation state of $+8$ because the four oxygen atoms contribute $-8$ . One formula unit represents one $\mathrm{M^{2+}}$ ion and two $\mathrm{M^{3+}}$ ions. The two M atoms initially at $+3$ each gain one electron, so one formula unit gains two electrons. $\mathrm{M_3O_4+8H^++2I^-\rightarrow3M^{2+}+I_2+4H_2O}$	Oxygen contributes $4(-2)=-8$ , so the oxidation states of the three M atoms must sum to $+8$ . If $a$ atoms are $+2$ and $3-a$ are $+3$ , then $2a+3(3-a)=8$ , giving $a=1$ . Converting the two $\mathrm{M^{3+}}$ centres to $\mathrm{M^{2+}}$ consumes two electrons; $\mathrm{2I^-\rightarrow I_2+2e^-}$ supplies them. Balance the four oxygen atoms with four water molecules and the eight hydrogen atoms with eight protons. The resulting equation has total charge $+6$ on both sides.	4
05.1	X has oxidation state $+7$ in $\mathrm{XO_4^-}$ . Sulfur changes from $+4$ in $\mathrm{SO_3^{2-}}$ to $+6$ in $\mathrm{SO_4^{2-}}$ , so each sulfite ion loses two electrons. The $1{:}2.50$ ratio transfers five electrons to each X atom, reducing X from $+7$ to $+2$ . $\mathrm{XO_4^-+8H^++5e^-\rightarrow X^{2+}+4H_2O}$ $\mathrm{2XO_4^-+5SO_3^{2-}+6H^+\rightarrow2X^{2+}+5SO_4^{2-}+3H_2O}$	Four oxygen atoms contribute $-8$ , so X is $+7$ in an ion with charge $-1$ . Sulfur is $+4$ in sulfite and $+6$ in sulfate, so $2.50$ sulfite ions release $5.00$ electrons per $\mathrm{XO_4^-}$ . X must therefore finish at $+2$ . Balance its reduction half-equation in acid as $\mathrm{XO_4^-+8H^++5e^-\rightarrow X^{2+}+4H_2O}$ . The oxidation half-equation is $\mathrm{SO_3^{2-}+H_2O\rightarrow SO_4^{2-}+2H^++2e^-}$ . Multiply the half-equations by $2$ and $5$ , add them, then cancel electrons, protons and water to obtain the overall equation; its total charge is $-6$ on each side.	5

3.1.8.1 · Born–Haber cycles (A-level only)

Tier 1 · Easy

Mark scheme for 3.1.8.1 Tier 1 · Easy
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01.1	The enthalpy change when one mole of $\mathrm{MgCl_2(s)}$ is formed from its gaseous ions under standard conditions. $\mathrm{Mg^{2+}(g)+2Cl^-(g)\rightarrow MgCl_2(s)}$	State formation of exactly one mole of the ionic solid and specify that the starting ions are gaseous. The formula requires one $\mathrm{Mg^{2+}}$ ion and two $\mathrm{Cl^-}$ ions.	2
02.1	$-2630\,\mathrm{kJ\,mol^{-1}}$ $\mathrm{Ca^{2+}(g)+2F^-(g)\rightarrow CaF_2(s)}$	Lattice formation is the reverse of lattice dissociation, so its enthalpy has the same magnitude and the opposite sign. Formation starts with the stoichiometric gaseous ions and produces one mole of the solid lattice.	3

Tier 2 · Standard

Mark scheme for 3.1.8.1 Tier 2 · Standard
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01.1	$-787\,\mathrm{kJ\,mol^{-1}}$	Hess's law gives $-411=108+121+496-349+\Delta_{latt}H^\circ$ . The non-lattice terms sum to $376\,\mathrm{kJ\,mol^{-1}}$ , so $\Delta_{latt}H^\circ=-411-376=-787\,\mathrm{kJ\,mol^{-1}}$ .	4
02.1	$+677\,\mathrm{kJ\,mol^{-1}}$ Dissociating the lattice requires energy to overcome the electrostatic attractions between oppositely charged ions, so the enthalpy change is positive.	The solution cycle is $\Delta_{sol}H^\circ=\Delta_{latt,diss}H^\circ+\Delta_{hyd}H^\circ(\mathrm{K^+})+\Delta_{hyd}H^\circ(\mathrm{Br^-})$ . Therefore $\Delta_{latt,diss}H^\circ=19.9-(-322)-(-335)=+676.9\,\mathrm{kJ\,mol^{-1}}$ , which is $+677\,\mathrm{kJ\,mol^{-1}}$ to three significant figures. Energy is absorbed when the ionic lattice is separated into gaseous ions.	4
03.1	$\Delta_{hyd}H(\mathrm{M^+})=-480\,\mathrm{kJ\,mol^{-1}}$	Dissolving reverses lattice formation, so the lattice-dissociation term is $+2350\,\mathrm{kJ\,mol^{-1}}$ . If the hydration enthalpy of $\mathrm{M^+}$ is $x$ , then $-90=2350+2x-1480$ . Therefore $2x=-960\,\mathrm{kJ\,mol^{-1}}$ and $x=-480\,\mathrm{kJ\,mol^{-1}}$ .	4

Tier 3 · Hard

Mark scheme for 3.1.8.1 Tier 3 · Hard
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01.1	$-122\,\mathrm{kJ\,mol^{-1}}$	Dissolving first reverses lattice formation, so the lattice-dissociation term is $+2526\,\mathrm{kJ\,mol^{-1}}$ . Hydration gives $-1920+2(-364)=-2648\,\mathrm{kJ\,mol^{-1}}$ . Hence $\Delta_{sol}H^\circ=2526-2648=-122\,\mathrm{kJ\,mol^{-1}}$ .	4
02.1	$IE_2(\mathrm{Mg})=+1451\,\mathrm{kJ\,mol^{-1}}$ The experimental lattice enthalpy is $141\,\mathrm{kJ\,mol^{-1}}$ more exothermic than the perfect-ionic-model value. The perfect ionic model assumes spherical ions with purely ionic bonding. The $\mathrm{Mg^{2+}}$ ion polarises the $\mathrm{O^{2-}}$ electron cloud, giving some covalent character and stronger attraction than the model predicts.	Apply the cycle in the formation direction: $-602=150+249+738+IE_2-141+742-3791$ . The known terms excluding $IE_2$ sum to $-2053\,\mathrm{kJ\,mol^{-1}}$ , so $IE_2=-602-(-2053)=+1451\,\mathrm{kJ\,mol^{-1}}$ . The experimental value is $-3791-(-3650)=-141\,\mathrm{kJ\,mol^{-1}}$ relative to the model, so its magnitude is $141\,\mathrm{kJ\,mol^{-1}}$ greater. This extra exothermicity indicates polarisation and partial covalent character not included in the perfect ionic model.	6
03.1	Bond dissociation enthalpy of $\mathrm{X_2}=+230\,\mathrm{kJ\,mol^{-1}}$ $\mathrm{X(g)+e^-\rightarrow X^-(g)}$	The formation cycle gives $-690=180+E+600+1200+2(-350)+(-2200)$ , where $E$ is the total atomisation enthalpy of the two moles of X atoms — which equals the bond dissociation enthalpy of one mole of $\mathrm{X_2}$ , since atomisation per mole of X atoms is half the bond enthalpy. The known terms total $-920\,\mathrm{kJ\,mol^{-1}}$ , so $E=-690-(-920)=+230\,\mathrm{kJ\,mol^{-1}}$ . The first electron affinity is the enthalpy change for one mole of gaseous atoms each gaining one electron: $\mathrm{X(g)+e^-\rightarrow X^-(g)}$ .	6
04.1	$\Delta_{latt}H^\circ(\text{formation})=-825\,\mathrm{kJ\,mol^{-1}}$ . $\Delta_{sol}H^\circ=+149\,\mathrm{kJ\,mol^{-1}}$ . $\mathrm{M^+(g)\rightarrow M^+(aq)}$	Apply Hess's law to the formation cycle: $-412=126+104+511-328+\Delta_{latt}H^\circ$ . The non-lattice terms total $+413\,\mathrm{kJ\,mol^{-1}}$ , so the lattice enthalpy of formation is $-412-413=-825\,\mathrm{kJ\,mol^{-1}}$ . Dissolving reverses lattice formation, so $\Delta_{sol}H^\circ=+825-304-372=+149\,\mathrm{kJ\,mol^{-1}}$ . Hydration converts one mole of gaseous cations to aqueous cations, as shown by $\mathrm{M^+(g)\rightarrow M^+(aq)}$ .	6

3.1.8.2 · Gibbs free-energy change, ΔG, and entropy change, ΔS (A-level only)

Tier 1 · Easy

Mark scheme for 3.1.8.2 Tier 1 · Easy
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01.1	$+73\,\mathrm{J\,K^{-1}\,mol^{-1}}$	Subtract the reactant total from the product total: $\Delta S^\circ=465-392=+73\,\mathrm{J\,K^{-1}\,mol^{-1}}$ .	2
02.1	The negative $\Delta G$ shows that the reaction is thermodynamically feasible under the stated conditions. It does not give the reaction rate; the reaction may still be slow because it has a large activation energy.	Separate thermodynamics from kinetics. The sign of $\Delta G$ predicts feasibility, whereas rate depends on the kinetic pathway and activation energy.	2

Tier 2 · Standard

Mark scheme for 3.1.8.2 Tier 2 · Standard
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01.1	$\Delta G=-5.25\,\mathrm{kJ\,mol^{-1}}$ The reaction is feasible at $350\,\mathrm{K}$ .	Convert $\Delta S$ to $0.135\,\mathrm{kJ\,K^{-1}\,mol^{-1}}$ . Then $\Delta G=42.0-350(0.135)=-5.25\,\mathrm{kJ\,mol^{-1}}$ . The negative value means the reaction is feasible under these conditions.	4
02.1	$\Delta S^\circ=-199\,\mathrm{J\,K^{-1}\,mol^{-1}}$ $\Delta G^\circ=-33.1\,\mathrm{kJ\,mol^{-1}}$ The reaction is feasible at $298\,\mathrm{K}$ .	$\Delta S^\circ=2(193)-[192+3(131)]=-199\,\mathrm{J\,K^{-1}\,mol^{-1}}$ . Convert this to $-0.199\,\mathrm{kJ\,K^{-1}\,mol^{-1}}$ before combining it with the enthalpy: $\Delta G^\circ=-92.4-298(-0.199)=-33.098\,\mathrm{kJ\,mol^{-1}}$ , which rounds to $-33.1\,\mathrm{kJ\,mol^{-1}}$ . Its negative sign shows thermodynamic feasibility.	5
03.1	Reaction P is thermodynamically feasible at all temperatures because both $\Delta H$ and $-T\Delta S$ are negative. Reaction Q is not thermodynamically feasible at any temperature because both $\Delta H$ and $-T\Delta S$ are positive.	For P, the enthalpy term is negative and positive entropy makes $-T\Delta S$ negative, so their sum remains negative at every positive temperature. For Q, the enthalpy term is positive and negative entropy makes $-T\Delta S$ positive, so their sum remains positive. These conclusions concern thermodynamic feasibility, not reaction rate.	4

Tier 3 · Hard

Mark scheme for 3.1.8.2 Tier 3 · Hard
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01.1	$\Delta H=+63.0\,\mathrm{kJ\,mol^{-1}}$ $\Delta S=+150\,\mathrm{J\,K^{-1}\,mol^{-1}}$ $T=420\,\mathrm{K}$	The gradient of the $\Delta G$ against $T$ line is $(-12.0-18.0)/(500-300)=-0.150\,\mathrm{kJ\,K^{-1}\,mol^{-1}}=-\Delta S$ , so $\Delta S=+0.150\,\mathrm{kJ\,K^{-1}\,mol^{-1}}$ . Using the $300\,\mathrm{K}$ value, $18.0=\Delta H-300(0.150)$ , giving $\Delta H=63.0\,\mathrm{kJ\,mol^{-1}}$ . At the threshold $\Delta G=0$ , so $T=63.0/0.150=420\,\mathrm{K}$ .	5
02.1	$\Delta S$ must be converted to $-0.180\,\mathrm{kJ\,K^{-1}\,mol^{-1}}$ before it is combined with $\Delta H$ . $\Delta G=+13.0\,\mathrm{kJ\,mol^{-1}}$ The reaction is not thermodynamically feasible at $600\,\mathrm{K}$ under the stated conditions; no conclusion about its rate follows from this calculation.	Use consistent energy units: $\Delta G=-95.0-600(-0.180)=-95.0+108=+13.0\,\mathrm{kJ\,mol^{-1}}$ . The positive value rules out thermodynamic feasibility under the stated conditions, while kinetics would require separate information about the reaction pathway and activation energy.	5
03.1	$\Delta S^\circ=-117\,\mathrm{J\,K^{-1}\,mol^{-1}}$ $S^\circ(\mathrm{C})=273\,\mathrm{J\,K^{-1}\,mol^{-1}}$ The reaction is feasible up to $615\,\mathrm{K}$ ; at the threshold $\Delta G=0$ .	Rearrange $\Delta G=\Delta H-T\Delta S$ to $\Delta S=(\Delta H-\Delta G)/T$ . Using kilojoules first gives $\Delta S=[-72.0-(-25.2)]/400=-0.117\,\mathrm{kJ\,K^{-1}\,mol^{-1}}=-117\,\mathrm{J\,K^{-1}\,mol^{-1}}$ . Also $\Delta S=S^\circ(\mathrm{C})-[180+210]$ , so $S^\circ(\mathrm{C})=390-117=273\,\mathrm{J\,K^{-1}\,mol^{-1}}$ . At the feasibility boundary, $0=-72.0-T(-0.117)$ , giving $T=615.38\ldots\,\mathrm{K}$ , or $615\,\mathrm{K}$ to three significant figures.	6
04.1	For P, the feasibility boundary is $T=400\,\mathrm{K}$ . P is feasible at temperatures at or above $400\,\mathrm{K}$ . For Q, the feasibility boundary is $T=480\,\mathrm{K}$ . Q is feasible at temperatures at or below $480\,\mathrm{K}$ . Both reactions are feasible for $400\,\mathrm{K}\leq T\leq480\,\mathrm{K}$ .	Convert the entropy changes to $+0.120$ and $-0.150\,\mathrm{kJ\,K^{-1}\,mol^{-1}}$ . At a boundary, $\Delta G=0$ . For P, $0=48.0-0.120T$ , so $T=400\,\mathrm{K}$ ; its positive entropy change makes $\Delta G$ fall as temperature rises. For Q, $0=-72.0-T(-0.150)$ , so $T=480\,\mathrm{K}$ ; its negative entropy change makes $\Delta G$ rise as temperature rises. Intersecting the two feasible intervals gives the stated closed range, with both boundaries included because $\Delta G=0$ is feasible.	5
05.1	$\Delta S$ must be positive because increasing temperature changes the reaction from not feasible to feasible. The observation at $300\,\mathrm{K}$ requires $\Delta S<120\,\mathrm{J\,K^{-1}\,mol^{-1}}$ . The observation at $450\,\mathrm{K}$ requires $\Delta S\geq80.0\,\mathrm{J\,K^{-1}\,mol^{-1}}$ . $80.0\leq\Delta S<120\,\mathrm{J\,K^{-1}\,mol^{-1}}$ .	Use $\Delta G=\Delta H-T\Delta S$ with $\Delta H=36000\,\mathrm{J\,mol^{-1}}$ . Not feasible at $300\,\mathrm{K}$ means $36000-300\Delta S>0$ , hence $\Delta S<120\,\mathrm{J\,K^{-1}\,mol^{-1}}$ ; the inequality is strict because the reaction was stated not to be feasible. Feasible at $450\,\mathrm{K}$ means $36000-450\Delta S\leq0$ , hence $\Delta S\geq80.0\,\mathrm{J\,K^{-1}\,mol^{-1}}$ ; equality is included because $\Delta G=0$ is feasible. Combining the bounds gives the allowed interval.	4

3.1.9.1 · Rate equations (A-level only)

Tier 1 · Easy

Mark scheme for 3.1.9.1 Tier 1 · Easy
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01.1	The rate is multiplied by $0.50$ ; it halves.	Halving $[A]$ contributes $(1/2)^2=1/4$ , while doubling $[B]$ contributes $2$ . The combined factor is $(1/4)(2)=1/2$ .	2
02.1	$n=2$ $\text{rate}=k[B]^2$	The units $\mathrm{dm^3\,mol^{-1}\,s^{-1}}$ correspond to an overall second-order rate equation. Since the order in $A$ is zero, the full order of two must come from $B$ , so $n=2$ .	3

Tier 2 · Standard

Mark scheme for 3.1.9.1 Tier 2 · Standard
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01.1	$k=0.960\,\mathrm{dm^6\,mol^{-2}\,s^{-1}}$	Rearrange to $k=\text{rate}/([A][B]^2)$ . Thus $k=(4.80\times10^{-4})/[0.200(0.0500)^2]=0.960$ . Since the overall order is $3$ , dividing $\mathrm{mol\,dm^{-3}\,s^{-1}}$ by $(\mathrm{mol\,dm^{-3}})^3$ gives $\mathrm{dm^6\,mol^{-2}\,s^{-1}}$ .	4
02.1	$E_a=68.1\,\mathrm{kJ\,mol^{-1}}$	For an Arrhenius plot, the gradient is $-E_a/R$ . Therefore $E_a=-R(\text{gradient})=-8.31(-8.20\times10^3)=68142\,\mathrm{J\,mol^{-1}}=68.1\,\mathrm{kJ\,mol^{-1}}$ .	3
03.1	$[A]=0.0894\,\mathrm{mol\,dm^{-3}}$	Rearrange the rate equation to $[A]=\sqrt{\text{rate}/(k[B])}$ . Substitution gives $[A]=\sqrt{(4.00\times10^{-4})/[0.250(0.200)]}=\sqrt{0.00800}=0.0894427\ldots\,\mathrm{mol\,dm^{-3}}$ , which is $0.0894\,\mathrm{mol\,dm^{-3}}$ to three significant figures.	4

Tier 3 · Hard

Mark scheme for 3.1.9.1 Tier 3 · Hard
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01.1	$E_a=54.6\,\mathrm{kJ\,mol^{-1}}$	Here $\ln(k_2/k_1)=\ln4=1.3863$ and $1/298-1/318=2.110\times10^{-4}\,\mathrm{K^{-1}}$ . Rearranging gives $E_a=8.31(1.3863)/(2.110\times10^{-4})=5.46\times10^4\,\mathrm{J\,mol^{-1}}=54.6\,\mathrm{kJ\,mol^{-1}}$ .	5
02.1	$A=2.86\times10^6\,\mathrm{s^{-1}}$ $k=2.21\times10^{-2}\,\mathrm{s^{-1}}$ at $335\,\mathrm{K}$	Convert $E_a$ to $52000\,\mathrm{J\,mol^{-1}}$ . Rearranging gives $A=k e^{E_a/(RT)}=(2.50\times10^{-3})e^{52000/[8.31(300)]}=2.861735\times10^6\,\mathrm{s^{-1}}$ . Then $k=(2.861735\times10^6)e^{-52000/[8.31(335)]}=0.0220989\,\mathrm{s^{-1}}$ , which is $2.21\times10^{-2}\,\mathrm{s^{-1}}$ to three significant figures.	6
03.1	$k_2/k_1=5.16$ $[X]_{330}=0.0792\,\mathrm{mol\,dm^{-3}}$	Convert $E_a$ to $45000\,\mathrm{J\,mol^{-1}}$ . Then $k_2/k_1=\exp\left[(45000/8.31)(1/300-1/330)\right]=5.1601\ldots$ . Equal rates require $k_1(0.180)^2=k_2[X]_{330}^2$ , so $[X]_{330}=0.180/\sqrt{5.1601\ldots}=0.0792397\ldots\,\mathrm{mol\,dm^{-3}}$ , which is $0.0792\,\mathrm{mol\,dm^{-3}}$ to three significant figures.	6
04.1	$\ln(k_2/k_1)=\ln8=2.079$ . $\dfrac{1}{T_2}=\dfrac{1}{295}-\dfrac{8.31}{62000}\ln8=3.111\times10^{-3}\,\mathrm{K^{-1}}$ . $T_2=321\,\mathrm{K}$ . The required temperature is higher than $295\,\mathrm{K}$ , consistent with a larger rate constant.	Use $\ln(k_2/k_1)=(E_a/R)(1/T_1-1/T_2)$ and convert $E_a$ to $62000\,\mathrm{J\,mol^{-1}}$ . Rearrangement gives $1/T_2=1/295-(8.31/62000)\ln8=0.003111118\,\mathrm{K^{-1}}$ . Taking the reciprocal gives $T_2=321.4278\ldots\,\mathrm{K}$ , which is $321\,\mathrm{K}$ to three significant figures. The positive activation energy means heating must increase $k$ , which checks the direction of the result.	4
05.1	$\dfrac{k_{cat}}{k_{uncat}}=\exp\left(\dfrac{E_{a,uncat}-E_{a,cat}}{RT}\right)$ . $\ln40.0=3.689$ . $E_{a,uncat}-E_{a,cat}=9.50\,\mathrm{kJ\,mol^{-1}}$ . $E_{a,cat}=68.5\,\mathrm{kJ\,mol^{-1}}$ .	Write the Arrhenius expression for each pathway at the same temperature. Dividing cancels the common Arrhenius constant and gives $k_{cat}/k_{uncat}=\exp[(E_{a,uncat}-E_{a,cat})/(RT)]$ . Therefore the reduction is $RT\ln40.0=8.31(310)(3.688879\ldots)=9502.92\,\mathrm{J\,mol^{-1}}=9.50\,\mathrm{kJ\,mol^{-1}}$ . Subtracting from $78.0\,\mathrm{kJ\,mol^{-1}}$ gives $68.5\,\mathrm{kJ\,mol^{-1}}$ .	4

3.1.9.2 · Determination of rate equation (A-level only)

Tier 1 · Easy

Mark scheme for 3.1.9.2 Tier 1 · Easy
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01.1	First order with respect to $X$ .	For $\text{rate}\propto[X]^m$ , the observation gives $2^m=2$ , so $m=1$ .	1
02.1	Measure the volume of carbon dioxide in a gas syringe at regular short time intervals. Keep a variable such as temperature constant between experiments.	Gas volume changes continuously as the reaction proceeds, so repeated gas-syringe readings produce a volume–time curve. A fair rate comparison requires other rate-affecting variables, especially temperature, to be held constant.	2

Tier 2 · Standard

Mark scheme for 3.1.9.2 Tier 2 · Standard
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01.1	$\text{rate}=k[A]^2$ $k=1.50\times10^{-2}\,\mathrm{dm^3\,mol^{-1}\,s^{-1}}$	From experiments 1 and 2, doubling $[A]$ multiplies the rate by $4$ , so the order in $A$ is $2$ . From experiments 2 and 3, doubling $[B]$ leaves the rate unchanged, so the order in $B$ is $0$ . Using experiment 2, $k=(6.00\times10^{-4})/(0.200)^2=1.50\times10^{-2}\,\mathrm{dm^3\,mol^{-1}\,s^{-1}}$ .	5
02.1	Second order with respect to $A$ and first order with respect to $B$ $\text{rate}=k[A]^2[B]$ $k=0.100\,\mathrm{dm^6\,mol^{-2}\,s^{-1}}$	From trials 1 to 2, $A$ doubles while $B$ halves and the rate doubles, so $2^m(0.5)^n=2$ . From trials 1 to 3, $A$ halves while $B$ quadruples and the rate is unchanged, so $(0.5)^m4^n=1$ . These relations give $m-n=1$ and $m=2n$ , hence $m=2$ and $n=1$ . Using trial 1, $k=(1.00\times10^{-4})/[(0.100)^2(0.100)]=0.100\,\mathrm{dm^6\,mol^{-2}\,s^{-1}}$ .	6
03.1	A plot of rate against $[X]^2$ gives a straight line through the origin. The reaction is second order with respect to $X$ . $k=0.200\,\mathrm{dm^3\,mol^{-1}\,s^{-1}}$	Calculate $\text{rate}/[X]^2$ for each result: $0.00200/0.0100=0.200$ , $0.00800/0.0400=0.200$ and $0.0180/0.0900=0.200$ . The constant ratio shows $\text{rate}=k[X]^2$ , so rate plotted against $[X]^2$ is linear and its gradient is $k=0.200\,\mathrm{dm^3\,mol^{-1}\,s^{-1}}$ .	4

Tier 3 · Hard

Mark scheme for 3.1.9.2 Tier 3 · Hard
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01.1	Mechanism II is consistent with the rate equation. Mechanism I predicts dependence on $[X][Y]$ , not $[X][Y]^2$ . For mechanism II, $[I]$ is proportional to $[X][Y]$ , so the slow-step rate $k'[I][Y]$ is proportional to $[X][Y]^2$ .	Write the concentration dependence of each proposed slow step. Mechanism I gives only one factor of $[Y]$ . In mechanism II the preceding fast equilibrium makes $[I]\propto[X][Y]$ ; substituting this into the slow-step expression gives $\text{rate}\propto[X][Y]^2$ , matching the experiment.	4
02.1	Second order with respect to $X$ $k=0.250\,\mathrm{dm^3\,mol^{-1}\,s^{-1}}$	Use the magnitudes of the negative tangent gradients as reaction rates. Halving $[X]$ from $0.0800$ to $0.0400\,\mathrm{mol\,dm^{-3}}$ reduces the rate by a factor of four, so the reaction is second order in $X$ . Thus $k=(1.60\times10^{-3})/(0.0800)^2=0.250\,\mathrm{dm^3\,mol^{-1}\,s^{-1}}$ ; the second point gives the same value.	5
03.1	$[X]$ decreases by $0.0162\,\mathrm{mol\,dm^{-3}}$ in every $45\,\mathrm{s}$ interval, so the concentration–time plot is a straight line with constant negative gradient. The reaction is zero order with respect to $X$ . $k=3.60\times10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$	Each $45\,\mathrm{s}$ interval decreases $[X]$ by $0.0162\,\mathrm{mol\,dm^{-3}}$ , so the data have a constant gradient and lie on a straight line. A constant rate while $[X]$ changes shows zero-order behaviour. The gradient is $(0.0612-0.1260)/(180-0)=-3.60\times10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ . For a zero-order reaction, $k$ is the magnitude of this gradient, so $k=3.60\times10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ .	6
04.1	The concentration halves in each successive $70\,\mathrm{s}$ interval. A constant halving time identifies first-order behaviour with respect to X. $k=9.90\times10^{-3}\,\mathrm{s^{-1}}$ . $\mathrm{rate}=k[X]$ . At the stated concentration, the rate is $2.97\times10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ .	The successive concentration ratios are all $1/2$ over equal $70\,\mathrm{s}$ intervals, which identifies first-order behaviour with respect to X. Use the magnitude of the tangent gradient as the instantaneous rate: for $\mathrm{rate}=k[X]$ , $k=(7.92\times10^{-4})/0.0800=9.90\times10^{-3}\,\mathrm{s^{-1}}$ . At $[X]=0.0300\,\mathrm{mol\,dm^{-3}}$ , the rate is $(9.90\times10^{-3})(0.0300)=2.97\times10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ .	5
05.1	The reaction is zero order with respect to A because changing $[A]$ by a factor of $2.50$ leaves the rate unchanged. The reaction is first order with respect to B because tripling $[B]$ triples the rate. $\mathrm{rate}=k[B]$ . $k=2.00\times10^{-3}\,\mathrm{s^{-1}}$ . The predicted rate is $4.40\times10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ .	Compare trials 1 and 2: only $[A]$ changes, by a factor of $2.50$ , while the rate is constant, so the order in A is zero. Compare trials 2 and 3: $[B]$ and the rate both triple, so the order in B is one. Thus $\mathrm{rate}=k[B]$ . Using trial 1, $k=(2.00\times10^{-4})/0.100=2.00\times10^{-3}\,\mathrm{s^{-1}}$ . The new value of $[A]$ has no effect, and the rate is $(2.00\times10^{-3})(0.220)=4.40\times10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ .	5

3.1.10 · Equilibrium constant Kp for homogeneous systems (A-level only)

Tier 1 · Easy

Mark scheme for 3.1.10 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$p_A=200\,\mathrm{kPa}$ $p_B=300\,\mathrm{kPa}$	The total amount is $5.00\,\mathrm{mol}$ . Hence $p_A=(2.00/5.00)(500)=200\,\mathrm{kPa}$ and $p_B=(3.00/5.00)(500)=300\,\mathrm{kPa}$ .	2
02.1	$\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}$ $\Delta n(\text{gas})=2-(1+3)=-2$ , so the units are $\mathrm{kPa^{-2}}$ .	For $\mathrm{N_2+3H_2\rightleftharpoons2NH_3}$ , the product pressure powers total $2$ and the reactant pressure powers total $4$ . The net pressure power is therefore $2-4=-2$ , giving $\mathrm{kPa^{-2}}$ . The sulfur dioxide equilibrium has $\Delta n=2-3=-1$ and would instead give $\mathrm{kPa^{-1}}$ .	3

Tier 2 · Standard

Mark scheme for 3.1.10 Tier 2 · Standard
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01.1	$K_p=152\,\mathrm{kPa}$	The total amount is $0.700\,\mathrm{mol}$ , so $p(\mathrm{N_2O_4})=(0.300/0.700)(200)=85.7\,\mathrm{kPa}$ and $p(\mathrm{NO_2})=(0.400/0.700)(200)=114.3\,\mathrm{kPa}$ . Then $K_p=p(\mathrm{NO_2})^2/p(\mathrm{N_2O_4})=(114.3)^2/85.7=152\,\mathrm{kPa}$ .	5
02.1	$p_A=150\,\mathrm{kPa}$ $p_B=200\,\mathrm{kPa}$ $K_p=267\,\mathrm{kPa}$	At equilibrium there are $0.600\,\mathrm{mol}$ of $A$ and $0.800\,\mathrm{mol}$ of $B$ , so the total is $1.400\,\mathrm{mol}$ . Hence $p_A=(0.600/1.400)(350)=150\,\mathrm{kPa}$ and $p_B=(0.800/1.400)(350)=200\,\mathrm{kPa}$ . Then $K_p=p_B^2/p_A=200^2/150=266.666\ldots\,\mathrm{kPa}$ , which is $267\,\mathrm{kPa}$ to three significant figures.	6
03.1	$p_B=200\,\mathrm{kPa}$ $P_{total}=250\,\mathrm{kPa}$	The equilibrium expression is $K_p=p_B/(p_A)^2$ . Therefore $p_B=K_p(p_A)^2=0.0800(50.0)^2=200\,\mathrm{kPa}$ . The total pressure is the sum of the two equilibrium partial pressures, $50.0+200=250\,\mathrm{kPa}$ .	4

Tier 3 · Hard

Mark scheme for 3.1.10 Tier 3 · Hard
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01.1	$K_p=9.00$ with no units Increasing pressure does not shift this equilibrium. $K_p$ is unchanged at constant temperature.	Formation of $1.20\,\mathrm{mol}$ of $\mathrm{HI}$ consumes $0.600\,\mathrm{mol}$ each of $\mathrm{H_2}$ and $\mathrm{I_2}$ , leaving $0.400\,\mathrm{mol}$ of each. The total remains $2.00\,\mathrm{mol}$ , so the partial pressures are $150$ , $50.0$ and $50.0\,\mathrm{kPa}$ for $\mathrm{HI}$ , $\mathrm{H_2}$ and $\mathrm{I_2}$ . Thus $K_p=150^2/(50.0\times50.0)=9.00$ . Both sides contain two moles of gas, so pressure causes no shift, and constant temperature means $K_p$ does not change.	6
02.1	$40.0\%$	Let the initial amount be $1$ and the dissociated fraction be $x$ . The equilibrium amounts are $1-x$ , $x$ and $x$ , with total $1+x$ . Therefore $p(\mathrm{PCl_5})=210(1-x)/(1+x)$ and each product pressure is $210x/(1+x)$ . Substitution gives $K_p=210x^2/(1-x^2)$ . Hence $40.0=210x^2/(1-x^2)$ , so $40.0=250x^2$ , $x=0.400$ and the percentage dissociation is $40.0\%$ .	6
03.1	The first mixture gives $K_p=30.0\,\mathrm{kPa}$ ; the reported second mixture gives $20.0\,\mathrm{kPa}$ , so it is inconsistent. The corrected second value is $p_B=49.0\,\mathrm{kPa}$ . At a fixed temperature, a pressure change can alter equilibrium position but does not change $K_p$ .	For this equation, $K_p=(p_B)^2/p_A$ . The first mixture gives $60.0^2/120=30.0\,\mathrm{kPa}$ , whereas the report gives $40.0^2/80.0=20.0\,\mathrm{kPa}$ . At the same temperature both must have the same $K_p$ . Hence $(p_B)^2/80.0=30.0$ , so $p_B=\sqrt{2400}=48.989\ldots\,\mathrm{kPa}$ , or $49.0\,\mathrm{kPa}$ . Pressure may shift the position of equilibrium, but only temperature changes the value of $K_p$ .	5
04.1	$K_p=8.00$ at $400\,\mathrm{K}$ . $K_p=0.640$ at $600\,\mathrm{K}$ . $K_p$ is dimensionless here because the two powers of pressure in the numerator and denominator cancel. The forward reaction is exothermic because increasing temperature decreases $K_p$ . A catalyst does not change $K_p$ because it increases the rates of the forward and reverse reactions equally, so equilibrium is reached faster without changing its position.	For $\mathrm{A+B\rightleftharpoons2C}$ , $K_p=(p_C)^2/(p_Ap_B)$ . Since $p_i=x_iP$ , $K_p=(x_CP)^2/[(x_AP)(x_BP)]=x_C^2/(x_Ax_B)$ : the total-pressure powers cancel because there are two gaseous moles on each side. Both mixtures contain $1.400\,\mathrm{mol}$ . At $400\,\mathrm{K}$ , $K_p=(0.800/1.400)^2/[(0.200/1.400)(0.400/1.400)]=8.00$ . At $600\,\mathrm{K}$ , $K_p=(0.400/1.400)^2/[(0.500/1.400)(0.500/1.400)]=0.640$ . Heating lowers $K_p$ , so it favours the endothermic reverse direction and the forward reaction is exothermic. A catalyst lowers the activation energy for both directions and changes neither equilibrium composition nor $K_p$ .	6
05.1	$K_p=\dfrac{x_CP}{(x_AP)(x_BP)^2}=\dfrac{x_C}{x_Ax_B^2P^2}$ . $P^2=5.00\times10^4\,\mathrm{kPa^2}$ . $P=224\,\mathrm{kPa}$ . $p_A=44.7\,\mathrm{kPa}$ and $p_B=112\,\mathrm{kPa}$ . $p_C=67.1\,\mathrm{kPa}$ .	Replace each partial pressure by $x_iP$ in $K_p=p_C/(p_Ap_B^2)$ . This gives $1.20\times10^{-4}=0.300/[0.200(0.500)^2P^2]$ . Therefore $P^2=0.300/[0.200(0.250)(1.20\times10^{-4})]=50000\,\mathrm{kPa^2}$ and $P=223.606\ldots\,\mathrm{kPa}$ , or $224\,\mathrm{kPa}$ . Multiplying by the mole fractions gives $44.721\ldots$ , $111.803\ldots$ and $67.082\ldots\,\mathrm{kPa}$ , which round to the stated values.	5

3.1.11.1 · Electrode potentials and cells (A-level only)

Tier 1 · Easy

Mark scheme for 3.1.11.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$E^\circ_{cell}=+1.10\,\mathrm{V}$ $\mathrm{Cu^{2+}}$ is reduced.	The copper couple has the more positive potential, so $\mathrm{Cu^{2+}}$ is reduced. Calculate $E^\circ_{cell}=0.34-(-0.76)=+1.10\,\mathrm{V}$ .	3
02.1	Neither member of the redox couple is a conducting solid that can act as an electrode. Platinum is an inert conductor that provides a surface for electron transfer without taking part in the reaction.	Identify that both iron species are aqueous. An inert solid conductor is therefore required to connect their electron transfer to the external circuit.	2

Tier 2 · Standard

Mark scheme for 3.1.11.1 Tier 2 · Standard
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01.1	$E^\circ_{cell}=+0.62\,\mathrm{V}$ $\mathrm{2Fe^{3+}+Sn^{2+}\rightarrow2Fe^{2+}+Sn^{4+}}$ $\mathrm{Pt(s)\|Sn^{2+}(aq),Sn^{4+}(aq)\|\|Fe^{3+}(aq),Fe^{2+}(aq)\|Pt(s)}$	$\mathrm{Fe^{3+}}$ is reduced because its potential is more positive; reverse the tin reduction half-equation so $\mathrm{Sn^{2+}}$ is oxidised. Balance two iron electrons against one tin half-equation to obtain the overall equation. The EMF is $0.77-0.15=+0.62\,\mathrm{V}$ , and inert platinum electrodes are required in both half-cells.	6
02.1	Connect the half-cells with a salt bridge and connect their electrodes to a high-resistance voltmeter, keeping both half-cells under standard conditions. The silver half-cell is the positive electrode and X is the negative electrode because electrons flow from X to silver. $E^\circ(\mathrm{X^{2+}/X})=-0.25\,\mathrm{V}$	Use a salt bridge to complete the ionic circuit and a high-resistance voltmeter to measure the open-circuit potential difference without drawing an appreciable current. Electron flow shows that silver is the positive reduction half-cell and X is the negative oxidation half-cell. Therefore $E^\circ_{cell}=E^\circ_{positive}-E^\circ_{negative}=0.80-E^\circ(\mathrm{X^{2+}/X})=1.05$ , so $E^\circ(\mathrm{X^{2+}/X})=0.80-1.05=-0.25\,\mathrm{V}$ .	6
03.1	$E^\circ_{cell}=+0.90\,\mathrm{V}$ $\mathrm{2A^{3+}+B\rightarrow2A^{2+}+B^{2+}}$ Electrode potentials are not multiplied when a half-equation is multiplied.	The more positive A couple is reduced, while B is oxidised by reversing its listed reduction equation. Multiply the A half-equation by two only to balance the two electrons: this does not change its potential. Therefore $E^\circ_{cell}=0.68-(-0.22)=+0.90\,\mathrm{V}$ . Adding the balanced half-equations gives $\mathrm{2A^{3+}+B\rightarrow2A^{2+}+B^{2+}}$ .	4

Tier 3 · Hard

Mark scheme for 3.1.11.1 Tier 3 · Hard
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01.1	$\mathrm{2Cu^+\rightarrow Cu^{2+}+Cu}$ $E^\circ_{cell}=+0.37\,\mathrm{V}$ Disproportionation is feasible under standard conditions.	One $\mathrm{Cu^+}$ ion is reduced by $\mathrm{Cu^++e^-\rightarrow Cu}$ , while another is oxidised by reversing $\mathrm{Cu^{2+}+e^-\rightarrow Cu^+}$ . Adding gives $\mathrm{2Cu^+\rightarrow Cu^{2+}+Cu}$ . The EMF is $0.52-0.15=+0.37\,\mathrm{V}$ , so the positive value supports feasibility.	5
02.1	Use the $\mathrm{Cl_2/Cl^-}$ and $\mathrm{Zn^{2+}/Zn}$ half-cells. $\mathrm{Zn(s)+Cl_2(g)\rightarrow Zn^{2+}(aq)+2Cl^-(aq)}$ $E^\circ_{cell}=+2.12\,\mathrm{V}$ $\mathrm{Zn(s)\|Zn^{2+}(aq)\|\|Cl^-(aq)\|Cl_2(g)\|Pt(s)}$	The largest separation is between the most positive reduction potential and the most negative reduction potential. Chlorine is reduced and zinc is oxidised, giving $E^\circ_{cell}=1.36-(-0.76)=+2.12\,\mathrm{V}$ . The oxidation half-cell is written on the left. The chlorine half-cell needs an inert platinum conductor because neither chlorine gas nor chloride solution is a conducting solid.	7
03.1	$\mathrm{2M^{3+}+M\rightarrow3M^{2+}}$ $E^\circ_{cell}=+0.80\,\mathrm{V}$ , so the reaction is thermodynamically feasible under standard conditions. $\mathrm{M(s)\|M^{2+}(aq)\|\|M^{3+}(aq),M^{2+}(aq)\|Pt(s)}$	$\mathrm{M^{3+}}$ is reduced by $\mathrm{M^{3+}+e^-\rightarrow M^{2+}}$ . Reverse the other listed half-equation to oxidise the metal: $\mathrm{M\rightarrow M^{2+}+2e^-}$ . Multiply the first equation by two, add and cancel electrons to obtain $\mathrm{2M^{3+}+M\rightarrow3M^{2+}}$ . The EMF is $0.70-(-0.10)=+0.80\,\mathrm{V}$ . Oxidation is written on the left of the cell representation, and the all-aqueous right half-cell needs inert platinum.	6
04.1	$\mathrm{Y+2X^+\rightarrow Y^{2+}+2X}$ $E^\circ_{cell}=+1.15\,\mathrm{V}$ , so Y reacts with $\mathrm{X^+}$ under standard conditions. $\mathrm{X^+}$ is the oxidising agent and Y is the reducing agent. For the hypothetical direction $\mathrm{Y+Z^{2+}\rightarrow Y^{2+}+Z}$ , $E^\circ_{cell}=-0.60\,\mathrm{V}$ . Y therefore does not react with $\mathrm{Z^{2+}}$ in that direction under standard conditions.	Y is oxidised by $\mathrm{Y\rightarrow Y^{2+}+2e^-}$ and each $\mathrm{X^+}$ gains one electron, so charge and electrons balance in $\mathrm{Y+2X^+\rightarrow Y^{2+}+2X}$ . Its standard EMF is $0.85-(-0.30)=+1.15\,\mathrm{V}$ , supporting the stated direction. If $\mathrm{Z^{2+}}$ were reduced by Y, the balanced equation would be $\mathrm{Y+Z^{2+}\rightarrow Y^{2+}+Z}$ , but $E^\circ_{cell}=-0.90-(-0.30)=-0.60\,\mathrm{V}$ , so that hypothetical direction is not feasible under standard conditions.	5
05.1	$E^\circ_{cell}=0.80-(-0.76)=+1.56\,\mathrm{V}$ . Raising $[\mathrm{Ag^+}]$ makes the $\mathrm{Ag^+/Ag}$ reduction potential more positive, so the EMF rises. Raising $[\mathrm{Zn^{2+}}]$ makes the $\mathrm{Zn^{2+}/Zn}$ reduction potential more positive, so the EMF falls because this negative-electrode potential is subtracted.	Silver has the more positive reduction potential, so it is the positive electrode and $E^\circ_{cell}=E^\circ(\mathrm{Ag^+/Ag})-E^\circ(\mathrm{Zn^{2+}/Zn})=+1.56\,\mathrm{V}$ . A higher concentration of the oxidised member of a half-cell makes its reduction more favourable. Raising $[\mathrm{Ag^+}]$ therefore raises the positive silver-electrode potential and increases their difference. Raising $[\mathrm{Zn^{2+}}]$ raises the zinc-electrode reduction potential; since this is the negative-electrode term in $E_{cell}=E_{positive}-E_{negative}$ , the potential difference decreases.	5

3.1.11.2 · Commercial applications of electrochemical cells (A-level only)

Tier 1 · Easy

Mark scheme for 3.1.11.2 Tier 1 · Easy
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01.1	A rechargeable cell has its electrode reactions reversed by an applied potential. A fuel cell is supplied continuously with fuel and oxidant and does not need electrical recharging.	Contrast how each cell is restored to operation: a rechargeable cell reverses its chemistry electrically, whereas a fuel cell continues when fresh reactants are supplied.	2
02.1	$\mathrm{Li}$ is oxidised. $\mathrm{Li^+[CoO_2]^-}$ is formed at the positive electrode.	Lithium loses an electron, so it is oxidised at the negative electrode. The positive-electrode equation consumes an electron and shows $\mathrm{Li^+[CoO_2]^-}$ as its product.	2

Tier 2 · Standard

Mark scheme for 3.1.11.2 Tier 2 · Standard
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01.1	$\mathrm{2H_2+O_2\rightarrow2H_2O}$ Oxidation releases electrons at the negative electrode and reduction consumes them at the positive electrode, so electrons flow through the external circuit.	Multiply the hydrogen half-equation by $2$ , add the two equations, then cancel $4e^-$ , $4\mathrm{OH^-}$ and two of the four water molecules. The separated electron release and consumption force electrons through the external circuit.	4
02.1	$\mathrm{Li+CoO_2\rightarrow Li^+[CoO_2]^-}$ Oxidation releases electrons at the lithium electrode and reduction consumes them at the positive electrode, so electrons travel through the external circuit between the separated electrodes.	Add the two electrode reactions and cancel one electron and one $\mathrm{Li^+}$ ion. Because electron release and electron consumption occur at different electrodes, the electrons must pass through the external conductor.	4
03.1	A positive standard EMF predicts that the cell reaction is feasible in the discharge direction under standard conditions; it does not show that the reaction is reversible. Rechargeability requires evidence that an external potential can drive the electrode reactions backwards and regenerate the original electrode substances. The products must remain available in a suitable arrangement, and material loss or irreversible side reactions must not prevent regeneration over repeated cycles.	Standard EMF is calculated for the spontaneous discharge reaction, so its sign supplies thermodynamic information about that direction only. Classification as rechargeable needs separate chemical evidence: applying an external potential must reverse both electrode processes and restore the starting materials. If products escape, electrode material is lost or competing reactions occur, a positive discharge EMF cannot make the cell rechargeable.	4

Tier 3 · Hard

Mark scheme for 3.1.11.2 Tier 3 · Hard
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01.1	$E_{cell}=+1.23\,\mathrm{V}$ Benefit: water is the only product at the point of use. Limitation: hydrogen production may require energy and may release carbon dioxide, depending on the source. Limitation: hydrogen is difficult to store and transport safely because it is flammable and has low volumetric energy density.	Calculate $E_{cell}=0.40-(-0.83)=+1.23\,\mathrm{V}$ . For the evaluation, distinguish the clean point-of-use reaction from whole-system issues such as how hydrogen is produced and the practical risks of storing a flammable gas.	5
02.1	Discharge reduction: $\mathrm{2P^{3+}+2e^-\rightarrow2P^{2+}}$ Discharge oxidation: $\mathrm{Q\rightarrow Q^{2+}+2e^-}$ Discharge overall: $\mathrm{2P^{3+}+Q\rightarrow2P^{2+}+Q^{2+}}$ $E^\circ_{cell}=+1.00\,\mathrm{V}$ Charging overall: $\mathrm{2P^{2+}+Q^{2+}\rightarrow2P^{3+}+Q}$	During spontaneous discharge, the more positive $P$ couple is reduced and the $Q$ reduction equation is reversed. Balance two electrons before addition. The EMF is $0.60-(-0.40)=+1.00\,\mathrm{V}$ . An external potential recharges the cell by driving both electrode reactions in the reverse directions.	7
03.1	Recommend the rechargeable cell because the available renewable electricity can reverse its electrode reactions overnight without a new fuel-distribution system. Nightly return to the depot provides regular charging time and avoids the need to store and transfer compressed flammable hydrogen. Rechargeable cells have a finite cycle life and their capacity may fall after repeated charge–discharge cycles. Their manufacture and disposal may use scarce or hazardous materials, so sourcing and recycling still require management.	Base the choice on the whole operating system. The fleet's schedule and existing electricity supply fit electrical recharging, while a fuel cell would require hydrogen manufacture, compression, storage and safe transfer that the depot lacks. The point-of-use product of a hydrogen–oxygen cell is water, but that does not remove the infrastructure constraint. A balanced recommendation also recognises the rechargeable cell's limited lifetime and the environmental consequences of its materials and disposal.	6
04.1	$6.00\,\mathrm{mol}$ of $\mathrm{H_2}$ is consumed. The hydrogen mass is $12.0\,\mathrm{g}$ . $3.00\,\mathrm{mol}$ of $\mathrm{O_2}$ is consumed. The oxygen mass is $96.0\,\mathrm{g}$ . $6.00\,\mathrm{mol}$ of $\mathrm{H_2O}$ is formed.	Hydrogen supplies two moles of electrons per mole, so its amount is $12.0/2=6.00\,\mathrm{mol}$ and its mass is $6.00(2.00)=12.0\,\mathrm{g}$ . Oxygen consumes four moles of electrons per mole, so its amount is $12.0/4=3.00\,\mathrm{mol}$ and its mass is $3.00(32.0)=96.0\,\mathrm{g}$ . Balancing the electron transfers gives the overall ratio $\mathrm{2H_2+O_2\rightarrow2H_2O}$ , so $6.00\,\mathrm{mol}$ of hydrogen forms $6.00\,\mathrm{mol}$ of water. The calculation also shows why continued operation requires a continuing gas supply rather than electrical recharging.	5
05.1	$\mathrm{Zn\rightarrow Zn^{2+}+2e^-}$ $\mathrm{2MnO_2+H_2O+2e^-\rightarrow Mn_2O_3+2OH^-}$ $\mathrm{Zn+2MnO_2+H_2O\rightarrow Zn^{2+}+Mn_2O_3+2OH^-}$ $E^\circ_{cell}=0.15-(-0.76)=+0.91\,\mathrm{V}$ . Oxidation at the zinc electrode releases electrons, which pass through the external circuit to the manganese oxide electrode and are consumed by reduction; ion movement through the electrolyte maintains charge balance.	The manganese dioxide couple has the more positive reduction potential, so its listed equation proceeds as reduction. Reverse the zinc reduction equation to give oxidation and two electrons. Adding the two electrode equations cancels the electrons and gives the stated overall equation; atoms and total charge are balanced. The EMF is $E^\circ_{positive}-E^\circ_{negative}=0.15-(-0.76)=+0.91\,\mathrm{V}$ . The separated redox processes force released electrons through the external circuit, producing current while electrolyte ions preserve electrical neutrality.	5

3.1.12.1 · Brønsted–Lowry acid–base equilibria in aqueous solution (A-level only)

Tier 1 · Easy

Mark scheme for 3.1.12.1 Tier 1 · Easy
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01.1	$\mathrm{H_2O}$ is the acid. $\mathrm{NH_3}$ is the base.	$\mathrm{H_2O}$ donates $\mathrm{H^+}$ to become $\mathrm{OH^-}$ , so it is the acid. $\mathrm{NH_3}$ accepts that proton to become $\mathrm{NH_4^+}$ , so it is the base.	2
02.1	$\mathrm{HCO_3^-}$ $\mathrm{OH^-}$	A conjugate acid is formed by adding one proton, so $\mathrm{CO_3^{2-}}$ becomes $\mathrm{HCO_3^-}$ . A conjugate base is formed by removing one proton, so $\mathrm{H_2O}$ becomes $\mathrm{OH^-}$ .	2

Tier 2 · Standard

Mark scheme for 3.1.12.1 Tier 2 · Standard
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01.1	$\mathrm{HSO_4^-/SO_4^{2-}}$ $\mathrm{H_3O^+/H_2O}$	$\mathrm{HSO_4^-}$ loses one proton to form $\mathrm{SO_4^{2-}}$ , so those form a pair. $\mathrm{H_2O}$ gains that proton to form $\mathrm{H_3O^+}$ , giving the second pair.	3
02.1	$\mathrm{H_2SO_3+NH_3\rightleftharpoons HSO_3^-+NH_4^+}$ $\mathrm{H_2SO_3}$ is the acid and $\mathrm{NH_3}$ is the base. The pairs are $\mathrm{H_2SO_3/HSO_3^-}$ and $\mathrm{NH_4^+/NH_3}$ .	$\mathrm{H_2SO_3}$ donates one proton to form $\mathrm{HSO_3^-}$ , while $\mathrm{NH_3}$ accepts it to form $\mathrm{NH_4^+}$ . Each conjugate pair differs by exactly one proton, and the total charge is zero on both sides.	5
03.1	Water donates the proton here, so its conjugate is $\mathrm{OH^-}$ ; $\mathrm{H_3O^+}$ is water's conjugate acid and cannot appear as the product of water acting as an acid. (Also accept: the proposed equation does not balance for hydrogen or for charge.) $\mathrm{H_2O+S^{2-}\rightleftharpoons OH^-+HS^-}$ The pairs are $\mathrm{H_2O/OH^-}$ and $\mathrm{HS^-/S^{2-}}$ .	Decide which species donates the proton: water gives one proton to sulfide, so water must form its conjugate base, $\mathrm{OH^-}$ . Writing $\mathrm{H_3O^+}$ treats water as if it had accepted a proton instead. The corrected equation balances atoms and has total charge $-2$ on each side, with pairs differing by exactly one proton.	4

Tier 3 · Hard

Mark scheme for 3.1.12.1 Tier 3 · Hard
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01.1	$\mathrm{H_2PO_4^-+H_2O\rightleftharpoons HPO_4^{2-}+H_3O^+}$ ; $\mathrm{H_2PO_4^-}$ is an acid. $\mathrm{H_2PO_4^-+H_2O\rightleftharpoons H_3PO_4+OH^-}$ ; $\mathrm{H_2PO_4^-}$ is a base.	To act as an acid, $\mathrm{H_2PO_4^-}$ donates $\mathrm{H^+}$ to water and forms $\mathrm{HPO_4^{2-}}$ . To act as a base, it accepts $\mathrm{H^+}$ from water and forms $\mathrm{H_3PO_4}$ . Both equations balance atoms and charge.	4
02.1	The two stated ions differ by two protons, so they are not a conjugate pair. The missing species is $\mathrm{HPO_4^{2-}}$ . Correct pairs are $\mathrm{H_2PO_4^-/HPO_4^{2-}}$ and $\mathrm{HPO_4^{2-}/PO_4^{3-}}$ .	Conjugate acid–base partners must differ by one $\mathrm{H^+}$ only. Removing one proton at a time gives $\mathrm{H_2PO_4^-\rightarrow HPO_4^{2-}\rightarrow PO_4^{3-}}$ , producing the two adjacent pairs.	4
03.1	The pairs are $\mathrm{H_3A/H_2A^-}$ , $\mathrm{H_2A^-/HA^{2-}}$ and $\mathrm{HA^{2-}/A^{3-}}$ . $\mathrm{H_2A^-}$ and $\mathrm{HA^{2-}}$ are amphoteric. $\mathrm{HA^{2-}+H_2O\rightleftharpoons H_2A^-+OH^-}$	Adjacent formulas in the sequence differ by one proton and therefore form conjugate pairs. Each middle species can either accept a proton to move left or donate a proton to move right, so both are amphoteric. To make $\mathrm{HA^{2-}}$ act as a base, let it accept a proton from water; water then becomes $\mathrm{OH^-}$ . The equation has charge $-2$ on each side and balances all atoms.	5

3.1.12.2 · Definition and determination of pH (A-level only)

Tier 1 · Easy

Mark scheme for 3.1.12.2 Tier 1 · Easy
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01.1	$\mathrm{pH}=2.49$	$\mathrm{pH}=-\log_{10}(3.2\times10^{-3})=2.49485$ , which is $2.49$ because the concentration has two significant figures.	2
02.1	$100:1$	The pH difference is $2.00$ . Since each pH unit represents a tenfold change in $[H^+]$ , the ratio is $10^{2.00}:1=100:1$ ; X has the larger hydrogen-ion concentration.	2

Tier 2 · Standard

Mark scheme for 3.1.12.2 Tier 2 · Standard
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01.1	$[H^+]=2.1\times10^{-4}\,\mathrm{mol\,dm^{-3}}$	$[H^+]=10^{-3.68}=2.089\times10^{-4}\,\mathrm{mol\,dm^{-3}}$ . Two decimal places in the pH correspond to two significant figures in the concentration, giving $2.1\times10^{-4}\,\mathrm{mol\,dm^{-3}}$ .	2
02.1	Dilution factor $=14.1$	For complete dissociation, the concentration ratio equals the hydrogen-ion concentration ratio. The dilution factor is $10^{2.45-1.30}=10^{1.15}=14.125\ldots$ , which is $14.1$ to three significant figures.	3
03.1	$[\mathrm{H_2X}]=8.9\times10^{-3}\,\mathrm{mol\,dm^{-3}}$ $n(\mathrm{H_2X})=1.3\times10^{-3}\,\mathrm{mol}$	$[H^+]=10^{-1.75}=0.0177828\,\mathrm{mol\,dm^{-3}}$ . Complete dissociation produces two moles of hydrogen ions per mole of acid, so $[\mathrm{H_2X}]=0.0177828/2=0.00889140\,\mathrm{mol\,dm^{-3}}$ . The amount in $0.150\,\mathrm{dm^3}$ is $0.00889140(0.150)=0.00133371\,\mathrm{mol}$ . The pH has two decimal places, so the answers are $8.9\times10^{-3}\,\mathrm{mol\,dm^{-3}}$ and $1.3\times10^{-3}\,\mathrm{mol}$ to two significant figures.	4

Tier 3 · Hard

Mark scheme for 3.1.12.2 Tier 3 · Hard
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01.1	$\mathrm{pH}=0.975$	Both acids are strong and monoprotic. Their amounts of $\mathrm{H^+}$ are $0.145(0.0275)=0.0039875\,\mathrm{mol}$ and $0.0730(0.0325)=0.0023725\,\mathrm{mol}$ . In the total volume $0.0600\,\mathrm{dm^3}$ , $[H^+]=0.006360/0.0600=0.106\,\mathrm{mol\,dm^{-3}}$ . Therefore $\mathrm{pH}=-\log_{10}(0.106)=0.975$ .	5
02.1	$10.5\,\mathrm{cm^3}$	The target concentration is $[H^+]=10^{-2.20}=0.00630957\,\mathrm{mol\,dm^{-3}}$ . The target amount in $0.250\,\mathrm{dm^3}$ is $0.00630957(0.250)=0.00157739\,\mathrm{mol}$ . Therefore the stock volume is $0.00157739/0.150=0.01051596\,\mathrm{dm^3}=10.51596\,\mathrm{cm^3}$ , which is $10.5\,\mathrm{cm^3}$ to three significant figures.	5
03.1	$[H^+]=0.102\,\mathrm{mol\,dm^{-3}}$ $\mathrm{pH}=0.991$	One cubic decimetre of solution has mass $1.02(1000)=1020\,\mathrm{g}$ . The acid mass is $0.00365(1020)=3.723\,\mathrm{g}$ , so its amount is $3.723/36.5=0.102\,\mathrm{mol}$ . The acid is strong and monoprotic, hence $[H^+]=0.102\,\mathrm{mol\,dm^{-3}}$ . Finally $\mathrm{pH}=-\log_{10}(0.102)=0.991399\ldots$ , giving $0.991$ .	5
04.1	$[H^+]_{mixture}=0.100\,\mathrm{mol\,dm^{-3}}$ . The mixture contains $5.00\times10^{-3}\,\mathrm{mol}$ of hydrogen ions. The hydrochloric acid contributes $3.60\times10^{-3}\,\mathrm{mol}$ of hydrogen ions. The unknown acid contributes $1.40\times10^{-3}\,\mathrm{mol}$ of hydrogen ions. The unknown acid concentration is $0.0467\,\mathrm{mol\,dm^{-3}}$ .	$[H^+]=10^{-1.000}=0.100\,\mathrm{mol\,dm^{-3}}$ . The total volume is exactly $0.0500\,\mathrm{dm^3}$ , so the total hydrogen-ion amount is $0.100(0.0500)=0.00500\,\mathrm{mol}$ . Hydrochloric acid supplies $0.180(0.0200)=0.00360\,\mathrm{mol}$ , leaving $0.00500-0.00360=0.00140\,\mathrm{mol}$ from the unknown acid. Complete monoprotic dissociation makes this the acid amount, so its concentration is $0.00140/0.0300=0.046666\ldots\,\mathrm{mol\,dm^{-3}}$ , or $0.0467\,\mathrm{mol\,dm^{-3}}$ to three significant figures.	5
05.1	The acid amount is $2.00\times10^{-2}\,\mathrm{mol}$ . The acid concentration is $4.00\times10^{-2}\,\mathrm{mol\,dm^{-3}}$ . $[H^+]=8.00\times10^{-2}\,\mathrm{mol\,dm^{-3}}$ . $n=2$ .	The amount of acid is $1.46/73.0=0.0200\,\mathrm{mol}$ , giving concentration $0.0200/0.500=0.0400\,\mathrm{mol\,dm^{-3}}$ . The measured pH gives $[H^+]=10^{-1.097}=0.07998\ldots\,\mathrm{mol\,dm^{-3}}$ , which is $0.0800\,\mathrm{mol\,dm^{-3}}$ to three significant figures. Since complete dissociation gives $[H^+]=n[\mathrm{H_nA}]$ , $n=0.0800/0.0400=2$ .	4

3.1.12.3 · The ionic product of water, Kw (A-level only)

Tier 1 · Easy

Mark scheme for 3.1.12.3 Tier 1 · Easy
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01.1	$\mathrm{pH}=12.398$	$\mathrm{NaOH}$ dissociates completely, so $[OH^-]=0.0250\,\mathrm{mol\,dm^{-3}}$ . Then $[H^+]=(1.00\times10^{-14})/0.0250=4.00\times10^{-13}\,\mathrm{mol\,dm^{-3}}$ , so $\mathrm{pH}=12.398$ .	3
02.1	$K_w=[H^+][OH^-]$ The concentration of pure liquid water is effectively constant and is incorporated into the value of $K_w$ .	Start from the equilibrium expression and combine the constant concentration of the pure liquid reactant with the equilibrium constant. Only the variable aqueous ion concentrations remain explicit.	3

Tier 2 · Standard

Mark scheme for 3.1.12.3 Tier 2 · Standard
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01.1	$\mathrm{pH}=11.845$	Each $\mathrm{Ba(OH)_2}$ gives two $\mathrm{OH^-}$ ions, so $[OH^-]=2(0.00350)=0.00700\,\mathrm{mol\,dm^{-3}}$ . Then $[H^+]=(1.00\times10^{-14})/0.00700=1.43\times10^{-12}\,\mathrm{mol\,dm^{-3}}$ , and $\mathrm{pH}=11.845$ .	4
02.1	$2.0\times10^{-2}\,\mathrm{mol\,dm^{-3}}$	$[H^+]=10^{-12.30}=5.01187\times10^{-13}\,\mathrm{mol\,dm^{-3}}$ . Hence $[OH^-]=(1.00\times10^{-14})/(5.01187\times10^{-13})=0.0199526\,\mathrm{mol\,dm^{-3}}$ . Complete $1{:}1$ dissociation makes the sodium hydroxide concentration equal to $[OH^-]$ , giving $2.0\times10^{-2}\,\mathrm{mol\,dm^{-3}}$ to two significant figures.	4
03.1	$K_w=1.4\times10^{-13}\,\mathrm{mol^2\,dm^{-6}}$ The temperature is above $298\,\mathrm{K}$ because $K_w$ is greater than $1.00\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}$ .	$[H^+]=10^{-6.42}=3.80189\times10^{-7}\,\mathrm{mol\,dm^{-3}}$ . In neutral water $[OH^-]=[H^+]$ , so $K_w=(3.80189\times10^{-7})^2=1.44544\times10^{-13}\,\mathrm{mol^2\,dm^{-6}}$ , which is $1.4\times10^{-13}\,\mathrm{mol^2\,dm^{-6}}$ to two significant figures. This is greater than the value $1.00\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}$ at $298\,\mathrm{K}$ , so the stated temperature is above $298\,\mathrm{K}$ .	4

Tier 3 · Hard

Mark scheme for 3.1.12.3 Tier 3 · Hard
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01.1	Neutral water: $\mathrm{pH}=6.699$ $0.0200\,\mathrm{mol\,dm^{-3}}$ $\mathrm{KOH}$ : $\mathrm{pH}=11.699$	For neutral water, $[H^+]=\sqrt{4.00\times10^{-14}}=2.00\times10^{-7}\,\mathrm{mol\,dm^{-3}}$ , so $\mathrm{pH}=6.699$ . For the strong base, $[OH^-]=0.0200\,\mathrm{mol\,dm^{-3}}$ and $[H^+]=(4.00\times10^{-14})/0.0200=2.00\times10^{-12}\,\mathrm{mol\,dm^{-3}}$ , giving $\mathrm{pH}=11.699$ .	5
02.1	At $298\,\mathrm{K}$ , neutral pH $=7.00$ . At $323\,\mathrm{K}$ , neutral pH $=6.63$ . The water is neutral because $[H^+]=[OH^-]$ , not because its pH must equal $7$ . Water dissociation is endothermic because increasing temperature increases $K_w$ and shifts the dissociation equilibrium towards the ions.	For neutral water, $[H^+]=[OH^-]=\sqrt{K_w}$ . At $298\,\mathrm{K}$ , $[H^+]=\sqrt{1.00\times10^{-14}}=1.00\times10^{-7}\,\mathrm{mol\,dm^{-3}}$ , so pH $=7.00$ . At $323\,\mathrm{K}$ , $[H^+]=\sqrt{5.48\times10^{-14}}=2.340939982\times10^{-7}\,\mathrm{mol\,dm^{-3}}$ , so pH $=6.630609721$ , which is $6.63$ to two decimal places. Neutrality still means equal hydrogen-ion and hydroxide-ion concentrations. Since heating increases $K_w$ , it favours the ions, so the forward dissociation is endothermic.	6
03.1	$n=2$ , so the hydroxide is $\mathrm{M(OH)_2}$ . After dilution, $\mathrm{pH}=11.079$ .	The measured pH gives $[H^+]=10^{-11.681}=2.08449\times10^{-12}\,\mathrm{mol\,dm^{-3}}$ . Hence $[OH^-]=K_w/[H^+]=0.0119933\,\mathrm{mol\,dm^{-3}}$ . The ratio $[OH^-]/[\mathrm{M(OH)_n}]=0.0119933/0.00600=1.999$ , so $n=2$ . Fourfold dilution makes $[OH^-]=0.0119933/4=0.00299833\,\mathrm{mol\,dm^{-3}}$ . Then $[H^+]=(2.50\times10^{-14})/0.00299833=8.338\times10^{-12}\,\mathrm{mol\,dm^{-3}}$ , giving $\mathrm{pH}=11.07894\ldots$ , or $11.079$ .	6
04.1	Sodium hydroxide supplies $2.00\times10^{-3}\,\mathrm{mol}$ of $\mathrm{OH^-}$ . Barium hydroxide supplies $1.50\times10^{-3}\,\mathrm{mol}$ of $\mathrm{OH^-}$ . $[OH^-]=0.0875\,\mathrm{mol\,dm^{-3}}$ in the mixture. $[H^+]=2.86\times10^{-13}\,\mathrm{mol\,dm^{-3}}$ . $\mathrm{pH}=12.544$ .	The sodium hydroxide contributes $0.0800(0.0250)=0.00200\,\mathrm{mol}$ of hydroxide ions. Each mole of $\mathrm{Ba(OH)_2}$ supplies two moles of hydroxide ions, giving $2(0.0500)(0.0150)=0.00150\,\mathrm{mol}$ . The total is $0.00350\,\mathrm{mol}$ in $0.0400\,\mathrm{dm^3}$ , so $[OH^-]=0.0875\,\mathrm{mol\,dm^{-3}}$ . Then $[H^+]=(2.50\times10^{-14})/0.0875=2.85714\ldots\times10^{-13}\,\mathrm{mol\,dm^{-3}}$ and pH $=12.544068\ldots$ , reported as $12.544$ because $[H^+]$ has three significant figures.	5
05.1	$[OH^-]=0.00500\,\mathrm{mol\,dm^{-3}}$ . $[H^+]=1.28\times10^{-11}\,\mathrm{mol\,dm^{-3}}$ . $K_w=6.40\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}$ . The temperature is above $298\,\mathrm{K}$ because this $K_w$ is greater than $1.00\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}$ .	Complete dissociation supplies two hydroxide ions per formula unit, so $[OH^-]=2(0.00250)=0.00500\,\mathrm{mol\,dm^{-3}}$ . From the measured pH, $[H^+]=10^{-10.893}=1.27938\ldots\times10^{-11}\,\mathrm{mol\,dm^{-3}}$ , which is $1.28\times10^{-11}\,\mathrm{mol\,dm^{-3}}$ . Therefore $K_w=[H^+][OH^-]=(1.27938\ldots\times10^{-11})(0.00500)=6.39691\ldots\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}$ , or $6.40\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}$ . Since water dissociation is endothermic and this exceeds the value at $298\,\mathrm{K}$ , the unknown temperature is higher.	4

3.1.12.4 · Weak acids and bases Ka for weak acids (A-level only)

Tier 1 · Easy

Mark scheme for 3.1.12.4 Tier 1 · Easy
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01.1	$pK_a=4.759$	$pK_a=-\log_{10}(1.74\times10^{-5})=4.759$ .	2
02.1	P is the stronger acid. Its $K_a$ is $7.9$ times that of Q.	The smaller $pK_a$ corresponds to the larger $K_a$ . The difference is $4.80-3.90=0.90$ , so $K_a(P)/K_a(Q)=10^{0.90}=7.943\ldots$ , which is $7.9$ to two significant figures.	3

Tier 2 · Standard

Mark scheme for 3.1.12.4 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$\mathrm{pH}=2.512$	$[H^+]\approx\sqrt{K_a[HA]}=\sqrt{(6.30\times10^{-5})(0.150)}=3.074\times10^{-3}\,\mathrm{mol\,dm^{-3}}$ . Therefore $\mathrm{pH}=-\log_{10}(3.074\times10^{-3})=2.512$ . The dissociation is about $2.05\%$ , so the approximation is reasonable.	4
02.1	$0.0995\,\mathrm{mol\,dm^{-3}}$	$[H^+]=10^{-2.70}=1.99526\times10^{-3}\,\mathrm{mol\,dm^{-3}}$ . From $[H^+]\approx\sqrt{K_ac}$ , $c=[H^+]^2/K_a=(1.99526\times10^{-3})^2/(4.00\times10^{-5})=0.0995268\,\mathrm{mol\,dm^{-3}}$ , which is $0.0995\,\mathrm{mol\,dm^{-3}}$ to three significant figures. The estimated dissociation is about $2.0\%$ , consistent with the approximation.	4
03.1	$[\mathrm{H^+}]=\sqrt{5.60\times10^{-5}\times0.200}=3.35\times10^{-3}\,\mathrm{mol\,dm^{-3}}$ Percentage dissociated $=(3.34664\times10^{-3}/0.200)\times100=1.67\%$	With small dissociation, $[\mathrm{H^+}]=\sqrt{K_a c}=\sqrt{5.60\times10^{-5}\times0.200}=3.34664\times10^{-3}\,\mathrm{mol\,dm^{-3}}$ . The fraction dissociated is $[\mathrm{H^+}]/c=3.34664\times10^{-3}/0.200=0.0167$ , so $1.67\%$ of the acid molecules are dissociated — consistent with the approximation that dissociation is small.	3

Tier 3 · Hard

Mark scheme for 3.1.12.4 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$K_a=2.1\times10^{-4}\,\mathrm{mol\,dm^{-3}}$ $pK_a=3.68$	$[H^+]=10^{-2.40}=3.981\times10^{-3}\,\mathrm{mol\,dm^{-3}}$ , so $[A^-]$ has the same value and $[HA]=0.0800-0.003981=0.076019\,\mathrm{mol\,dm^{-3}}$ . Thus $K_a=(3.981\times10^{-3})^2/0.076019=2.085\times10^{-4}\,\mathrm{mol\,dm^{-3}}$ . Using the precision of the pH data gives $K_a=2.1\times10^{-4}\,\mathrm{mol\,dm^{-3}}$ and $pK_a=3.68$ .	5
02.1	The approximation predicts $x=0.00200\,\mathrm{mol\,dm^{-3}}$ , which is $20.0\%$ of the initial concentration and is not small. $\mathrm{pH}=2.742$	The approximation gives $x=\sqrt{(4.00\times10^{-4})(0.0100)}=0.00200\,\mathrm{mol\,dm^{-3}}$ , or $20.0\%$ dissociation, so it is unsuitable. Rearrangement of the exact expression gives $x^2+(4.00\times10^{-4})x-4.00\times10^{-6}=0$ . The positive root is $x=0.00180998\,\mathrm{mol\,dm^{-3}}$ , so $\mathrm{pH}=-\log_{10}(0.00180998)=2.74233$ , which is $2.742$ to three decimal places.	6
03.1	$[H^+]=[A^-]=3.16\times10^{-3}\,\mathrm{mol\,dm^{-3}}$ $[HA]_{eq}=0.625\,\mathrm{mol\,dm^{-3}}$ and the initial concentration is $0.628\,\mathrm{mol\,dm^{-3}}$ . $M_r=97.1$	$[H^+]=10^{-2.500}=0.00316228\,\mathrm{mol\,dm^{-3}}$ , and $[A^-]$ is equal to this. From $K_a=[H^+][A^-]/[HA]$ , $[HA]_{eq}=(0.00316228)^2/(1.60\times10^{-5})=0.625\,\mathrm{mol\,dm^{-3}}$ . The initial concentration includes the dissociated acid, so it is $0.625+0.00316228=0.628162\,\mathrm{mol\,dm^{-3}}$ . Therefore $M_r=61.0/0.628162=97.108\ldots$ , which is $97.1$ to three significant figures.	6
04.1	Assume the amount dissociated is small, so $[HA]_{eq}\approx c$ and $[H^+]\approx\sqrt{K_ac}$ . Before dilution, $[H^+]=1.118\times10^{-3}\,\mathrm{mol\,dm^{-3}}$ , pH $=2.952$ and dissociation $=2.24\%$ . After tenfold dilution, $[H^+]=3.536\times10^{-4}\,\mathrm{mol\,dm^{-3}}$ , pH $=3.452$ and dissociation $=7.07\%$ . Dilution shifts the dissociation equilibrium towards the ions, so a larger percentage of acid molecules dissociates even though $[H^+]$ falls.	For $\mathrm{HA\rightleftharpoons H^++A^-}$ , let $[H^+]=x$ and use the stated small-dissociation approximation $c-x\approx c$ , giving $K_a\approx x^2/c$ and $x\approx\sqrt{K_ac}$ . Initially, $x=\sqrt{(2.50\times10^{-5})(0.0500)}=1.11803\ldots\times10^{-3}\,\mathrm{mol\,dm^{-3}}$ , so pH $=2.95154\ldots=2.952$ and percentage dissociation $=100x/0.0500=2.236\ldots\%=2.24\%$ . After tenfold dilution, $c=0.00500\,\mathrm{mol\,dm^{-3}}$ and $x=3.53553\ldots\times10^{-4}\,\mathrm{mol\,dm^{-3}}$ , giving pH $=3.45154\ldots=3.452$ and $100x/c=7.071\ldots\%=7.07\%$ . The hydrogen-ion concentration falls, but the dissociated fraction rises as dilution favours the side with more dissolved particles.	5
05.1	$[H^+]=[A^-]=0.0800c$ and $[HA]=0.920c$ . $c=0.0250\,\mathrm{mol\,dm^{-3}}$ . $[H^+]=2.00\times10^{-3}\,\mathrm{mol\,dm^{-3}}$ . $\mathrm{pH}=2.699$ .	Let the initial concentration be $c$ . At $8.00\%$ dissociation, $[H^+]=[A^-]=0.0800c$ and $[HA]=0.920c$ . Therefore $K_a=(0.0800c)^2/(0.920c)=0.0064c/0.920$ , so $c=K_a(0.920)/0.0064=(1.74\times10^{-4})(0.920)/0.0064=0.0250125\,\mathrm{mol\,dm^{-3}}$ , or $0.0250\,\mathrm{mol\,dm^{-3}}$ . Then $[H^+]=0.0800c=0.002001\,\mathrm{mol\,dm^{-3}}$ and pH $=-\log_{10}(0.002001)=2.69875\ldots$ , or $2.699$ .	5

3.1.12.5 · pH curves, titrations and indicators (A-level only)

Tier 1 · Easy

Mark scheme for 3.1.12.5 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Methyl orange. Its transition range lies within the steep pH change; the phenolphthalein range does not.	Compare each transition range directly with the near-vertical section of the pH curve. Only methyl orange changes colour during that rapid pH change.	2
02.1	The acid is weak because the buffer region shows that the acid and its conjugate base coexist before equivalence. The base is strong because the equivalence point is above pH $7$ , as expected when a weak acid is neutralised by a strong base and its conjugate base hydrolyses.	A buffer region before equivalence is evidence that a weak acid and its conjugate base are both present. An equivalence point above pH $7$ is caused by the basic conjugate base, consistent with titration of a weak acid by a strong base.	3

Tier 2 · Standard

Mark scheme for 3.1.12.5 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Equivalence volume $=24.0\,\mathrm{cm^3}$ pH at equivalence $\approx7$	The acid amount is $0.120(0.0250)=0.00300\,\mathrm{mol}$ . The reaction is $1{:}1$ , so the same amount of $\mathrm{NaOH}$ is needed: $V=0.00300/0.125=0.0240\,\mathrm{dm^3}=24.0\,\mathrm{cm^3}$ . A strong acid–strong base equivalence mixture is approximately neutral.	3
02.1	Calibrate the pH probe with suitable buffer solutions before collecting the titration data. Add much smaller measured portions of base near the rapid pH change, swirling and recording a stable pH after each addition.	Calibration reduces systematic error in the measured pH. Smaller volume intervals near equivalence provide enough closely spaced points to locate and draw the steep section rather than stepping across it in one addition.	4
03.1	The diluted curve has a higher initial pH because the initial hydrogen ion concentration is lower. The volume of base at equivalence is unchanged because the amount of acid is unchanged. The pH at equivalence remains approximately $7$ at the same temperature. After the same fixed excess volume of base, the diluted mixture has the lower pH because the excess hydroxide ions occupy a larger total volume.	Adding water lowers the acid concentration but does not change its amount, so the initial pH rises while the base amount needed for equivalence stays the same. A strong acid–strong base mixture is approximately neutral at equivalence. Beyond equivalence, the amount of excess hydroxide is the same at the stated base volume, but the added water makes the total solution volume larger; therefore $[OH^-]$ is lower and the pH is lower.	4

Tier 3 · Hard

Mark scheme for 3.1.12.5 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Half-equivalence volume $=12.0\,\mathrm{cm^3}$ $\mathrm{pH}=4.602$ The equivalence pH is above $7$ .	The acid amount is $0.150(0.0200)=0.00300\,\mathrm{mol}$ , so equivalence needs $0.00300/0.125=0.0240\,\mathrm{dm^3}=24.0\,\mathrm{cm^3}$ of base. Half-equivalence is therefore at $12.0\,\mathrm{cm^3}$ . There $[HA]=[A^-]$ , so $\mathrm{pH}=pK_a=-\log_{10}(2.50\times10^{-5})=4.602$ . At equivalence the conjugate base hydrolyses, making the solution alkaline.	5
02.1	Acid concentration $=0.225\,\mathrm{mol\,dm^{-3}}$ $K_a=1.74\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ The acid is weak. Phenolphthalein is suitable because its transition range lies within the rapid pH rise; methyl orange is not suitable.	At equivalence, the acid amount equals the hydroxide amount: $n=0.150(0.0300)=0.00450\,\mathrm{mol}$ . Therefore the acid concentration is $0.00450/0.0200=0.225\,\mathrm{mol\,dm^{-3}}$ . At $15.0\,\mathrm{cm^3}$ the titration is at half-equivalence, so $\mathrm{pH}=pK_a=4.76$ and $K_a=10^{-4.76}=1.737800\ldots\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ , which is $1.74\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ to three significant figures. The half-equivalence behaviour and the alkaline equivalence region identify a weak acid. Only the phenolphthalein range lies inside the stated rapid rise.	7
03.1	After neutralisation, $n(\mathrm{HA})=2.23\times10^{-3}\,\mathrm{mol}$ and $n(\mathrm{A^-})=7.60\times10^{-4}\,\mathrm{mol}$ . $\mathrm{pH}=4.190$ ; this is the pre-equivalence buffer region. Phenolphthalein should be used because its transition range lies within the rapid pH change; the methyl orange range does not.	Initially, $n(\mathrm{HA})=0.0260(0.115)=2.990\times10^{-3}\,\mathrm{mol}$ and $n(\mathrm{OH^-})=0.00950(0.0800)=7.60\times10^{-4}\,\mathrm{mol}$ . Neutralisation forms $7.60\times10^{-4}\,\mathrm{mol}$ of $\mathrm{A^-}$ and leaves $2.230\times10^{-3}\,\mathrm{mol}$ of $\mathrm{HA}$ . The additive total volume is $0.0355\,\mathrm{dm^3}$ , giving $[\mathrm{A^-}]=0.02141\,\mathrm{mol\,dm^{-3}}$ and $[\mathrm{HA}]=0.06282\,\mathrm{mol\,dm^{-3}}$ . Thus $[H^+]=K_a[\mathrm{HA}]/[\mathrm{A^-}]=6.455\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ and pH $=4.1901\ldots$ , or $4.190$ to three decimal places, matching the three significant figures of $[\mathrm{H^+}]$ . Both weak acid and conjugate base are present, so this is the buffer region before equivalence. Phenolphthalein changes over pH $8.3$ – $10.0$ , entirely within the rapid rise from $7.7$ to $10.5$ ; methyl orange changes below that steep region.	6
04.1	The initial hydrogen chloride amount is $3.60\times10^{-3}\,\mathrm{mol}$ . The equivalence volume is $24.0\,\mathrm{cm^3}$ . The added hydroxide amount is $3.00\times10^{-3}\,\mathrm{mol}$ , leaving $6.00\times10^{-4}\,\mathrm{mol}$ of $\mathrm{H^+}$ . $[H^+]=0.0120\,\mathrm{mol\,dm^{-3}}$ . $\mathrm{pH}=1.921$ ; the mixture is on the acidic pre-equivalence section of the curve.	The acid amount is $0.120(0.0300)=0.00360\,\mathrm{mol}$ , so a $1{:}1$ neutralisation needs $0.00360/0.150=0.0240\,\mathrm{dm^3}=24.0\,\mathrm{cm^3}$ of base. At $20.0\,\mathrm{cm^3}$ , the hydroxide amount is $0.150(0.0200)=0.00300\,\mathrm{mol}$ , leaving exactly $0.000600\,\mathrm{mol}$ of hydrogen ions. The total volume is $0.0500\,\mathrm{dm^3}$ , so $[H^+]=0.000600/0.0500=0.0120\,\mathrm{mol\,dm^{-3}}$ . Therefore pH $=-\log_{10}(0.0120)=1.920818\ldots$ , or $1.921$ to three decimal places, before equivalence.	5
05.1	The initial pH is $0.921$ . The equivalence volume is $30.0\,\mathrm{cm^3}$ . The equivalence pH is below $7$ because $\mathrm{BH^+}$ , the conjugate acid of the weak base, reacts with water to form $\mathrm{H^+}$ . The buffer region occurs after equivalence, when excess weak base B and its conjugate acid $\mathrm{BH^+}$ are both present. Methyl orange is suitable; phenolphthalein is not because its transition range lies above the steep acidic equivalence region.	Complete dissociation of hydrochloric acid gives $[H^+]=0.120\,\mathrm{mol\,dm^{-3}}$ , so pH $=-\log_{10}(0.120)=0.920818\ldots=0.921$ . The acid amount is $0.120(0.0250)=0.00300\,\mathrm{mol}$ . Since $\mathrm{B+H^+\rightarrow BH^+}$ is $1{:}1$ , equivalence requires $V=0.00300/0.100=0.0300\,\mathrm{dm^3}=30.0\,\mathrm{cm^3}$ . At equivalence the solution contains $\mathrm{BH^+}$ , which donates protons to water, so the pH is below $7$ . Before equivalence, strong acid is in excess; only after equivalence do excess B and the $\mathrm{BH^+}$ already formed coexist as a weak-base/conjugate-acid buffer. The acidic steep section includes methyl orange's range, whereas phenolphthalein changes colour too far into the alkaline region.	6

3.1.12.6 · Buffer action (A-level only)

Tier 1 · Easy

Mark scheme for 3.1.12.6 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$\mathrm{pH}=4.745$	Equal concentrations make $[A^-]/[HA]=1$ , so $\log_{10}1=0$ and $\mathrm{pH}=pK_a$ . Therefore $\mathrm{pH}=-\log_{10}(1.80\times10^{-5})=4.745$ .	3
02.1	Its pH remains approximately unchanged because the conjugate-base to weak-acid concentration ratio is unchanged. Its ability to resist pH change decreases because both buffer-component concentrations are lower after dilution.	Dilution multiplies both component concentrations by the same factor, so their ratio and hence the pH are approximately unchanged. The smaller concentrations mean fewer moles are available per unit volume to remove added acid or base.	2

Tier 2 · Standard

Mark scheme for 3.1.12.6 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$\mathrm{A^-+H^+\rightarrow HA}$ removes added acid. $\mathrm{HA+OH^-\rightarrow A^-+H_2O}$ removes added base.	Use the conjugate base component to accept added protons and the weak acid component to neutralise added hydroxide ions. Because each added reagent is converted mainly into a weak buffer component, $[H^+]$ changes only slightly.	4
02.1	$\mathrm{pH}=4.27$ (2 d.p.); accept $4.26$ if $pK_a$ was rounded to $4.74$ before use	Initially, $n(\mathrm{HA})=0.200(0.100)=0.0200\,\mathrm{mol}$ and $n(\mathrm{OH^-})=0.100(0.0500)=0.00500\,\mathrm{mol}$ . Partial neutralisation leaves $0.0150\,\mathrm{mol}$ of $\mathrm{HA}$ and forms $0.00500\,\mathrm{mol}$ of $\mathrm{A^-}$ . With $pK_a=4.744727\ldots$ , $\mathrm{pH}=pK_a+\log_{10}(0.00500/0.0150)=4.267606\ldots$ , which is $4.27$ to two decimal places.	5
03.1	$pK_a=4.70$ $K_a=2.0\times10^{-5}\,\mathrm{mol\,dm^{-3}}$	Because both components are in the same solution, their concentration ratio equals their amount ratio: $[A^-]/[HA]=0.0750/0.0300=2.50$ . Rearranging $\mathrm{pH}=pK_a+\log_{10}([A^-]/[HA])$ gives $pK_a=5.10-\log_{10}(2.50)=4.70206\ldots$ , or $4.70$ to two decimal places. Therefore $K_a=10^{-4.70206\ldots}=1.986\ldots\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ , or $2.0\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ to two significant figures.	4

Tier 3 · Hard

Mark scheme for 3.1.12.6 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$\mathrm{pH}=4.548$	Added $\mathrm{H^+}$ reacts with $\mathrm{A^-}$ , so the new amounts are $0.0700\,\mathrm{mol}$ of $\mathrm{A^-}$ and $0.110\,\mathrm{mol}$ of $\mathrm{HA}$ . The common volume cancels in their concentration ratio. Thus $\mathrm{pH}=pK_a+\log_{10}(0.0700/0.110)=4.7447-0.1963=4.548$ .	5
02.1	After dilution, $\mathrm{pH}=4.92$ (2 d.p.). After adding sodium hydroxide, $\mathrm{pH}=5.01$ (2 d.p.).	Dilution changes both concentrations by the same factor, so $\mathrm{pH}=pK_a+\log_{10}(0.0600/0.0400)=4.744727\ldots+\log_{10}(1.5)=4.920818\ldots$ , which is $4.92$ to two decimal places. The added hydroxide amount is $0.500(0.0100)=0.00500\,\mathrm{mol}$ and it reacts by $\mathrm{HA+OH^-\rightarrow A^-+H_2O}$ . The new amounts are $0.0350\,\mathrm{mol}$ of $\mathrm{HA}$ and $0.0650\,\mathrm{mol}$ of $\mathrm{A^-}$ . The common final volume cancels, so $\mathrm{pH}=4.744727\ldots+\log_{10}(0.0650/0.0350)=5.013572\ldots$ , which is $5.01$ to two decimal places.	7
03.1	Maximum $\mathrm{HCl}=0.021\,\mathrm{mol}$ . Maximum $\mathrm{NaOH}=0.0066\,\mathrm{mol}$ . The upper-pH limit is reached with less added base because the initial buffer already contains more $\mathrm{A^-}$ than HA.	If $x$ moles of acid are added, the amounts become $0.0600-x$ of $\mathrm{A^-}$ and $0.0400+x$ of HA. At pH $4.60$ , their ratio is $10^{4.60-4.80}=0.630957$ , so $(0.0600-x)/(0.0400+x)=0.630957$ and $x=0.0213137\,\mathrm{mol}$ , or $0.021\,\mathrm{mol}$ to two significant figures. If $y$ moles of base are added, the amounts become $0.0600+y$ and $0.0400-y$ . At pH $5.10$ , the ratio is $10^{5.10-4.80}=1.99526$ , so $(0.0600+y)/(0.0400-y)=1.99526$ and $y=0.00661394\,\mathrm{mol}$ , or $0.0066\,\mathrm{mol}$ to two significant figures. Thus the permitted base addition is smaller.	6
04.1	$[A^-]/[HA]=10^{3.90-4.20}=0.5012$ . $n(HA)+n(A^-)=0.100\,\mathrm{mol}$ . $n(HA)=0.0666\,\mathrm{mol}$ . $n(A^-)=0.0334\,\mathrm{mol}$ . $0.0334\,\mathrm{mol}$ of sodium hydroxide is required for preparation by partial neutralisation.	The Henderson–Hasselbalch relation gives $[A^-]/[HA]=10^{\mathrm{pH}-pK_a}=10^{-0.30}=0.5011872336\ldots$ . Use this unrounded ratio in the amount calculation. Because both components share one solution, let $n(HA)=x$ and $n(A^-)=0.5011872336\ldots x$ . Then $x(1+0.5011872336\ldots)=0.100$ , so $n(HA)=0.0666139\,\mathrm{mol}$ and $n(A^-)=0.0333861\,\mathrm{mol}$ . Each mole of hydroxide converts one mole of HA into one mole of $A^-$ , so $0.0334\,\mathrm{mol}$ of sodium hydroxide is required.	5
05.1	Buffer 1 contributes $0.0200\,\mathrm{mol}$ HA and $0.0100\,\mathrm{mol}$ $\mathrm{A^-}$ . Buffer 2 contributes $0.0120\,\mathrm{mol}$ HA and $0.0360\,\mathrm{mol}$ $\mathrm{A^-}$ . The mixed amounts are $0.0320\,\mathrm{mol}$ HA and $0.0460\,\mathrm{mol}$ $\mathrm{A^-}$ . $[A^-]/[HA]=0.0460/0.0320=1.4375$ . $\mathrm{pH}=4.758$ .	Calculate component amounts before combining the solutions. Buffer 1 supplies $0.200(0.100)=0.0200\,\mathrm{mol}$ HA and $0.200(0.0500)=0.0100\,\mathrm{mol}$ $A^-$ . Buffer 2 supplies $0.300(0.0400)=0.0120\,\mathrm{mol}$ HA and $0.300(0.120)=0.0360\,\mathrm{mol}$ $A^-$ . Their totals are $0.0320$ and $0.0460\,\mathrm{mol}$ . Both share the same final volume, so the concentration ratio equals $0.0460/0.0320=1.4375$ . Hence pH $=4.600+\log_{10}(1.4375)=4.757607\ldots$ , or $4.758$ .	5