3.1 Physical chemistry — printable coverage pack

3.1.1.1 · Fundamental particles

Knowledge of atomic structure has evolved as new evidence produced improved models. The current model has a tiny nucleus containing protons and neutrons, with electrons occupying the space around it.
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 $1/1836$ .
For a neutral atom, the numbers of protons and electrons are equal. An ion forms when electrons are lost or gained; the nucleus is unchanged.
When finding an ion's charge, subtract the number of electrons from the number of protons. A common error is to include neutrons, which carry no charge.

Tier 1 · Easy

1. State the relative charge and relative mass of an electron.[2 marks]
Answer
- Relative charge $=-1$
- Relative mass $\approx1/1836$
Method: 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.

Tier 2 · Standard

1. A particle contains 15 protons, 16 neutrons and 18 electrons. State its overall charge and identify which particles are in its nucleus.[3 marks]
Answer
- Overall charge $=3-$
- The nucleus contains the 15 protons and 16 neutrons
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 13 protons
- Ion charge $=2-$
- 15 electrons
Method: 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 $2-$ .

3.1.1.2 · Mass number and isotopes

The atomic number $Z$ is the number of protons, while the mass number $A$ is the total number of protons and neutrons; therefore neutrons $=A-Z$ .
Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. Their chemical properties are similar because they have the same electron configuration.
In a TOF mass spectrometer, gaseous particles are ionised, accelerated to the same kinetic energy, separated by flight time and detected. For mononuclear $1+$ ions, peak position gives relative isotopic mass and peak height gives relative abundance.
Calculate $A_r$ as $\sum(\text{isotopic mass}\times\text{fractional abundance})$ ; an isotope pattern and its weighted mean can identify an element, while a molecular-ion peak can give $M_r$ . A common error is to use percentage abundances without dividing by $100$ .

Tier 1 · Easy

1. For the ion $^{81}_{35}\mathrm{Br^-}$ , give its proton count, neutron count and electron count.[3 marks]
Answer
- 35 protons
- 46 neutrons
- 36 electrons
Method: 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.

Tier 2 · Standard

1. 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]
Answer
- $A_r=50.54$
Method: Use the weighted mean: $A_r=(50\times73.0+52\times27.0)/100=(3650+1404)/100=50.54$ .

Tier 3 · Hard

1. 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 $t\propto\sqrt{m/z}$ to deduce the mass number of the second isotope.[4 marks]
Answer
- Mass number $=70$
Method: 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 $^{70}\mathrm{X}$ .

3.1.1.3 · Electron configuration

For atoms up to $Z=36$ , fill sub-shells in the order $1s$ , $2s$ , $2p$ , $3s$ , $3p$ , $4s$ , $3d$ , $4p$ , while recognising $\mathrm{Cr:[Ar]3d^5\,4s^1}$ and $\mathrm{Cu:[Ar]3d^{10}\,4s^1}$ as exceptions.
When transition-metal ions form, remove electrons from $4s$ before $3d$ . Check that the superscripts total the species' electron count.
First ionisation energy is the energy required to remove one electron from each atom in one mole of gaseous atoms. A successive equation has the form $\mathrm{X^{(n-1)+}(g)\rightarrow X^{n+}(g)+e^-}$ .
A large jump between successive ionisation energies shows that removal has begun from a shell closer to the nucleus. A common error is to explain a sub-shell anomaly using nuclear charge alone while ignoring distance, shielding and sub-shell energy.

Tier 1 · Easy

1. Write the electron configuration, in sub-shell notation, of $\mathrm{S^{2-}}$ . Sulfur has atomic number 16.[1 mark]
Answer
- $1s^2\,2s^2\,2p^6\,3s^2\,3p^6$
Method: A sulfur atom has 16 electrons and the $2-$ ion has gained two, giving 18. Filling the sub-shells in order gives $1s^2\,2s^2\,2p^6\,3s^2\,3p^6$ .

Tier 2 · Standard

1. Define first ionisation energy and write an equation, including state symbols, for the first ionisation of magnesium.[3 marks]
Answer
- Energy required to remove one electron from each atom in one mole of gaseous atoms
- $\mathrm{Mg(g)\rightarrow Mg^+(g)+e^-}$
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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 $3s^2\,3p^2$
- $\mathrm{Si^{4+}(g)\rightarrow Si^{5+}(g)+e^-}$
Method: 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: $\mathrm{Si^{4+}(g)\rightarrow Si^{5+}(g)+e^-}$ .

3.1.2.1 · Relative atomic mass and relative molecular mass

Relative atomic mass, $A_r$ , is the weighted mean mass of an atom of an element relative to $1/12$ of the mass of a carbon-12 atom.
Relative molecular mass, $M_r$ , is the mean mass of a molecule relative to $1/12$ of the mass of a carbon-12 atom; both relative quantities have no units.
Find $M_r$ by multiplying each element's $A_r$ by its subscript and summing. Use the term relative formula mass for ionic compounds because they do not contain discrete molecules.
Include every atom inside brackets, waters of crystallisation and repeated groups. A common error is to attach $\mathrm{g\,mol^{-1}}$ to $M_r$ rather than to molar mass.

Tier 1 · Easy

1. 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]
Answer
- $M_r=60.0$
Method: Urea contains one C, one O, two N and four H atoms. Therefore $M_r=12.0+16.0+2(14.0)+4(1.0)=60.0$ .

Tier 2 · Standard

1. 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]
Answer
- Relative formula mass $=246.4$
Method: 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 $120.4+126.0=246.4$ .

Tier 3 · Hard

1. 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]
Answer
- $A_r(\mathrm{X})=48.05$
- Abundance of $^{47}\mathrm{X}=47.5\%$
Method: 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 $47.5\%$ .

3.1.2.2 · The mole and the Avogadro constant

One mole contains the Avogadro constant, $N_A$ , of specified entities; the entity may be an atom, molecule, ion, electron or formula unit.
Use $n=m/M$ for mass, $N=nN_A$ for particle count, and $n=cV$ for solutions when $V$ is in $\mathrm{dm^3}$ .
A reliable route is to convert the given quantity to moles, apply any particle or formula-unit multiplier, then convert to the requested quantity.
State the entity being counted and convert $\mathrm{cm^3}$ to $\mathrm{dm^3}$ by dividing by $1000$ . A common error is to multiply the concentration by a volume still in $\mathrm{cm^3}$ .

Tier 1 · Easy

1. Calculate the amount, in moles, in $5.30\,\mathrm{g}$ of $\mathrm{Na_2CO_3}$ . Use $M_r=106.0$ .[2 marks]
Answer
- $0.0500\,\mathrm{mol}$
Method: Use $n=m/M$ : $n=5.30/106.0=0.0500\,\mathrm{mol}$ .

Tier 2 · Standard

1. 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]
Answer
- $7.53\times10^{21}$ molecules
Method: 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.

Tier 3 · Hard

1. $18.0\,\mathrm{cm^3}$ 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]
Answer
- $8.13\times10^{21}$ sulfate ions
Method: 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.

3.1.2.3 · The ideal gas equation

The ideal gas equation is $pV=nRT$ , with pressure 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 $1000$ , $\mathrm{cm^3}$ to $\mathrm{m^3}$ by multiplying by $10^{-6}$ , and $\mathrm{dm^3}$ to $\mathrm{m^3}$ by multiplying by $10^{-3}$ .
Rearrange symbolically before substituting; for example $n=pV/(RT)$ and $M=mRT/(pV)$ when using gas data to find molar mass.
Use absolute temperature, $T/\mathrm{K}=\theta/^{\circ}\mathrm{C}+273$ . A common error is to substitute a Celsius temperature or mix litres with SI pressure.

Tier 1 · Easy

1. $0.100\,\mathrm{mol}$ of gas occupies $2.49\times10^{-3}\,\mathrm{m^3}$ at $300\,\mathrm{K}$ . Calculate its pressure. Take $R=8.31$ .[3 marks]
Answer
- $1.00\times10^5\,\mathrm{Pa}$
Method: Rearrange to $p=nRT/V$ . Then $p=[0.100\times8.31\times300]/(2.49\times10^{-3})=1.00\times10^5\,\mathrm{Pa}$ .

Tier 2 · Standard

1. 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]
Answer
- $0.873\,\mathrm{dm^3}$
Method: 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 $0.873\,\mathrm{dm^3}$ .

Tier 3 · Hard

1. 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]
Answer
- $118\,\mathrm{g\,mol^{-1}}$
Method: 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 $M=m/n=0.318/(2.687\times10^{-3})=118\,\mathrm{g\,mol^{-1}}$ .

3.1.2.4 · Empirical and molecular formula

An empirical formula gives the simplest whole-number ratio of atoms of each element, whereas a molecular formula gives the actual number of each type of atom in a molecule.
Convert each mass or percentage to moles by dividing by $A_r$ , divide all mole values by the smallest, then scale to whole numbers when necessary.
Calculate the empirical formula mass and use $M_r/(\text{empirical formula mass})$ to find the integer multiplier for the molecular formula.
Do not round a ratio such as $1.5$ directly to $2$ ; multiply every ratio by the same small integer. For combustion data, remember that one mole of water contains two moles of hydrogen atoms.

Tier 1 · Easy

1. 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]
Answer
- $\mathrm{Al_2O_3}$
Method: 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 $\mathrm{Al_2O_3}$ .

Tier 2 · Standard

1. 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]
Answer
- Empirical formula $\mathrm{C_2H_4O}$
- Molecular formula $\mathrm{C_4H_8O_2}$
Method: 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 $\mathrm{C_4H_8O_2}$ .

Tier 3 · Hard

1. 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]
Answer
- Empirical formula $\mathrm{CH_2O}$
- Molecular formula $\mathrm{C_5H_{10}O_5}$
Method: 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 $\mathrm{C_5H_{10}O_5}$ .

3.1.2.5 · Balanced equations and associated calculations

Balance full and ionic equations by conserving atoms and total charge, changing coefficients only and leaving chemical formulae unchanged.
For reacting quantities, convert the known amount to moles, use the balanced-equation ratio, then convert the required amount to mass, gas volume, concentration or solution volume.
Percentage yield is $100\times\text{actual yield}/\text{theoretical yield}$ , while atom economy is $100\times M_r(\text{desired product})/\sum M_r(\text{reactants})$ using coefficients. High atom economy reduces waste and raw-material demand, bringing economic, environmental and ethical advantages.
Check for a limiting reactant before calculating theoretical yield. A common error is to use a mass ratio directly instead of the stoichiometric mole ratio.

Tier 1 · Easy

1. Balance the equation $\mathrm{C_3H_8+O_2\rightarrow CO_2+H_2O}$ using the smallest whole-number coefficients.[1 mark]
Answer
- $\mathrm{C_3H_8+5O_2\rightarrow3CO_2+4H_2O}$
Method: 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 $5\mathrm{O_2}$ .

Tier 2 · Standard

1. $25.0\,\mathrm{cm^3}$ 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]
Answer
- $0.188\,\mathrm{g}$
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- Ammonia is the limiting reactant
- Maximum urea mass $=150\,\mathrm{kg}$
- Atom economy $=76.9\%$
- Percentage yield $=84.0\%$
Method: 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 $(126/150)\times100=84.0\%$ .

3.1.3.1 · Ionic bonding

Ionic bonding is the electrostatic attraction between oppositely charged ions throughout a giant lattice.
Predict simple-ion charges from Periodic Table groups, then combine ions in the smallest ratio that makes the total charge zero.
Keep a compound ion intact when balancing charge; for example two $\mathrm{Al^{3+}}$ ions require three $\mathrm{SO_4^{2-}}$ ions, giving $\mathrm{Al_2(SO_4)_3}$ .
Brackets are required around more than one compound ion. A common error is to describe ionic bonding as electron transfer; transfer forms the ions, while attraction between ions is the bond.

Tier 1 · Easy

1. Write the formula of aluminium sulfate from $\mathrm{Al^{3+}}$ and $\mathrm{SO_4^{2-}}$ ions.[1 mark]
Answer
- $\mathrm{Al_2(SO_4)_3}$
Method: 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 $\mathrm{Al_2(SO_4)_3}$ .

Tier 2 · Standard

1. Explain the nature of ionic bonding in magnesium oxide.[3 marks]
Answer
- 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
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- $\mathrm{X^{3+}}$
- $\mathrm{X(NO_3)_3}$
- $\mathrm{X(OH)_3}$
Method: 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 $\mathrm{X(OH)_3}$ .

3.1.3.2 · Nature of covalent and dative covalent bonds

A single covalent bond is a shared pair of electrons; double and triple bonds contain two and three shared pairs respectively.
Represent an ordinary covalent bond with a line between atoms, showing one shared pair without implying that either atom supplied both electrons.
In a co-ordinate or dative covalent bond, both electrons in the shared pair come from one atom; draw an arrow from the lone-pair donor to the electron-pair acceptor.
After a dative bond forms it behaves as a covalent bond. A common error is to point the arrow towards the donor rather than away from it.

Tier 1 · Easy

1. State what is meant by a single covalent bond.[1 mark]
Answer
- A shared pair of electrons
Method: A single covalent bond consists of one electron pair shared between two atoms.

Tier 2 · Standard

1. $\mathrm{BCl_3}$ 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]
Answer
- $\mathrm{Cl_3B\leftarrow NH_3}$
- The nitrogen atom donates the electron pair
Method: Nitrogen has the lone pair and boron accepts it. The arrow must therefore start at N and point towards B, represented as $\mathrm{Cl_3B\leftarrow NH_3}$ .

Tier 3 · Hard

1. 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]
Answer
- $\mathrm{NH_3+H^+\rightarrow NH_4^+}$
- 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
Method: 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.

3.1.3.3 · Metallic bonding

Metallic bonding is the electrostatic attraction between positive ions arranged in a lattice and delocalised electrons.
The delocalised electrons move through the structure, so metals conduct electricity in both solid and liquid states.
Metallic bonding is non-directional: layers of ions can slide while remaining attracted to the electron sea, making many metals malleable and ductile.
Stronger metallic bonding generally follows greater ionic charge, smaller ion size or more delocalised electrons per atom. A common error is to describe discrete metal molecules or electron pairs between particular atoms.

Tier 1 · Easy

1. Complete the definition of metallic bonding.[2 marks]
Answer
- Electrostatic attraction between positive ions and delocalised electrons
Method: Identify both components of the lattice and the force between them: positive metal ions are held by electrostatic attraction to delocalised electrons.

Tier 2 · Standard

1. Explain why a metal conducts electricity when solid and remains conductive when molten.[3 marks]
Answer
- The metal contains delocalised electrons
- These electrons are mobile and carry charge through the structure
- They remain delocalised when the lattice melts
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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
Method: 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.

3.1.3.4 · Bonding and physical properties

The four crystal types are ionic, metallic, macromolecular (giant covalent) and molecular. When drawing one, show the specified number and type of particles in a representative repeating arrangement, including charges for ions.
Melting a substance overcomes attractions between particles but does not normally break covalent bonds within simple molecules. Stronger or more extensive attractions require more energy.
Ionic substances conduct only when ions can move; metals and graphite conduct through delocalised electrons; simple molecular substances usually lack mobile charged particles.
Use evidence such as melting point and conductivity together before assigning a structure. A common error is to say that graphite conducts because whole carbon atoms move.

Tier 1 · Easy

1. Iodine forms crystals containing discrete $\mathrm{I_2}$ molecules. State the type of crystal structure and the attractions overcome when iodine melts.[2 marks]
Answer
- Molecular crystal
- Intermolecular forces between $\mathrm{I_2}$ molecules are overcome
Method: 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.

Tier 2 · Standard

1. 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]
Answer
- 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
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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
Method: 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.

3.1.3.5 · Shapes of simple molecules and ions

Bonding pairs and lone pairs are electron charge clouds that repel and arrange as far apart as possible around the central atom.
Count all electron pairs, minimise repulsion, then name the shape from atom positions. With five pairs, lone pairs prefer equatorial sites: one, two and three lone pairs give seesaw, T-shaped and linear derivatives respectively.
Repulsion decreases in the order lone pair-lone pair $>$ lone pair-bond pair $>$ bond pair-bond pair, so lone pairs compress adjacent bond angles.
State both shape and bond angle: linear $180^{\circ}$ , trigonal planar $120^{\circ}$ , tetrahedral $109.5^{\circ}$ , trigonal bipyramidal $90^{\circ}$ , $120^{\circ}$ and $180^{\circ}$ , and octahedral $90^{\circ}$ and $180^{\circ}$ are the starting geometries. A common error is to count a multiple bond as several charge clouds.

Tier 1 · Easy

1. State the shape and bond angle of $\mathrm{BF_3}$ around the boron atom.[2 marks]
Answer
- Trigonal planar
- $120^{\circ}$
Method: 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.

Tier 2 · Standard

1. 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]
Answer
- 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 $107^{\circ}$
Method: 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 $107^{\circ}$ .

Tier 3 · Hard

1. 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]
Answer
- Two lone pairs on Br
- Six electron pairs in an octahedral arrangement
- Square planar shape
- Bond angles $90^{\circ}$ and $180^{\circ}$
Method: 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.

3.1.3.6 · Bond polarity

Electronegativity is the power of an atom to attract the bonding pair of electrons in a covalent bond.
A difference in electronegativity produces an unsymmetrical electron distribution: the more electronegative atom is $\delta^-$ and the other atom is $\delta^+$ .
To decide whether a molecule has a permanent dipole, identify its polar bonds and use the molecular shape to determine whether their dipoles cancel.
A common error is to assume that every molecule containing polar bonds is polar; symmetrically arranged bond dipoles can have a zero resultant.

Tier 1 · Easy

1. 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]
Answer
- $\mathrm{H^{\delta+}-Cl^{\delta-}}$
Method: Chlorine has the greater electronegativity, so it attracts the bonding pair more strongly. Chlorine is therefore $\delta^-$ and hydrogen is $\delta^+$ .

Tier 2 · Standard

1. Both $\mathrm{CO_2}$ and $\mathrm{SO_2}$ contain polar bonds. Explain why only one of these molecules has a permanent dipole.[4 marks]
Answer
- $\mathrm{CO_2}$ 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.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

3.1.3.7 · Forces between molecules

Induced dipole–dipole forces act between all atoms and molecules; their strength generally increases with the number of electrons and molecular surface contact.
Permanent dipole–dipole attractions occur between polar molecules, while hydrogen bonding requires hydrogen bonded to nitrogen, oxygen or fluorine and a lone pair on such an atom in another molecule.
When comparing melting or boiling points, identify all intermolecular forces and compare their overall strength; stronger attractions require more energy to overcome.
Hydrogen bonds hold water molecules in an open lattice in ice. A common error is to say that hydrogen bonds are stronger in ice; the lower density is caused by the more open arrangement.

Tier 1 · Easy

1. State the strongest type of intermolecular force between $\mathrm{CH_3Cl}$ molecules.[1 mark]
Answer
- Permanent dipole–dipole forces.
Method: $\mathrm{CH_3Cl}$ 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.

Tier 2 · Standard

1. Explain why ethane, $\mathrm{C_2H_6}$ , has a higher boiling point than methane, $\mathrm{CH_4}$ .[3 marks]
Answer
- 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.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

3.1.4.1 · Enthalpy change

Enthalpy change, $\Delta H$ , is the heat energy change at constant pressure; an exothermic change has $\Delta H<0$ and an endothermic change has $\Delta H>0$ .
A standard enthalpy change applies at $100\,\text{kPa}$ and a stated temperature, with each substance in its standard state.
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 for forming one mole of a compound from its elements in their standard states; a common error is to define it for forming any stated amount.

Tier 1 · Easy

1. 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]
Answer
- Exothermic; $\Delta H$ is negative.
Method: Energy leaving the reacting chemicals warms the surroundings, so the reaction is exothermic. The products have lower enthalpy than the reactants, giving $\Delta H<0$ .

Tier 2 · Standard

1. Define standard enthalpy of formation and standard enthalpy of combustion. Include the amount of substance and the required conditions in each definition.[4 marks]
Answer
- 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.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- $\Delta_\mathrm{c}H^\circ=-888\,\text{kJ mol}^{-1}$
Method: The energy released per mole is $222/0.250=888\,\text{kJ mol}^{-1}$ . Combustion is exothermic, so attach a negative sign: $\Delta_\mathrm{c}H^\circ=-888\,\text{kJ mol}^{-1}$ .

3.1.4.2 · Calorimetry

Use $q=mc\Delta T$ , with the mass of the substance undergoing the temperature change, its specific heat capacity and a consistently signed temperature change.
Convert the measured heat change into a molar enthalpy using $\Delta H=-q/n$ when a temperature rise shows that the reaction released heat to the surroundings.
For solution calorimetry, the mass is usually estimated from total solution volume and density; the reacting amount is found from the limiting reagent.
Heat loss, incomplete combustion and heating the apparatus commonly make an experimental combustion enthalpy less exothermic than the accepted value; do not invent a correction without evidence.

Tier 1 · Easy

1. 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]
Answer
- $q=2.7\times10^3\,\text{J}$ (or $2.7\,\text{kJ}$ )
Method: Substitute into $q=mc\Delta T$ : $q=100\times4.18\times6.5=2717\,\text{J}$ . To two significant figures this is $2.7\times10^3\,\text{J}$ .

Tier 2 · Standard

1. 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]
Answer
- $\Delta H=-56.8\,\text{kJ mol}^{-1}$
Method: 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 $\Delta H=-2.8424/0.0500=-56.8\,\text{kJ mol}^{-1}$ .

Tier 3 · Hard

1. 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]
Answer
- $\Delta_\mathrm{c}H=-1.60\times10^3\,\text{kJ mol}^{-1}$ ; for example, heat is lost to the surroundings or used to warm the apparatus.
Method: 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.

3.1.4.3 · Applications of Hess's law

Hess's law states that the enthalpy change for a reaction is independent of the route taken, provided the initial and final states are the same.
Using formation enthalpies, calculate $\Delta_\mathrm{r}H^\circ=\sum\Delta_\mathrm{f}H^\circ(\text{products})-\sum\Delta_\mathrm{f}H^\circ(\text{reactants})$ , including equation coefficients.
Using combustion enthalpies to a common set of products, calculate the total for the reactants minus the total for the products.
Reversing an equation reverses the sign of its enthalpy change, and multiplying an equation multiplies its enthalpy change; a common error is to alter one without the other.

Tier 1 · Easy

1. 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]
Answer
- $-55\,\text{kJ mol}^{-1}$
Method: Add the enthalpy changes for the two consecutive routes: $+25+(-80)=-55\,\text{kJ mol}^{-1}$ .

Tier 2 · Standard

1. 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]
Answer
- $\Delta_\mathrm{r}H^\circ=-99\,\text{kJ mol}^{-1}$
Method: Apply products minus reactants. The standard formation enthalpy of $\mathrm{O_2(g)}$ is zero, so $\Delta_\mathrm{r}H^\circ=-396-[-297+\tfrac12(0)]=-99\,\text{kJ mol}^{-1}$ .

Tier 3 · Hard

1. 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]
Answer
- $\Delta H=-124\,\text{kJ mol}^{-1}$
Method: All three substances can be combusted to the same final products. Therefore use combustion of the reactants minus combustion of the product: $\Delta H=[-2058+(-286)]-(-2220)=-2344+2220=-124\,\text{kJ mol}^{-1}$ .

3.1.4.4 · Bond enthalpies

Mean bond enthalpy is the mean energy required to break one mole of a specified covalent bond in gaseous molecules.
Estimate a gaseous reaction enthalpy with $\Delta H=\sum E(\text{bonds broken})-\sum E(\text{bonds formed})$ ; breaking bonds is endothermic and forming bonds is exothermic.
A reliable method is to count every bond on both sides, multiply by the equation coefficients and cancel unchanged bonds only after the count is correct.
Values are approximate because a tabulated mean averages the same bond across different molecular environments; it is not the exact bond enthalpy in a particular molecule.

Tier 1 · Easy

1. Define the term mean bond enthalpy.[2 marks]
Answer
- The mean energy required to break one mole of a specified covalent bond in gaseous molecules.
Method: 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.

Tier 2 · Standard

1. 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]
Answer
- $\Delta H=-183\,\text{kJ mol}^{-1}$
Method: Break one $\mathrm{H-H}$ and one $\mathrm{Cl-Cl}$ bond, then form two $\mathrm{H-Cl}$ bonds. Thus $\Delta H=(436+243)-2(431)=679-862=-183\,\text{kJ mol}^{-1}$ .

Tier 3 · Hard

1. 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]
Answer
- $\Delta H=-126\,\text{kJ mol}^{-1}$ ; mean bond enthalpies are averages from bonds in different molecular environments.
Method: 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.

3.1.5.1 · Collision theory

Particles must collide before they can react, and a collision leads to reaction only if it has sufficient energy and a suitable orientation where required.
Activation energy is the minimum energy required for a reaction to occur following a collision between particles.
Use collision theory by separating collision frequency from the fraction of collisions that are successful; both can affect rate.
Most collisions do not cause reaction because their energy is below the activation energy; a common error is to claim that particles did not collide at all.

Tier 1 · Easy

1. Define activation energy.[2 marks]
Answer
- The minimum energy required for a reaction to occur following a collision between particles.
Method: The definition must state both minimum energy and that this energy allows colliding particles to react.

Tier 2 · Standard

1. Explain why most collisions between reactant particles do not produce a reaction.[3 marks]
Answer
- Most colliding particles have energy below the activation energy; for some reactions, collisions must also have a suitable orientation.
Method: A collision is necessary but not sufficient. Compare the collision energy with $E_\mathrm{a}$ and, where bond arrangement matters, check that the particles meet in an orientation capable of rearranging bonds.

Tier 3 · Hard

1. A reacting mixture undergoes $2.0\times10^{10}$ molecular collisions each second. A fraction $4.0\times10^{-6}$ have at least the activation energy, and one quarter of these have a suitable orientation. Calculate the number of successful collisions per second and explain the role of the activation energy.[4 marks]
Answer
- $2.0\times10^4$ successful collisions per second; the activation energy is the minimum collision energy needed for reaction.
Method: Energetic collisions occur at $(2.0\times10^{10})(4.0\times10^{-6})=8.0\times10^4\,\text{s}^{-1}$ . Multiplying by the suitable-orientation fraction gives $(8.0\times10^4)(1/4)=2.0\times10^4\,\text{s}^{-1}$ . Collisions below $E_\mathrm{a}$ cannot cross the reaction's energy barrier.

3.1.5.2 · Maxwell–Boltzmann distribution

A Maxwell–Boltzmann curve for a gas plots number of molecules against molecular energy; it starts at the origin, rises to a maximum and approaches the energy axis without meeting it.
The area under the curve represents the total number of molecules, so equal samples at different temperatures have equal areas.
At higher temperature the peak is lower and farther right, the distribution is broader, and a greater area lies beyond a fixed activation energy.
A common sketching error is to draw a maximum molecular energy or let the curve cross the energy axis; the distribution has a long high-energy tail.

Tier 1 · Easy

1. State the quantity represented on each axis of a Maxwell–Boltzmann distribution for a gas.[2 marks]
Answer
- Horizontal axis: molecular energy; vertical axis: number (or fraction) of molecules.
Method: 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.

Tier 2 · Standard

1. Describe how to add a higher-temperature curve to a Maxwell–Boltzmann distribution while keeping the number of gas molecules constant.[4 marks]
Answer
- 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.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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 $T_2$ .

3.1.5.3 · Effect of temperature on reaction rate

Rate of reaction is the change in concentration of a reactant or product per unit time; another measured quantity may be used when it tracks reaction progress.
Raising temperature increases particle speeds and collision frequency, but this is not the main reason for the often large rate increase.
On a Maxwell–Boltzmann distribution, higher temperature gives a much larger area beyond $E_\mathrm{a}$ , so a greater fraction of collisions can react.
A common error is to say that temperature lowers activation energy; without a catalyst, $E_\mathrm{a}$ is unchanged and the energy distribution changes.

Tier 1 · Easy

1. State the effect of increasing temperature on the rate of a reaction, with all other conditions unchanged.[1 mark]
Answer
- The rate increases.
Method: At a higher temperature, collisions occur more frequently and a greater fraction of them have sufficient energy, so successful collisions occur more often.

Tier 2 · Standard

1. Use a Maxwell–Boltzmann distribution to explain why a small temperature increase can cause a large increase in reaction rate.[4 marks]
Answer
- 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.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

3.1.5.4 · Effect of concentration and pressure

Increasing the concentration of a solution places more reactant particles in a given volume, so collisions occur more frequently.
Increasing the pressure of reacting gases by decreasing their volume also raises the number of particles per unit volume and the collision frequency.
At constant temperature, a concentration or pressure change does not alter the Maxwell–Boltzmann energy distribution or the activation energy.
A common error is to claim that higher pressure makes each gas particle move faster; temperature controls the energy distribution, while pressure here changes particle spacing.

Tier 1 · Easy

1. State why increasing the concentration of a dissolved reactant usually increases reaction rate.[2 marks]
Answer
- There are more reactant particles per unit volume, so collision frequency increases.
Method: 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.

Tier 2 · Standard

1. A gaseous reacting mixture is compressed to half its original volume at constant temperature. Explain why its reaction rate increases.[3 marks]
Answer
- The pressure and number of particles per unit volume increase, so particles are closer together and collide more frequently.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

3.1.5.5 · Catalysts

A catalyst increases reaction rate without being changed in chemical composition or amount by the overall reaction.
It provides an alternative reaction route with a lower activation energy; it does not lower the energy of every molecule.
At fixed temperature the Maxwell–Boltzmann curve is unchanged, but moving the $E_\mathrm{a}$ threshold left increases the area representing molecules able to react.
A catalyst does not change $\Delta H$ or the equilibrium position; it speeds both forward and reverse reactions and allows equilibrium to be reached sooner.

Tier 1 · Easy

1. Define a catalyst.[2 marks]
Answer
- A substance that increases reaction rate without being changed in chemical composition or amount by the overall reaction.
Method: 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.

Tier 2 · Standard

1. Use a Maxwell–Boltzmann distribution to explain how a catalyst increases the rate of a gas-phase reaction at constant temperature.[4 marks]
Answer
- 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.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

3.1.6.1 · Chemical equilibria and Le Chatelier's principle

At dynamic equilibrium in a closed system, forward and reverse reactions continue at equal rates and reactant and product concentrations remain constant.
Le Chatelier's principle predicts that an equilibrium shifts in the direction that opposes a change in concentration, pressure or temperature.
Higher pressure favours the side with fewer moles of gas, while higher temperature favours the endothermic direction; pressure has no positional effect when gaseous mole totals are equal.
A catalyst does not change equilibrium position. Industrial conditions often compromise between equilibrium yield, reaction rate, safety and cost rather than maximising one factor alone.

Tier 1 · Easy

1. State two features of a reversible reaction at dynamic equilibrium.[2 marks]
Answer
- The forward and reverse reactions have equal rates; reactant and product concentrations remain constant.
Method: Equilibrium is dynamic because both reactions continue. Equal rates create no net concentration change, but the concentrations need not be equal to each other.

Tier 2 · Standard

1. 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]
Answer
- Higher pressure increases the ammonia yield; higher temperature decreases it; a catalyst has no effect on the equilibrium yield.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

3.1.6.2 · Equilibrium constant Kc for homogeneous systems

For $aA+bB\rightleftharpoons cC+dD$ , write $K_\mathrm{c}=[C]^c[D]^d/([A]^a[B]^b)$ using equilibrium concentrations in $\text{mol dm}^{-3}$ .
Calculate equilibrium concentrations before substituting, using stoichiometry to relate the changes and dividing equilibrium amounts by the stated volume.
Derive the units of $K_\mathrm{c}$ from the powers in the expression; they may cancel completely for some equations.
At a fixed temperature, changing concentration or adding a catalyst does not change $K_\mathrm{c}$ . Temperature can change it: heating increases $K_\mathrm{c}$ for an endothermic forward reaction and decreases it for an exothermic one.

Tier 1 · Easy

1. Write the expression for $K_\mathrm{c}$ for $\mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)}$ .[1 mark]
Answer
- $K_\mathrm{c}=\dfrac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}$
Method: 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 $[\mathrm{HI}]$ .

Tier 2 · Standard

1. 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]
Answer
- $K_\mathrm{c}=24.0$ ; no units
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- $K_\mathrm{c}=1.30$ with no units; increasing temperature decreases $K_\mathrm{c}$ .
Method: 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.

3.1.7 · Oxidation, reduction and redox equations

Oxidation is loss of electrons and reduction is gain of electrons; an oxidising agent accepts electrons and is reduced, while a reducing agent donates electrons and is oxidised.
For a neutral compound, oxidation states sum to zero; for an ion, they sum to the ionic charge. Uncombined elements have oxidation state zero.
Build a half-equation by balancing the changing species and charge with electrons; in acidic solution, use $\mathrm{H_2O}$ and $\mathrm{H^+}$ to balance oxygen and hydrogen where needed.
Before adding half-equations, multiply them so that the numbers of electrons are equal. A common error is to cancel electrons without first balancing both charge and atoms.

Tier 1 · Easy

1. Determine the oxidation state of manganese in $\mathrm{MnO_4^-}$ .[2 marks]
Answer
- $+7$
Method: Let the oxidation state of manganese be $x$ . Oxygen is $-2$ , so $x+4(-2)=-1$ . Therefore $x=+7$ .

Tier 2 · Standard

1. 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]
Answer
- $\mathrm{2Fe^{2+}+Cl_2\rightarrow2Fe^{3+}+2Cl^-}$
Method: Multiply the iron half-equation by $2$ so that it releases $2e^-$ . Add the half-equations and cancel the electrons to obtain $\mathrm{2Fe^{2+}+Cl_2\rightarrow2Fe^{3+}+2Cl^-}$ .

Tier 3 · Hard

1. In acidic solution, dichromate(VI) ions oxidise $\mathrm{Sn^{2+}}$ ions to $\mathrm{Sn^{4+}}$ . Construct the overall ionic equation.[5 marks]
Answer
- $\mathrm{Cr_2O_7^{2-}+14H^++3Sn^{2+}\rightarrow2Cr^{3+}+7H_2O+3Sn^{4+}}$
Method: 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.

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

Lattice enthalpy of formation is the enthalpy change when one mole of an ionic solid forms from its gaseous ions; lattice dissociation has the same magnitude and the opposite sign.
A Born–Haber cycle links enthalpy of formation to atomisation, bond dissociation, ionisation energy, electron affinity and lattice enthalpy using Hess's law.
For solution cycles, lattice dissociation separates the solid into gaseous ions and hydration enthalpies then convert those ions into aqueous ions; follow every stoichiometric multiplier in the formula unit.
A substantial difference between a Born–Haber lattice enthalpy and a perfect-ionic-model value is evidence of covalent character. Common calculation errors are reversing the lattice sign or omitting a repeated ion term.

Tier 1 · Easy

1. Define the lattice enthalpy of formation of $\mathrm{MgCl_2}$ .[2 marks]
Answer
- The enthalpy change when one mole of $\mathrm{MgCl_2(s)}$ is formed from its gaseous ions under standard conditions.
- $\mathrm{Mg^{2+}(g)+2Cl^-(g)\rightarrow MgCl_2(s)}$
Method: 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.

Tier 2 · Standard

1. 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]
Answer
- $-787\,\mathrm{kJ\,mol^{-1}}$
Method: 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 $\Delta_{latt}H^\circ=-411-376=-787\,\mathrm{kJ\,mol^{-1}}$ .

Tier 3 · Hard

1. 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]
Answer
- $-122\,\mathrm{kJ\,mol^{-1}}$
Method: 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 $\Delta_{sol}H^\circ=2526-2648=-122\,\mathrm{kJ\,mol^{-1}}$ .

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

Entropy measures the dispersal of energy and matter; calculate a reaction entropy using $\Delta S^\circ=\sum S^\circ(\text{products})-\sum S^\circ(\text{reactants})$ , including coefficients.
Use $\Delta G=\Delta H-T\Delta S$ with temperature in kelvin and consistent energy units; entropy values in $\mathrm{J\,K^{-1}\,mol^{-1}}$ usually need converting to $\mathrm{kJ\,K^{-1}\,mol^{-1}}$ .
Under the stated conditions a change is thermodynamically feasible when $\Delta G$ is zero or negative, but feasibility does not imply that the reaction is fast.
On a graph of $\Delta G$ against $T$ , the intercept is $\Delta H$ and the gradient is $-\Delta S$ . A common error is to give the gradient as $+\Delta S$ .

Tier 1 · Easy

1. 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]
Answer
- $+73\,\mathrm{J\,K^{-1}\,mol^{-1}}$
Method: Subtract the reactant total from the product total: $\Delta S^\circ=465-392=+73\,\mathrm{J\,K^{-1}\,mol^{-1}}$ .

Tier 2 · Standard

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

Tier 3 · Hard

1. 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]
Answer
- $\Delta H=+63.0\,\mathrm{kJ\,mol^{-1}}$
- $\Delta S=+150\,\mathrm{J\,K^{-1}\,mol^{-1}}$
- $T=420\,\mathrm{K}$
Method: 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 $T=63.0/0.150=420\,\mathrm{K}$ .

3.1.9.1 · Rate equations (A-level only)

A rate equation has the form $\text{rate}=k[A]^m[B]^n$ ; each order is determined experimentally and is restricted here to $0$ , $1$ or $2$ .
The overall order is the sum of the individual orders. Derive the units of $k$ by rearranging the specific rate equation rather than memorising one set of units.
Increasing temperature increases $k$ . The Arrhenius equation is $k=Ae^{-E_a/(RT)}$ , and $\ln k=-E_a/(RT)+\ln A$ gives a straight line against $1/T$ .
Orders are powers in the experimental rate equation, not balancing numbers from the overall chemical equation; copying stoichiometric coefficients is a common error.

Tier 1 · Easy

1. 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]
Answer
- The rate is multiplied by $0.50$ ; it halves.
Method: Halving $[A]$ contributes $(1/2)^2=1/4$ , while doubling $[B]$ contributes $2$ . The combined factor is $(1/4)(2)=1/2$ .

Tier 2 · Standard

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

Tier 3 · Hard

1. 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]
Answer
- $E_a=54.6\,\mathrm{kJ\,mol^{-1}}$
Method: Here $\ln(k_2/k_1)=\ln4=1.3863$ and $1/298-1/318=2.110\times10^{-4}\,\mathrm{K^{-1}}$ . Rearranging gives $E_a=8.31(1.3863)/(2.110\times10^{-4})=5.46\times10^4\,\mathrm{J\,mol^{-1}}=54.6\,\mathrm{kJ\,mol^{-1}}$ .

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

Compare experiments in which only one initial concentration changes: an unchanged rate indicates zero order, a proportional change first order, and a squared change second order.
A concentration–time curve gives an instantaneous rate from the gradient of a tangent; initial-rate methods use the tangent at the start or a clock time whose reciprocal is proportional to rate.
For a first-order reactant, successive half-lives are constant. A zero-order concentration–time graph is a straight line whose negative gradient has magnitude $k$ .
The experimental rate equation constrains the rate-determining step and any preceding fast equilibrium, but agreement with a proposed mechanism supports rather than proves that mechanism.

Tier 1 · Easy

1. At constant temperature, doubling $[X]$ while all other concentrations remain unchanged doubles the initial rate. Deduce the order with respect to $X$ .[1 mark]
Answer
- First order with respect to $X$ .
Method: For $\text{rate}\propto[X]^m$ , the observation gives $2^m=2$ , so $m=1$ .

Tier 2 · Standard

1. 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]
Answer
- $\text{rate}=k[A]^2$
- $k=1.50\times10^{-2}\,\mathrm{dm^3\,mol^{-1}\,s^{-1}}$
Method: 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, $k=(6.00\times10^{-4})/(0.200)^2=1.50\times10^{-2}\,\mathrm{dm^3\,mol^{-1}\,s^{-1}}$ .

Tier 3 · Hard

1. 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]
Answer
- 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 $[X][Y]^2$ .
Method: 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.

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

For a gas mixture, $p_i=x_iP_{total}$ , where the mole fraction is $x_i=n_i/n_{total}$ ; use equilibrium amounts to find equilibrium partial pressures.
Construct $K_p$ from gaseous products over gaseous reactants, with each partial pressure raised to its balancing number. All species in this homogeneous system are gases.
At a fixed temperature, changing pressure may change the equilibrium composition but does not change $K_p$ ; only a temperature change changes its value.
A catalyst speeds attainment of equilibrium but changes neither the equilibrium position nor $K_p$ . Keep pressure units consistent because the units of $K_p$ follow from the expression.

Tier 1 · Easy

1. 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]
Answer
- $p_A=200\,\mathrm{kPa}$
- $p_B=300\,\mathrm{kPa}$
Method: The total amount is $5.00\,\mathrm{mol}$ . Hence $p_A=(2.00/5.00)(500)=200\,\mathrm{kPa}$ and $p_B=(3.00/5.00)(500)=300\,\mathrm{kPa}$ .

Tier 2 · Standard

1. 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]
Answer
- $K_p=152\,\mathrm{kPa}$
Method: 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 $K_p=p(\mathrm{NO_2})^2/p(\mathrm{N_2O_4})=(114.3)^2/85.7=152\,\mathrm{kPa}$ .

Tier 3 · Hard

1. 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]
Answer
- $K_p=9.00$ with no units
- Increasing pressure does not shift this equilibrium.
- $K_p$ is unchanged at constant temperature.
Method: 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.

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

Standard electrode potentials are measured relative to the standard hydrogen electrode at $298\,\mathrm{K}$ , $100\,\mathrm{kPa}$ and ion concentrations of $1.00\,\mathrm{mol\,dm^{-3}}$ .
Tables write electrode half-equations as reductions. The more positive potential is the reduction at the positive electrode; reverse the less positive half-equation for oxidation.
Calculate $E^\circ_{cell}=E^\circ_{positive}-E^\circ_{negative}$ . A positive value supports thermodynamic feasibility under standard conditions, but it gives no information about reaction rate.
In conventional cell notation, put the oxidation half-cell on the left and reduction half-cell on the right, use a single line for a phase boundary and a double line for the salt bridge; include platinum when no conducting solid is present.

Tier 1 · Easy

1. 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]
Answer
- $E^\circ_{cell}=+1.10\,\mathrm{V}$
- $\mathrm{Cu^{2+}}$ is reduced.
Method: The copper couple has the more positive potential, so $\mathrm{Cu^{2+}}$ is reduced. Calculate $E^\circ_{cell}=0.34-(-0.76)=+1.10\,\mathrm{V}$ .

Tier 2 · Standard

1. 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]
Answer
- $E^\circ_{cell}=+0.62\,\mathrm{V}$
- $\mathrm{2Fe^{3+}+Sn^{2+}\rightarrow2Fe^{2+}+Sn^{4+}}$
- $\mathrm{Pt(s)|Sn^{2+}(aq),Sn^{4+}(aq)||Fe^{3+}(aq),Fe^{2+}(aq)|Pt(s)}$
Method: $\mathrm{Fe^{3+}}$ 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.

Tier 3 · Hard

1. 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]
Answer
- $\mathrm{2Cu^+\rightarrow Cu^{2+}+Cu}$
- $E^\circ_{cell}=+0.37\,\mathrm{V}$
- Disproportionation is feasible under standard conditions.
Method: 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.

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

Commercial cells may be non-rechargeable, rechargeable or fuel cells. Rechargeable cells use an external potential to drive the electrode reactions in reverse.
In the simplified lithium cell, the negative reaction is $\mathrm{Li\rightarrow Li^++e^-}$ and the positive reaction is $\mathrm{Li^++CoO_2+e^-\rightarrow Li^+[CoO_2]^-}$ .
An alkaline hydrogen–oxygen fuel cell is supplied continuously with reactants; its overall reaction is $\mathrm{2H_2+O_2\rightarrow2H_2O}$ , so it is refuelled rather than electrically recharged.
Evaluate benefits and risks across the full system: operating emissions, energy density and continuous supply must be balanced against manufacture, fuel production, storage, safety, lifetime and disposal.

Tier 1 · Easy

1. State one difference between a rechargeable cell and a hydrogen–oxygen fuel cell.[2 marks]
Answer
- 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.
Method: 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.

Tier 2 · Standard

1. 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]
Answer
- $\mathrm{2H_2+O_2\rightarrow2H_2O}$
- Oxidation releases electrons at the negative electrode and reduction consumes them at the positive electrode, so electrons flow through the external circuit.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- $E_{cell}=+1.23\,\mathrm{V}$
- 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.
Method: 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.

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

A Brønsted–Lowry acid donates a proton and a Brønsted–Lowry base accepts a proton; acid–base equilibria are proton-transfer reactions.
A conjugate acid–base pair differs by one $\mathrm{H^+}$ ; removing a proton from an acid gives its conjugate base, while adding one to a base gives its conjugate acid.
Some species are amphoteric and can donate or accept a proton depending on the reaction partner, so identify roles from the equation rather than from a fixed label.
Show charge as well as atoms when writing proton-transfer equations. A common error is to name a conjugate pair whose formulas differ by more than one proton.

Tier 1 · Easy

1. In $\mathrm{NH_3+H_2O\rightleftharpoons NH_4^++OH^-}$ , identify the acid and the base on the left-hand side.[2 marks]
Answer
- $\mathrm{H_2O}$ is the acid.
- $\mathrm{NH_3}$ is the base.
Method: $\mathrm{H_2O}$ 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.

Tier 2 · Standard

1. For $\mathrm{HSO_4^-+H_2O\rightleftharpoons SO_4^{2-}+H_3O^+}$ , identify both conjugate acid–base pairs.[3 marks]
Answer
- $\mathrm{HSO_4^-/SO_4^{2-}}$
- $\mathrm{H_3O^+/H_2O}$
Method: $\mathrm{HSO_4^-}$ 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.

Tier 3 · Hard

1. Use two equations with water to show that $\mathrm{H_2PO_4^-}$ is amphoteric. Identify its role in each equation.[4 marks]
Answer
- $\mathrm{H_2PO_4^-+H_2O\rightleftharpoons HPO_4^{2-}+H_3O^+}$ ; $\mathrm{H_2PO_4^-}$ is an acid.
- $\mathrm{H_2PO_4^-+H_2O\rightleftharpoons H_3PO_4+OH^-}$ ; $\mathrm{H_2PO_4^-}$ is a base.
Method: 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.

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

The pH scale is logarithmic: $\mathrm{pH}=-\log_{10}[H^+]$ , with $[H^+]$ in $\mathrm{mol\,dm^{-3}}$ .
Reverse the logarithm with $[H^+]=10^{-\mathrm{pH}}$ . A change of one pH unit represents a factor of ten in hydrogen-ion concentration.
For a strong monoprotic acid, complete dissociation makes $[H^+]$ equal to the acid concentration after any dilution or mixing calculation.
Retain calculator precision until the end. Conventionally, the number of decimal places in pH matches the number of significant figures in $[H^+]$ .

Tier 1 · Easy

1. Calculate the pH of a solution for which $[H^+]=3.2\times10^{-3}\,\mathrm{mol\,dm^{-3}}$ .[2 marks]
Answer
- $\mathrm{pH}=2.49$
Method: $\mathrm{pH}=-\log_{10}(3.2\times10^{-3})=2.49485$ , which is $2.49$ because the concentration has two significant figures.

Tier 2 · Standard

1. A solution has pH $3.68$ . Calculate $[H^+]$ in $\mathrm{mol\,dm^{-3}}$ .[2 marks]
Answer
- $[H^+]=2.1\times10^{-4}\,\mathrm{mol\,dm^{-3}}$
Method: $[H^+]=10^{-3.68}=2.089\times10^{-4}\,\mathrm{mol\,dm^{-3}}$ . Two decimal places in the pH correspond to two significant figures in the concentration, giving $2.1\times10^{-4}\,\mathrm{mol\,dm^{-3}}$ .

Tier 3 · Hard

1. $27.5\,\mathrm{cm^3}$ 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]
Answer
- $\mathrm{pH}=0.975$
Method: 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 $\mathrm{pH}=-\log_{10}(0.106)=0.975$ .

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

Water dissociates slightly and its ionic product is $K_w=[H^+][OH^-]$ ; the value of $K_w$ depends on temperature.
For a strong base, first use its formula and complete dissociation to find $[OH^-]$ , then calculate $[H^+]=K_w/[OH^-]$ and hence pH.
In a neutral solution $[H^+]=[OH^-]=\sqrt{K_w}$ . Neutral pH is $7.00$ only at a temperature where $K_w=1.00\times10^{-14}$ .
Do not take the negative logarithm of $[OH^-]$ and report it directly as pH; that value is pOH unless it is converted using the appropriate $K_w$ .

Tier 1 · Easy

1. 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]
Answer
- $\mathrm{pH}=12.398$
Method: $\mathrm{NaOH}$ 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 $\mathrm{pH}=12.398$ .

Tier 2 · Standard

1. 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]
Answer
- $\mathrm{pH}=11.845$
Method: 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 $\mathrm{pH}=11.845$ .

Tier 3 · Hard

1. 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]
Answer
- Neutral water: $\mathrm{pH}=6.699$
- $0.0200\,\mathrm{mol\,dm^{-3}}$ $\mathrm{KOH}$ : $\mathrm{pH}=11.699$
Method: 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 $\mathrm{pH}=11.699$ .

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

A weak acid dissociates only slightly: $\mathrm{HA\rightleftharpoons H^++A^-}$ , with $K_a=[H^+][A^-]/[HA]$ .
For a weak monoprotic acid of initial concentration $c$ , the approximation $[H^+]\approx\sqrt{K_ac}$ is valid only when dissociation is small compared with $c$ .
Acid strength may be expressed as $pK_a=-\log_{10}K_a$ ; a larger $K_a$ and smaller $pK_a$ mean a stronger acid.
When pH is given, use $[H^+]=10^{-\mathrm{pH}}$ , take $[A^-]=[H^+]$ , and subtract the dissociated amount from the equilibrium $[HA]$ if an exact $K_a$ is required.

Tier 1 · Easy

1. A weak acid has $K_a=1.74\times10^{-5}\,\mathrm{mol\,dm^{-3}}$ . Calculate its $pK_a$ .[2 marks]
Answer
- $pK_a=4.759$
Method: $pK_a=-\log_{10}(1.74\times10^{-5})=4.759$ .

Tier 2 · Standard

1. 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]
Answer
- $\mathrm{pH}=2.512$
Method: $[H^+]\approx\sqrt{K_a[HA]}=\sqrt{(6.30\times10^{-5})(0.150)}=3.074\times10^{-3}\,\mathrm{mol\,dm^{-3}}$ . Therefore $\mathrm{pH}=-\log_{10}(3.074\times10^{-3})=2.512$ . The dissociation is about $2.05\%$ , so the approximation is reasonable.

Tier 3 · Hard

1. 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]
Answer
- $K_a=2.1\times10^{-4}\,\mathrm{mol\,dm^{-3}}$
- $pK_a=3.68$
Method: $[H^+]=10^{-2.40}=3.981\times10^{-3}\,\mathrm{mol\,dm^{-3}}$ , 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 $pK_a=3.68$ .

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

Know the characteristic curves for strong acid–strong base, strong acid–weak base, weak acid–strong base and weak acid–weak base titrations of monoprotic species.
At equivalence, the reacting acid and base amounts are stoichiometrically equal; the equivalence pH is about $7$ only for a strong acid–strong base titration.
Choose an indicator whose transition range lies within the steep section around the equivalence volume. An indicator is unsuitable when its range falls outside that rapid pH change.
In a weak acid–strong base titration, $\mathrm{pH}=pK_a$ at half-equivalence. Do not confuse half-equivalence with the equivalence point.

Tier 1 · Easy

1. 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]
Answer
- Methyl orange.
- Its transition range lies within the steep pH change; the phenolphthalein range does not.
Method: Compare each transition range directly with the near-vertical section of the pH curve. Only methyl orange changes colour during that rapid pH change.

Tier 2 · Standard

1. 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]
Answer
- Equivalence volume $=24.0\,\mathrm{cm^3}$
- pH at equivalence $\approx7$
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- Half-equivalence volume $=12.0\,\mathrm{cm^3}$
- $\mathrm{pH}=4.602$
- The equivalence pH is above $7$ .
Method: 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.

3.1.12.6 · Buffer action (A-level only)

An acidic buffer contains a weak acid and its salt, while a basic buffer contains a weak base and its salt; each pair provides both a proton donor and a proton acceptor.
In an acidic buffer, $\mathrm{A^-}$ removes added $\mathrm{H^+}$ and $\mathrm{HA}$ removes added $\mathrm{OH^-}$ ; in a basic buffer, the weak base removes $\mathrm{H^+}$ and its conjugate acid removes $\mathrm{OH^-}$ .
For an acidic buffer, $[H^+]=K_a[HA]/[A^-]$ , equivalently $\mathrm{pH}=pK_a+\log_{10}([A^-]/[HA])$ .
Before calculating pH after acid or base is added, adjust the amounts of both buffer components stoichiometrically. Dilution alone leaves their ratio, and hence pH approximately, unchanged.

Tier 1 · Easy

1. 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]
Answer
- $\mathrm{pH}=4.745$
Method: Equal concentrations make $[A^-]/[HA]=1$ , so $\log_{10}1=0$ and $\mathrm{pH}=pK_a$ . Therefore $\mathrm{pH}=-\log_{10}(1.80\times10^{-5})=4.745$ .

Tier 2 · Standard

1. An acidic buffer contains $\mathrm{HA}$ and $\mathrm{A^-}$ . Explain, using equations, how it resists small additions of acid and base.[4 marks]
Answer
- $\mathrm{A^-+H^+\rightarrow HA}$ removes added acid.
- $\mathrm{HA+OH^-\rightarrow A^-+H_2O}$ removes added base.
Method: 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.

Tier 3 · Hard

1. $0.500\,\mathrm{dm^3}$ 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]
Answer
- $\mathrm{pH}=4.548$
Method: 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 $\mathrm{pH}=pK_a+\log_{10}(0.0700/0.110)=4.7447-0.1963=4.548$ .