AQA A-level Chemistry coverage

Physical chemistry

Section 3.1
40 spec leafs

Notes and three levels of exam-style practice for each registered specification leaf in this section.

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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+1 and relative mass 11; a neutron has charge 00 and relative mass 11; an electron has charge 1-1 and relative mass about 1/18361/1836.
  • 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

2 marks
ORIGINAL

State the relative charge and relative mass of an electron.

Tier 2 · Standard

3 marks
ORIGINAL

A particle contains 15 protons, 16 neutrons and 18 electrons. State its overall charge and identify which particles are in its nucleus.

Tier 3 · Hard

4 marks
ORIGINAL

The nucleus of an atom has charge 2.08×1018C2.08\times10^{-18}\,\mathrm{C}. Use the proton charge 1.60×1019C1.60\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.

3.1.1.2

Mass number and isotopes

  • The atomic number ZZ is the number of protons, while the mass number AA is the total number of protons and neutrons; therefore neutrons =AZ=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+1+ ions, peak position gives relative isotopic mass and peak height gives relative abundance.
  • Calculate ArA_r as (isotopic mass×fractional abundance)\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 MrM_r. A common error is to use percentage abundances without dividing by 100100.

Tier 1 · Easy

3 marks
ORIGINAL

For the ion 3581Br^{81}_{35}\mathrm{Br^-}, give its proton count, neutron count and electron count.

Tier 2 · Standard

3 marks
ORIGINAL

An element has two isotopes, 50Q^{50}\mathrm{Q} with abundance 73.0%73.0\% and 52Q^{52}\mathrm{Q} with abundance 27.0%27.0\%. Calculate the relative atomic mass of Q.

Tier 3 · Hard

4 marks
ORIGINAL

In a TOF mass spectrometer, singly charged 64X+^{64}\mathrm{X^+} ions take 1.280×105s1.280\times10^{-5}\,\mathrm{s} to reach the detector. Another singly charged isotope of X takes 1.339×105s1.339\times10^{-5}\,\mathrm{s}. All ions receive the same kinetic energy. Use tm/zt\propto\sqrt{m/z} to deduce the mass number of the second isotope.

3.1.1.3

Electron configuration

  • For atoms up to Z=36Z=36, fill sub-shells in the order 1s1s, 2s2s, 2p2p, 3s3s, 3p3p, 4s4s, 3d3d, 4p4p, while recognising Cr:[Ar]3d54s1\mathrm{Cr:[Ar]3d^5\,4s^1} and Cu:[Ar]3d104s1\mathrm{Cu:[Ar]3d^{10}\,4s^1} as exceptions.
  • When transition-metal ions form, remove electrons from 4s4s before 3d3d. 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 X(n1)+(g)Xn+(g)+e\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 mark
ORIGINAL

Write the electron configuration, in sub-shell notation, of S2\mathrm{S^{2-}}. Sulfur has atomic number 16.

Tier 2 · Standard

3 marks
ORIGINAL

Define first ionisation energy and write an equation, including state symbols, for the first ionisation of magnesium.

Tier 3 · Hard

5 marks
ORIGINAL

Element X is in Period 3. Its first five ionisation energies, in kJmol1\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.

3.1.2.1

Relative atomic mass and relative molecular mass

  • Relative atomic mass, ArA_r, is the weighted mean mass of an atom of an element relative to 1/121/12 of the mass of a carbon-12 atom.
  • Relative molecular mass, MrM_r, is the mean mass of a molecule relative to 1/121/12 of the mass of a carbon-12 atom; both relative quantities have no units.
  • Find MrM_r by multiplying each element's ArA_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 gmol1\mathrm{g\,mol^{-1}} to MrM_r rather than to molar mass.

Tier 1 · Easy

2 marks
ORIGINAL

Calculate the relative molecular mass of urea, CO(NH2)2\mathrm{CO(NH_2)_2}. Use ArA_r: H =1.0=1.0, C =12.0=12.0, N =14.0=14.0, O =16.0=16.0.

Tier 2 · Standard

3 marks
ORIGINAL

Calculate the relative formula mass of MgSO47H2O\mathrm{MgSO_4\cdot7H_2O}. Use ArA_r: H =1.0=1.0, O =16.0=16.0, Mg =24.3=24.3, S =32.1=32.1.

Tier 3 · Hard

4 marks
ORIGINAL

An ionic oxide has formula X2O3\mathrm{X_2O_3} and relative formula mass 144.1. Element X contains only isotopes 47X^{47}\mathrm{X} and 49X^{49}\mathrm{X}. Calculate Ar(X)A_r(\mathrm{X}) and the percentage abundance of 47X^{47}\mathrm{X}. Use Ar(O)=16.0A_r(\mathrm{O})=16.0.

3.1.2.2

The mole and the Avogadro constant

  • One mole contains the Avogadro constant, NAN_A, of specified entities; the entity may be an atom, molecule, ion, electron or formula unit.
  • Use n=m/Mn=m/M for mass, N=nNAN=nN_A for particle count, and n=cVn=cV for solutions when VV is in dm3\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 cm3\mathrm{cm^3} to dm3\mathrm{dm^3} by dividing by 10001000. A common error is to multiply the concentration by a volume still in cm3\mathrm{cm^3}.

Tier 1 · Easy

2 marks
ORIGINAL

Calculate the amount, in moles, in 5.30g5.30\,\mathrm{g} of Na2CO3\mathrm{Na_2CO_3}. Use Mr=106.0M_r=106.0.

Tier 2 · Standard

2 marks
ORIGINAL

A sample contains 0.0125mol0.0125\,\mathrm{mol} of oxygen molecules. Calculate the number of O2\mathrm{O_2} molecules. Use NA=6.02×1023mol1N_A=6.02\times10^{23}\,\mathrm{mol^{-1}}.

Tier 3 · Hard

4 marks
ORIGINAL

18.0cm318.0\,\mathrm{cm^3} of 0.250moldm30.250\,\mathrm{mol\,dm^{-3}} aluminium sulfate solution contains fully dissociated Al2(SO4)3\mathrm{Al_2(SO_4)_3}. Calculate the number of sulfate ions present. Use NA=6.02×1023mol1N_A=6.02\times10^{23}\,\mathrm{mol^{-1}}.

3.1.2.3

The ideal gas equation

  • The ideal gas equation is pV=nRTpV=nRT, with pressure in Pa\mathrm{Pa}, volume in m3\mathrm{m^3}, temperature in K\mathrm{K} and amount in mol\mathrm{mol}.
  • Convert kPa\mathrm{kPa} to Pa\mathrm{Pa} by multiplying by 10001000, cm3\mathrm{cm^3} to m3\mathrm{m^3} by multiplying by 10610^{-6}, and dm3\mathrm{dm^3} to m3\mathrm{m^3} by multiplying by 10310^{-3}.
  • Rearrange symbolically before substituting; for example n=pV/(RT)n=pV/(RT) and M=mRT/(pV)M=mRT/(pV) when using gas data to find molar mass.
  • Use absolute temperature, T/K=θ/C+273T/\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

3 marks
ORIGINAL

0.100mol0.100\,\mathrm{mol} of gas occupies 2.49×103m32.49\times10^{-3}\,\mathrm{m^3} at 300K300\,\mathrm{K}. Calculate its pressure. Take R=8.31R=8.31.

Tier 2 · Standard

4 marks
ORIGINAL

Calculate the volume occupied by 0.0350mol0.0350\,\mathrm{mol} of an ideal gas at 315K315\,\mathrm{K} and 105kPa105\,\mathrm{kPa}. Give the answer in dm3\mathrm{dm^3}. Take R=8.31R=8.31.

Tier 3 · Hard

5 marks
ORIGINAL

A 0.318g0.318\,\mathrm{g} sample of a volatile liquid forms 85.0cm385.0\,\mathrm{cm^3} of vapour at 98.0kPa98.0\,\mathrm{kPa} and 373K373\,\mathrm{K}. Calculate its molar mass. Take R=8.31R=8.31.

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 ArA_r, divide all mole values by the smallest, then scale to whole numbers when necessary.
  • Calculate the empirical formula mass and use Mr/(empirical formula mass)M_r/(\text{empirical formula mass}) to find the integer multiplier for the molecular formula.
  • Do not round a ratio such as 1.51.5 directly to 22; 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

3 marks
ORIGINAL

A compound contains 2.70g2.70\,\mathrm{g} of aluminium and 2.40g2.40\,\mathrm{g} of oxygen. Determine its empirical formula. Use ArA_r: Al =27.0=27.0, O =16.0=16.0.

Tier 2 · Standard

5 marks
ORIGINAL

A compound is 54.5%54.5\% carbon, 9.1%9.1\% hydrogen and 36.4%36.4\% oxygen by mass. Its MrM_r is 88. Determine its empirical and molecular formulae. Use ArA_r: H =1.0=1.0, C =12.0=12.0, O =16.0=16.0.

Tier 3 · Hard

7 marks
ORIGINAL

Complete combustion of 0.900g0.900\,\mathrm{g} of a compound containing only carbon, hydrogen and oxygen produces 1.320g1.320\,\mathrm{g} of CO2\mathrm{CO_2} and 0.540g0.540\,\mathrm{g} of H2O\mathrm{H_2O}. The compound has Mr=150M_r=150. Determine its molecular formula. Use ArA_r: H =1.0=1.0, C =12.0=12.0, O =16.0=16.0.

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×actual yield/theoretical yield100\times\text{actual yield}/\text{theoretical yield}, while atom economy is 100×Mr(desired product)/Mr(reactants)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 mark
ORIGINAL

Balance the equation C3H8+O2CO2+H2O\mathrm{C_3H_8+O_2\rightarrow CO_2+H_2O} using the smallest whole-number coefficients.

Tier 2 · Standard

4 marks
ORIGINAL

25.0cm325.0\,\mathrm{cm^3} of 0.150moldm30.150\,\mathrm{mol\,dm^{-3}} hydrochloric acid reacts with excess calcium carbonate: CaCO3+2HClCaCl2+CO2+H2O\mathrm{CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O}. Calculate the maximum mass of calcium carbonate that can react. Use Mr(CaCO3)=100.1M_r(\mathrm{CaCO_3})=100.1.

Tier 3 · Hard

7 marks
ORIGINAL

Urea is made by 2NH3+CO2CO(NH2)2+H2O\mathrm{2NH_3+CO_2\rightarrow CO(NH_2)_2+H_2O}. A process uses 85.0kg85.0\,\mathrm{kg} of ammonia and 132kg132\,\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 126kg126\,\mathrm{kg} is obtained. Use molar masses in kgkmol1\mathrm{kg\,kmol^{-1}}: NH3=17.0\mathrm{NH_3}=17.0, CO2=44.0\mathrm{CO_2}=44.0, urea =60.0=60.0, H2O=18.0\mathrm{H_2O}=18.0.

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 Al3+\mathrm{Al^{3+}} ions require three SO42\mathrm{SO_4^{2-}} ions, giving Al2(SO4)3\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 mark
ORIGINAL

Write the formula of aluminium sulfate from Al3+\mathrm{Al^{3+}} and SO42\mathrm{SO_4^{2-}} ions.

Tier 2 · Standard

3 marks
ORIGINAL

Explain the nature of ionic bonding in magnesium oxide.

Tier 3 · Hard

4 marks
ORIGINAL

An element X forms the ionic compound X2(CO3)3\mathrm{X_2(CO_3)_3}. Deduce the charge on the X ion, then write the formulae of its nitrate and hydroxide. Nitrate is NO3\mathrm{NO_3^-} and hydroxide is OH\mathrm{OH^-}.

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 mark
ORIGINAL

State what is meant by a single covalent bond.

Tier 2 · Standard

3 marks
ORIGINAL

BCl3\mathrm{BCl_3} accepts a lone pair from NH3\mathrm{NH_3}. Represent the dative covalent bond in the product and state which atom donates the electron pair.

Tier 3 · Hard

4 marks
ORIGINAL

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.

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

2 marks
ORIGINAL

Complete the definition of metallic bonding.

Tier 2 · Standard

3 marks
ORIGINAL

Explain why a metal conducts electricity when solid and remains conductive when molten.

Tier 3 · Hard

4 marks
ORIGINAL

Magnesium has a higher melting point than sodium. Explain this difference using metallic bonding and the electron configurations Na:[Ne]3s1\mathrm{Na:[Ne]3s^1} and Mg:[Ne]3s2\mathrm{Mg:[Ne]3s^2}.

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

2 marks
ORIGINAL

Iodine forms crystals containing discrete I2\mathrm{I_2} molecules. State the type of crystal structure and the attractions overcome when iodine melts.

Tier 2 · Standard

4 marks
ORIGINAL

Graphite has a high melting point and conducts electricity parallel to its layers. Explain both properties in terms of its structure and bonding.

Tier 3 · Hard

6 marks
ORIGINAL

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.

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 180180^{\circ}, trigonal planar 120120^{\circ}, tetrahedral 109.5109.5^{\circ}, trigonal bipyramidal 9090^{\circ}, 120120^{\circ} and 180180^{\circ}, and octahedral 9090^{\circ} and 180180^{\circ} are the starting geometries. A common error is to count a multiple bond as several charge clouds.

Tier 1 · Easy

2 marks
ORIGINAL

State the shape and bond angle of BF3\mathrm{BF_3} around the boron atom.

Tier 2 · Standard

4 marks
ORIGINAL

Use electron-pair repulsion theory to explain the shape of NH3\mathrm{NH_3} and its H-N-H bond angle of 107107^{\circ}.

Tier 3 · Hard

5 marks
ORIGINAL

The ion BrF4\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.

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 δ\delta^- and the other atom is δ+\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 mark
ORIGINAL

The electronegativities of hydrogen and chlorine are 2.22.2 and 3.23.2, respectively. Add the correct partial charge to each atom in an HCl\mathrm{H-Cl} bond.

Tier 2 · Standard

4 marks
ORIGINAL

Both CO2\mathrm{CO_2} and SO2\mathrm{SO_2} contain polar bonds. Explain why only one of these molecules has a permanent dipole.

Tier 3 · Hard

5 marks
ORIGINAL

Fluorine is more electronegative than carbon. Compare the polarity of tetrahedral CF4\mathrm{CF_4} and tetrahedral CH3F\mathrm{CH_3F}, referring to bond dipoles and molecular shape.

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 mark
ORIGINAL

State the strongest type of intermolecular force between CH3Cl\mathrm{CH_3Cl} molecules.

Tier 2 · Standard

3 marks
ORIGINAL

Explain why ethane, C2H6\mathrm{C_2H_6}, has a higher boiling point than methane, CH4\mathrm{CH_4}.

Tier 3 · Hard

5 marks
ORIGINAL

Explain why water has a much higher boiling point than H2S\mathrm{H_2S} and why solid water is less dense than liquid water.

3.1.4.1

Enthalpy change

  • Enthalpy change, ΔH\Delta H, is the heat energy change at constant pressure; an exothermic change has ΔH<0\Delta H<0 and an endothermic change has ΔH>0\Delta H>0.
  • A standard enthalpy change applies at 100kPa100\,\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

2 marks
ORIGINAL

A reaction transfers 38kJ38\,\text{kJ} of heat from the reacting chemicals to the surroundings. State whether the reaction is exothermic or endothermic and give the sign of ΔH\Delta H.

Tier 2 · Standard

4 marks
ORIGINAL

Define standard enthalpy of formation and standard enthalpy of combustion. Include the amount of substance and the required conditions in each definition.

Tier 3 · Hard

3 marks
ORIGINAL

Complete combustion of 0.250mol0.250\,\text{mol} of a liquid releases 222kJ222\,\text{kJ} under standard conditions. Calculate its standard enthalpy of combustion.

3.1.4.2

Calorimetry

  • Use q=mcΔTq=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 ΔH=q/n\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

2 marks
ORIGINAL

A 100g100\,\text{g} sample of water warms by 6.5K6.5\,\text{K}. Use c=4.18J g1K1c=4.18\,\text{J g}^{-1}\text{K}^{-1} to calculate the heat gained by the water.

Tier 2 · Standard

5 marks
ORIGINAL

Equal 50.0cm350.0\,\text{cm}^3 portions of hydrochloric acid and sodium hydroxide are combined. Each solution has concentration 1.00mol dm31.00\,\text{mol dm}^{-3}, and the temperature rises by 6.80K6.80\,\text{K}. Assume density =1.00g cm3=1.00\,\text{g cm}^{-3} and c=4.18J g1K1c=4.18\,\text{J g}^{-1}\text{K}^{-1}. Calculate the enthalpy change per mole of water formed.

Tier 3 · Hard

6 marks
ORIGINAL

Burning 0.720g0.720\,\text{g} of propan-1-ol, C3H8O\mathrm{C_3H_8O}, heats 250g250\,\text{g} of water by 18.4K18.4\,\text{K}. Use c=4.18J g1K1c=4.18\,\text{J g}^{-1}\text{K}^{-1} and Mr(C3H8O)=60.0M_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.

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 ΔrH=ΔfH(products)ΔfH(reactants)\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 mark
ORIGINAL

For the changes AB\mathrm{A\rightarrow B} and BC\mathrm{B\rightarrow C}, the enthalpy changes are +25kJ mol1+25\,\text{kJ mol}^{-1} and 80kJ mol1-80\,\text{kJ mol}^{-1}. Calculate the enthalpy change for AC\mathrm{A\rightarrow C}.

Tier 2 · Standard

2 marks
ORIGINAL

Use ΔfH[SO2(g)]=297kJ mol1\Delta_\mathrm{f}H^\circ[\mathrm{SO_2(g)}]=-297\,\text{kJ mol}^{-1} and ΔfH[SO3(g)]=396kJ mol1\Delta_\mathrm{f}H^\circ[\mathrm{SO_3(g)}]=-396\,\text{kJ mol}^{-1} to calculate ΔrH\Delta_\mathrm{r}H^\circ for SO2(g)+12O2(g)SO3(g)\mathrm{SO_2(g)+\tfrac12O_2(g)\rightarrow SO_3(g)}.

Tier 3 · Hard

3 marks
ORIGINAL

The standard enthalpies of combustion of C3H6(g)\mathrm{C_3H_6(g)}, H2(g)\mathrm{H_2(g)} and C3H8(g)\mathrm{C_3H_8(g)} are 2058-2058, 286-286 and 2220kJ mol1-2220\,\text{kJ mol}^{-1}, respectively. Use Hess's law to calculate the enthalpy change for C3H6(g)+H2(g)C3H8(g)\mathrm{C_3H_6(g)+H_2(g)\rightarrow C_3H_8(g)}.

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 ΔH=E(bonds broken)E(bonds formed)\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

2 marks
ORIGINAL

Define the term mean bond enthalpy.

Tier 2 · Standard

3 marks
ORIGINAL

Use the mean bond enthalpies E(HH)=436E(\mathrm{H-H})=436, E(ClCl)=243E(\mathrm{Cl-Cl})=243 and E(HCl)=431kJ mol1E(\mathrm{H-Cl})=431\,\text{kJ mol}^{-1} to estimate ΔH\Delta H for H2(g)+Cl2(g)2HCl(g)\mathrm{H_2(g)+Cl_2(g)\rightarrow2HCl(g)}.

Tier 3 · Hard

4 marks
ORIGINAL

Estimate the enthalpy change for C2H4(g)+H2(g)C2H6(g)\mathrm{C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)} using E(C=C)=612E(\mathrm{C{=}C})=612, E(HH)=436E(\mathrm{H-H})=436, E(CC)=348E(\mathrm{C-C})=348 and E(CH)=413kJ mol1E(\mathrm{C-H})=413\,\text{kJ mol}^{-1}. Explain why a value obtained from standard formation enthalpies may differ.

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

2 marks
ORIGINAL

Define activation energy.

Tier 2 · Standard

3 marks
ORIGINAL

Explain why most collisions between reactant particles do not produce a reaction.

Tier 3 · Hard

4 marks
ORIGINAL

A reacting mixture undergoes 2.0×10102.0\times10^{10} molecular collisions each second. A fraction 4.0×1064.0\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.

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

2 marks
ORIGINAL

State the quantity represented on each axis of a Maxwell–Boltzmann distribution for a gas.

Tier 2 · Standard

4 marks
ORIGINAL

Describe how to add a higher-temperature curve to a Maxwell–Boltzmann distribution while keeping the number of gas molecules constant.

Tier 3 · Hard

5 marks
ORIGINAL

Two equal samples of the same gas are at temperatures T1T_1 and T2T_2, where T2>T1T_2>T_1. A fixed energy EE^* 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 EE^*.

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 EaE_\mathrm{a}, so a greater fraction of collisions can react.
  • A common error is to say that temperature lowers activation energy; without a catalyst, EaE_\mathrm{a} is unchanged and the energy distribution changes.

Tier 1 · Easy

1 mark
ORIGINAL

State the effect of increasing temperature on the rate of a reaction, with all other conditions unchanged.

Tier 2 · Standard

4 marks
ORIGINAL

Use a Maxwell–Boltzmann distribution to explain why a small temperature increase can cause a large increase in reaction rate.

Tier 3 · Hard

5 marks
ORIGINAL

In the first 10.0s10.0\,\text{s} of a reaction, 8.4cm38.4\,\text{cm}^3 of gas forms at 25C25\,^{\circ}\text{C} and 14.1cm314.1\,\text{cm}^3 forms at 35C35\,^{\circ}\text{C}. Calculate the factor by which the average rate over this interval increases, then explain the change using molecular energies.

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

2 marks
ORIGINAL

State why increasing the concentration of a dissolved reactant usually increases reaction rate.

Tier 2 · Standard

3 marks
ORIGINAL

A gaseous reacting mixture is compressed to half its original volume at constant temperature. Explain why its reaction rate increases.

Tier 3 · Hard

5 marks
ORIGINAL

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.

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 EaE_\mathrm{a} threshold left increases the area representing molecules able to react.
  • A catalyst does not change ΔH\Delta H or the equilibrium position; it speeds both forward and reverse reactions and allows equilibrium to be reached sooner.

Tier 1 · Easy

2 marks
ORIGINAL

Define a catalyst.

Tier 2 · Standard

4 marks
ORIGINAL

Use a Maxwell–Boltzmann distribution to explain how a catalyst increases the rate of a gas-phase reaction at constant temperature.

Tier 3 · Hard

5 marks
ORIGINAL

An exothermic reaction has ΔH=35kJ mol1\Delta H=-35\,\text{kJ mol}^{-1} and an uncatalysed forward activation energy of 145kJ mol1145\,\text{kJ mol}^{-1}. A catalyst lowers the forward activation energy to 82kJ mol182\,\text{kJ mol}^{-1}. Calculate the reverse activation energy for each route and state why ΔH\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

2 marks
ORIGINAL

State two features of a reversible reaction at dynamic equilibrium.

Tier 2 · Standard

3 marks
ORIGINAL

For N2(g)+3H2(g)2NH3(g)\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.

Tier 3 · Hard

6 marks
ORIGINAL

Sulfur trioxide is made by the exothermic equilibrium 2SO2(g)+O2(g)2SO3(g)\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.

3.1.6.2

Equilibrium constant Kc for homogeneous systems

  • For aA+bBcC+dDaA+bB\rightleftharpoons cC+dD, write Kc=[C]c[D]d/([A]a[B]b)K_\mathrm{c}=[C]^c[D]^d/([A]^a[B]^b) using equilibrium concentrations in mol dm3\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 KcK_\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 KcK_\mathrm{c}. Temperature can change it: heating increases KcK_\mathrm{c} for an endothermic forward reaction and decreases it for an exothermic one.

Tier 1 · Easy

1 mark
ORIGINAL

Write the expression for KcK_\mathrm{c} for H2(g)+I2(g)2HI(g)\mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)}.

Tier 2 · Standard

3 marks
ORIGINAL

At equilibrium, [H2]=0.200mol dm3[\mathrm{H_2}]=0.200\,\text{mol dm}^{-3}, [I2]=0.300mol dm3[\mathrm{I_2}]=0.300\,\text{mol dm}^{-3} and [HI]=1.20mol dm3[\mathrm{HI}]=1.20\,\text{mol dm}^{-3} for H2(g)+I2(g)2HI(g)\mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)}. Calculate KcK_\mathrm{c} and state its units.

Tier 3 · Hard

6 marks
ORIGINAL

For the exothermic equilibrium A(g)+B(g)2C(g)\mathrm{A(g)+B(g)\rightleftharpoons2C(g)}, a 2.00dm32.00\,\text{dm}^3 vessel initially contains 0.800mol0.800\,\text{mol} of A, 0.600mol0.600\,\text{mol} of B and no C. At equilibrium it contains 0.500mol0.500\,\text{mol} of C. Calculate KcK_\mathrm{c} and predict the effect of increasing temperature on its value.

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 H2O\mathrm{H_2O} and H+\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

2 marks
ORIGINAL

Determine the oxidation state of manganese in MnO4\mathrm{MnO_4^-}.

Tier 2 · Standard

3 marks
ORIGINAL

Combine the half-equations Fe2+Fe3++e\mathrm{Fe^{2+}\rightarrow Fe^{3+}+e^-} and Cl2+2e2Cl\mathrm{Cl_2+2e^-\rightarrow2Cl^-} to give the overall redox equation.

Tier 3 · Hard

5 marks
ORIGINAL

In acidic solution, dichromate(VI) ions oxidise Sn2+\mathrm{Sn^{2+}} ions to Sn4+\mathrm{Sn^{4+}}. Construct the overall ionic 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

2 marks
ORIGINAL

Define the lattice enthalpy of formation of MgCl2\mathrm{MgCl_2}.

Tier 2 · Standard

4 marks
ORIGINAL

For NaCl(s)\mathrm{NaCl(s)}, ΔfH=411kJmol1\Delta_fH^\circ=-411\,\mathrm{kJ\,mol^{-1}}. The atomisation enthalpy of sodium is +108kJmol1+108\,\mathrm{kJ\,mol^{-1}}, half the chlorine bond enthalpy is +121kJmol1+121\,\mathrm{kJ\,mol^{-1}}, the first ionisation energy of sodium is +496kJmol1+496\,\mathrm{kJ\,mol^{-1}} and the first electron affinity of chlorine is 349kJmol1-349\,\mathrm{kJ\,mol^{-1}}. Calculate the lattice enthalpy of formation of NaCl\mathrm{NaCl}.

Tier 3 · Hard

4 marks
ORIGINAL

The lattice enthalpy of formation of MgCl2\mathrm{MgCl_2} is 2526kJmol1-2526\,\mathrm{kJ\,mol^{-1}}. The hydration enthalpies of Mg2+\mathrm{Mg^{2+}} and Cl\mathrm{Cl^-} are 1920-1920 and 364kJmol1-364\,\mathrm{kJ\,mol^{-1}} respectively. Calculate the enthalpy of solution of MgCl2\mathrm{MgCl_2}.

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 ΔS=S(products)S(reactants)\Delta S^\circ=\sum S^\circ(\text{products})-\sum S^\circ(\text{reactants}), including coefficients.
  • Use ΔG=ΔHTΔS\Delta G=\Delta H-T\Delta S with temperature in kelvin and consistent energy units; entropy values in JK1mol1\mathrm{J\,K^{-1}\,mol^{-1}} usually need converting to kJK1mol1\mathrm{kJ\,K^{-1}\,mol^{-1}}.
  • Under the stated conditions a change is thermodynamically feasible when ΔG\Delta G is zero or negative, but feasibility does not imply that the reaction is fast.
  • On a graph of ΔG\Delta G against TT, the intercept is ΔH\Delta H and the gradient is ΔS-\Delta S. A common error is to give the gradient as +ΔS+\Delta S.

Tier 1 · Easy

2 marks
ORIGINAL

For a reaction, the total standard entropy of the products is 465JK1mol1465\,\mathrm{J\,K^{-1}\,mol^{-1}} and that of the reactants is 392JK1mol1392\,\mathrm{J\,K^{-1}\,mol^{-1}}. Calculate ΔS\Delta S^\circ.

Tier 2 · Standard

4 marks
ORIGINAL

A reaction has ΔH=+42.0kJmol1\Delta H=+42.0\,\mathrm{kJ\,mol^{-1}} and ΔS=+135JK1mol1\Delta S=+135\,\mathrm{J\,K^{-1}\,mol^{-1}}. Calculate ΔG\Delta G at 350K350\,\mathrm{K} and state whether the reaction is feasible at this temperature.

Tier 3 · Hard

5 marks
ORIGINAL

For a reaction, ΔG=+18.0kJmol1\Delta G=+18.0\,\mathrm{kJ\,mol^{-1}} at 300K300\,\mathrm{K} and ΔG=12.0kJmol1\Delta G=-12.0\,\mathrm{kJ\,mol^{-1}} at 500K500\,\mathrm{K}. Assume ΔH\Delta H and ΔS\Delta S are constant. Determine ΔH\Delta H, ΔS\Delta S and the temperature at which the reaction first becomes feasible.

3.1.9.1

Rate equations (A-level only)

  • A rate equation has the form rate=k[A]m[B]n\text{rate}=k[A]^m[B]^n; each order is determined experimentally and is restricted here to 00, 11 or 22.
  • The overall order is the sum of the individual orders. Derive the units of kk by rearranging the specific rate equation rather than memorising one set of units.
  • Increasing temperature increases kk. The Arrhenius equation is k=AeEa/(RT)k=Ae^{-E_a/(RT)}, and lnk=Ea/(RT)+lnA\ln k=-E_a/(RT)+\ln A gives a straight line against 1/T1/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

2 marks
ORIGINAL

A reaction has rate equation rate=k[A]2[B]\text{rate}=k[A]^2[B]. State the factor by which the rate changes when [A][A] is halved and [B][B] is doubled at constant temperature.

Tier 2 · Standard

4 marks
ORIGINAL

For rate=k[A][B]2\text{rate}=k[A][B]^2, the rate is 4.80×104moldm3s14.80\times10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}} when [A]=0.200moldm3[A]=0.200\,\mathrm{mol\,dm^{-3}} and [B]=0.0500moldm3[B]=0.0500\,\mathrm{mol\,dm^{-3}}. Calculate kk and give its units.

Tier 3 · Hard

5 marks
ORIGINAL

For this calculation use R=8.31R=8.31. A reaction has rate constants 1.80×103s11.80\times10^{-3}\,\mathrm{s^{-1}} at 298K298\,\mathrm{K} and 7.20×103s17.20\times10^{-3}\,\mathrm{s^{-1}} at 318K318\,\mathrm{K}. Calculate EaE_a in kJmol1\mathrm{kJ\,mol^{-1}} from ln(k2/k1)=(Ea/R)(1/T21/T1)\ln(k_2/k_1)=-(E_a/R)(1/T_2-1/T_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 kk.
  • 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 mark
ORIGINAL

At constant temperature, doubling [X][X] while all other concentrations remain unchanged doubles the initial rate. Deduce the order with respect to XX.

Tier 2 · Standard

5 marks
ORIGINAL

Three invented kinetic trials gave these values. Trial 1: [A]=0.100[A]=0.100, [B]=0.200[B]=0.200, rate =1.50×104=1.50\times10^{-4}; trial 2: [A]=0.200[A]=0.200, [B]=0.200[B]=0.200, rate =6.00×104=6.00\times10^{-4}; trial 3: [A]=0.200[A]=0.200, [B]=0.400[B]=0.400, rate =6.00×104=6.00\times10^{-4}. Concentrations are in moldm3\mathrm{mol\,dm^{-3}} and rates in moldm3s1\mathrm{mol\,dm^{-3}\,s^{-1}}. Deduce the rate equation and calculate kk with units.

Tier 3 · Hard

4 marks
ORIGINAL

The experimental rate equation for a reaction is rate=k[X][Y]2\text{rate}=k[X][Y]^2. Mechanism I has a single slow step X+Yproducts\mathrm{X+Y\rightarrow products}. Mechanism II has a fast equilibrium X+YI\mathrm{X+Y\rightleftharpoons I} followed by the slow step I+Yproducts\mathrm{I+Y\rightarrow products}. Determine which mechanism is consistent with the rate equation and explain your choice.

3.1.10

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

  • For a gas mixture, pi=xiPtotalp_i=x_iP_{total}, where the mole fraction is xi=ni/ntotalx_i=n_i/n_{total}; use equilibrium amounts to find equilibrium partial pressures.
  • Construct KpK_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 KpK_p; only a temperature change changes its value.
  • A catalyst speeds attainment of equilibrium but changes neither the equilibrium position nor KpK_p. Keep pressure units consistent because the units of KpK_p follow from the expression.

Tier 1 · Easy

2 marks
ORIGINAL

A gas mixture contains 2.00mol2.00\,\mathrm{mol} of AA and 3.00mol3.00\,\mathrm{mol} of BB at a total pressure of 500kPa500\,\mathrm{kPa}. Calculate the partial pressure of each gas.

Tier 2 · Standard

5 marks
ORIGINAL

At equilibrium for N2O4(g)2NO2(g)\mathrm{N_2O_4(g)\rightleftharpoons2NO_2(g)}, there are 0.300mol0.300\,\mathrm{mol} of N2O4\mathrm{N_2O_4} and 0.400mol0.400\,\mathrm{mol} of NO2\mathrm{NO_2} at a total pressure of 2.00×102kPa2.00\times10^2\,\mathrm{kPa}. Calculate KpK_p, including units.

Tier 3 · Hard

6 marks
ORIGINAL

Initially, a vessel contains 1.00mol1.00\,\mathrm{mol} of H2\mathrm{H_2} and 1.00mol1.00\,\mathrm{mol} of I2\mathrm{I_2} for H2(g)+I2(g)2HI(g)\mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)}. At equilibrium there are 1.20mol1.20\,\mathrm{mol} of HI\mathrm{HI} and the total pressure is 250kPa250\,\mathrm{kPa}. Calculate KpK_p. State the effect of increasing the total pressure at constant temperature on the equilibrium position and on KpK_p.

3.1.11.1

Electrode potentials and cells (A-level only)

  • Standard electrode potentials are measured relative to the standard hydrogen electrode at 298K298\,\mathrm{K}, 100kPa100\,\mathrm{kPa} and ion concentrations of 1.00moldm31.00\,\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 Ecell=EpositiveEnegativeE^\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

3 marks
ORIGINAL

Given E(Cu2+/Cu)=+0.34VE^\circ(\mathrm{Cu^{2+}/Cu})=+0.34\,\mathrm{V} and E(Zn2+/Zn)=0.76VE^\circ(\mathrm{Zn^{2+}/Zn})=-0.76\,\mathrm{V}, calculate EcellE^\circ_{cell} and identify the species reduced.

Tier 2 · Standard

6 marks
ORIGINAL

The standard potentials are E(Fe3+/Fe2+)=+0.77VE^\circ(\mathrm{Fe^{3+}/Fe^{2+}})=+0.77\,\mathrm{V} and E(Sn4+/Sn2+)=+0.15VE^\circ(\mathrm{Sn^{4+}/Sn^{2+}})=+0.15\,\mathrm{V}. Calculate the EMF, write the overall reaction and give the conventional cell representation.

Tier 3 · Hard

5 marks
ORIGINAL

Use E(Cu+/Cu)=+0.52VE^\circ(\mathrm{Cu^+/Cu})=+0.52\,\mathrm{V} and E(Cu2+/Cu+)=+0.15VE^\circ(\mathrm{Cu^{2+}/Cu^+})=+0.15\,\mathrm{V} to determine whether Cu+\mathrm{Cu^+} ions can disproportionate under standard conditions. Give the equation and calculate EcellE^\circ_{cell}.

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 LiLi++e\mathrm{Li\rightarrow Li^++e^-} and the positive reaction is Li++CoO2+eLi+[CoO2]\mathrm{Li^++CoO_2+e^-\rightarrow Li^+[CoO_2]^-}.
  • An alkaline hydrogen–oxygen fuel cell is supplied continuously with reactants; its overall reaction is 2H2+O22H2O\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

2 marks
ORIGINAL

State one difference between a rechargeable cell and a hydrogen–oxygen fuel cell.

Tier 2 · Standard

4 marks
ORIGINAL

The alkaline fuel-cell half-equations are H2+2OH2H2O+2e\mathrm{H_2+2OH^-\rightarrow2H_2O+2e^-} and O2+2H2O+4e4OH\mathrm{O_2+2H_2O+4e^-\rightarrow4OH^-}. Deduce the overall equation and explain how the reactions generate a current.

Tier 3 · Hard

5 marks
ORIGINAL

An alkaline hydrogen–oxygen fuel cell uses couples with electrode potentials +0.40V+0.40\,\mathrm{V} and 0.83V-0.83\,\mathrm{V}. Calculate its EMF. Then give one benefit and two risks or limitations of using hydrogen fuel cells in vehicles.

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 H+\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

2 marks
ORIGINAL

In NH3+H2ONH4++OH\mathrm{NH_3+H_2O\rightleftharpoons NH_4^++OH^-}, identify the acid and the base on the left-hand side.

Tier 2 · Standard

3 marks
ORIGINAL

For HSO4+H2OSO42+H3O+\mathrm{HSO_4^-+H_2O\rightleftharpoons SO_4^{2-}+H_3O^+}, identify both conjugate acid–base pairs.

Tier 3 · Hard

4 marks
ORIGINAL

Use two equations with water to show that H2PO4\mathrm{H_2PO_4^-} is amphoteric. Identify its role in each equation.

3.1.12.2

Definition and determination of pH (A-level only)

  • The pH scale is logarithmic: pH=log10[H+]\mathrm{pH}=-\log_{10}[H^+], with [H+][H^+] in moldm3\mathrm{mol\,dm^{-3}}.
  • Reverse the logarithm with [H+]=10pH[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+][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+][H^+].

Tier 1 · Easy

2 marks
ORIGINAL

Calculate the pH of a solution for which [H+]=3.2×103moldm3[H^+]=3.2\times10^{-3}\,\mathrm{mol\,dm^{-3}}.

Tier 2 · Standard

2 marks
ORIGINAL

A solution has pH 3.683.68. Calculate [H+][H^+] in moldm3\mathrm{mol\,dm^{-3}}.

Tier 3 · Hard

5 marks
ORIGINAL

27.5cm327.5\,\mathrm{cm^3} of 0.145moldm30.145\,\mathrm{mol\,dm^{-3}} hydrochloric acid is mixed with 32.5cm332.5\,\mathrm{cm^3} of 0.0730moldm30.0730\,\mathrm{mol\,dm^{-3}} nitric acid. Calculate the pH of the mixture. Assume volumes are additive.

3.1.12.3

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

  • Water dissociates slightly and its ionic product is Kw=[H+][OH]K_w=[H^+][OH^-]; the value of KwK_w depends on temperature.
  • For a strong base, first use its formula and complete dissociation to find [OH][OH^-], then calculate [H+]=Kw/[OH][H^+]=K_w/[OH^-] and hence pH.
  • In a neutral solution [H+]=[OH]=Kw[H^+]=[OH^-]=\sqrt{K_w}. Neutral pH is 7.007.00 only at a temperature where Kw=1.00×1014K_w=1.00\times10^{-14}.
  • Do not take the negative logarithm of [OH][OH^-] and report it directly as pH; that value is pOH unless it is converted using the appropriate KwK_w.

Tier 1 · Easy

3 marks
ORIGINAL

At 298K298\,\mathrm{K}, Kw=1.00×1014mol2dm6K_w=1.00\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}. Calculate the pH of 0.0250moldm30.0250\,\mathrm{mol\,dm^{-3}} sodium hydroxide.

Tier 2 · Standard

4 marks
ORIGINAL

At 298K298\,\mathrm{K}, calculate the pH of 0.00350moldm30.00350\,\mathrm{mol\,dm^{-3}} barium hydroxide. Use Kw=1.00×1014mol2dm6K_w=1.00\times10^{-14}\,\mathrm{mol^2\,dm^{-6}} and assume complete dissociation.

Tier 3 · Hard

5 marks
ORIGINAL

At a higher temperature, Kw=4.00×1014mol2dm6K_w=4.00\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}. Calculate the pH of neutral water and the pH of 0.0200moldm30.0200\,\mathrm{mol\,dm^{-3}} potassium hydroxide at this temperature.

3.1.12.4

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

  • A weak acid dissociates only slightly: HAH++A\mathrm{HA\rightleftharpoons H^++A^-}, with Ka=[H+][A]/[HA]K_a=[H^+][A^-]/[HA].
  • For a weak monoprotic acid of initial concentration cc, the approximation [H+]Kac[H^+]\approx\sqrt{K_ac} is valid only when dissociation is small compared with cc.
  • Acid strength may be expressed as pKa=log10KapK_a=-\log_{10}K_a; a larger KaK_a and smaller pKapK_a mean a stronger acid.
  • When pH is given, use [H+]=10pH[H^+]=10^{-\mathrm{pH}}, take [A]=[H+][A^-]=[H^+], and subtract the dissociated amount from the equilibrium [HA][HA] if an exact KaK_a is required.

Tier 1 · Easy

2 marks
ORIGINAL

A weak acid has Ka=1.74×105moldm3K_a=1.74\times10^{-5}\,\mathrm{mol\,dm^{-3}}. Calculate its pKapK_a.

Tier 2 · Standard

4 marks
ORIGINAL

Calculate the pH of a 0.150moldm30.150\,\mathrm{mol\,dm^{-3}} solution of a weak monoprotic acid with Ka=6.30×105moldm3K_a=6.30\times10^{-5}\,\mathrm{mol\,dm^{-3}}. Use the small-dissociation approximation.

Tier 3 · Hard

5 marks
ORIGINAL

A 0.0800moldm30.0800\,\mathrm{mol\,dm^{-3}} solution of a weak monoprotic acid has pH 2.402.40. Calculate KaK_a without assuming that the equilibrium acid concentration equals its initial concentration, and hence calculate pKapK_a.

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 77 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, pH=pKa\mathrm{pH}=pK_a at half-equivalence. Do not confuse half-equivalence with the equivalence point.

Tier 1 · Easy

2 marks
ORIGINAL

A strong acid–weak base titration has a steep pH change from 3.83.8 to 6.56.5. Methyl orange changes colour from pH 3.13.1 to 4.44.4, while phenolphthalein changes from pH 8.38.3 to 10.010.0. Select the suitable indicator and explain your choice.

Tier 2 · Standard

3 marks
ORIGINAL

At 298K298\,\mathrm{K}, a flask holds 25.0cm325.0\,\mathrm{cm^3} of 0.120moldm30.120\,\mathrm{mol\,dm^{-3}} hydrochloric acid. A burette supplies 0.125moldm30.125\,\mathrm{mol\,dm^{-3}} sodium hydroxide. Calculate the equivalence volume and state the approximate pH at equivalence.

Tier 3 · Hard

5 marks
ORIGINAL

A flask contains 20.0cm320.0\,\mathrm{cm^3} of 0.150moldm30.150\,\mathrm{mol\,dm^{-3}} weak monoprotic acid with Ka=2.50×105moldm3K_a=2.50\times10^{-5}\,\mathrm{mol\,dm^{-3}}. Sodium hydroxide of concentration 0.125moldm30.125\,\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 77.

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, A\mathrm{A^-} removes added H+\mathrm{H^+} and HA\mathrm{HA} removes added OH\mathrm{OH^-}; in a basic buffer, the weak base removes H+\mathrm{H^+} and its conjugate acid removes OH\mathrm{OH^-}.
  • For an acidic buffer, [H+]=Ka[HA]/[A][H^+]=K_a[HA]/[A^-], equivalently pH=pKa+log10([A]/[HA])\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

3 marks
ORIGINAL

An acidic buffer contains equal concentrations of a weak acid and its conjugate base. The acid has Ka=1.80×105moldm3K_a=1.80\times10^{-5}\,\mathrm{mol\,dm^{-3}}. Calculate the pH.

Tier 2 · Standard

4 marks
ORIGINAL

An acidic buffer contains HA\mathrm{HA} and A\mathrm{A^-}. Explain, using equations, how it resists small additions of acid and base.

Tier 3 · Hard

5 marks
ORIGINAL

0.500dm30.500\,\mathrm{dm^3} of a buffer contains 0.100mol0.100\,\mathrm{mol} of HA\mathrm{HA} and 0.0800mol0.0800\,\mathrm{mol} of A\mathrm{A^-}. The acid has Ka=1.80×105moldm3K_a=1.80\times10^{-5}\,\mathrm{mol\,dm^{-3}}. Calculate the pH after adding 0.0100mol0.0100\,\mathrm{mol} of hydrochloric acid. Assume the volume is unchanged.