By /
IAL A2 Chemistry Recap – Topics 11, 12 and 13 | Alvina (2026)

Kinetics, Entropy, Energetics and Equilibrium

Pearson Edexcel IAL A2 Chemistry — Topics 11, 12 and 13

Alvina (2026)

For use on Wednesday, 2 September 2026

01

Topic 11 — Kinetics

Initial Rates and Reaction Mechanisms

Reactants A and B react in solution according to the overall equation below. A series of initial-rate experiments is carried out at constant temperature.

2A(aq) + B(aq) ⟶ products
Experiment Initial [A]
/ mol dm−3		 Initial [B]
/ mol dm−3		 Initial rate
/ mol dm−3 s−1
1 0.100 0.100 2.40 × 10−4
2 0.200 0.100 9.60 × 10−4
3 0.200 0.300 2.88 × 10−3
4 0.300 0.200 To be calculated
  1. Use Experiments 1 and 2 to determine the order with respect to A. Explain how the data support your answer.
  2. Use Experiments 2 and 3 to determine the order with respect to B. Explain how the data support your answer.
  3. Write the rate equation and state the overall order.
  4. Calculate the value of the rate constant, k, including its units.
  5. Calculate the initial rate in Experiment 4.
  6. Consider the proposed mechanism:
    Step 1: A + B ⇌ I     fast equilibrium
    Step 2: I + A ⟶ products     slow
    Explain how this mechanism is consistent with:
    1. the experimentally determined rate equation;
    2. the overall equation.
  7. Explain why the orders in an experimentally determined rate equation cannot usually be predicted from the coefficients in the overall equation.
16 marks
02

Topic 11 — Extended A2 Kinetics

Arrhenius Analysis and Heterogeneous Catalysis

The reaction between gaseous ethene and hydrogen has the rate equation shown below.

C2H4(g) + H2(g) ⟶ C2H6(g)

Rate = k[C2H4][H2]

Experimental values of the rate constant for the uncatalysed reaction are shown below.

Temperature, T / K Rate constant, k
/ dm3 mol−1 s−1
298 1.25 × 10−4
318 1.10 × 10−3
Extended-response question: use precise A2 terminology. Your explanation of heterogeneous catalysis must describe the complete sequence occurring at the catalyst surface.
  1. State the overall order of the reaction and explain the units of the rate constant.
  2. The Arrhenius equation may be written as:
    ln k = −Ea / RT + ln A
    State what is represented by:
    1. Ea;
    2. A.
  3. Calculate the activation energy for the uncatalysed reaction using the two values of k. Use:
    ln(k2 / k1) = −Ea / R (1/T2 − 1/T1)
    R = 8.31 J K−1 mol−1
  4. If the concentrations of both reactants remain constant, calculate the factor by which the reaction rate increases when the temperature changes from 298 K to 318 K.
  5. A graph of ln k against 1/T is a straight line. State:
    1. the expression for its gradient;
    2. the expression for its intercept.
  6. Nickel can act as a heterogeneous catalyst for the reaction. Explain in detail how the nickel catalyst increases the rate. Your answer should include:
    1. adsorption at active sites;
    2. the effect on bonds in the reactants;
    3. the alternative reaction pathway;
    4. formation and desorption of the product;
    5. regeneration of active sites.
  7. Explain why nickel changes the rate of attainment of equilibrium but does not change the equilibrium composition or the equilibrium constant.
18 marks
03

Topic 12B — Lattice Energy

Born–Haber Cycle, Polarisation and Solution

The data below may be used to determine the experimental lattice energy of magnesium chloride, MgCl2(s).

Enthalpy change Value / kJ mol−1
ΔfH°[MgCl2(s)] −642
ΔatH°[Mg(s)] +148
First ionisation energy of Mg +738
Second ionisation energy of Mg +1451
Bond dissociation enthalpy of Cl2 +243
First electron affinity of Cl −349

Electrostatic theory predicts a theoretical lattice energy of −2326 kJ mol−1 for MgCl2. The following hydration enthalpies are also available.

Ion ΔhydH° / kJ mol−1
Mg2+(g) −1920
Cl−(g) −364
  1. Define lattice energy using the convention in which lattice formation is exothermic.
  2. Construct a Born–Haber cycle and calculate the experimental lattice energy of MgCl2.
  3. Calculate the difference between the experimental lattice energy and the theoretical lattice energy.
  4. Explain what the difference between these two values indicates about the bonding in MgCl2.
  5. Explain, in terms of charge density and polarisation, why the difference between experimental and theoretical lattice energies is greater for MgCl2 than for NaCl.
  6. Calculate the standard enthalpy change of solution of MgCl2.
  7. Explain why both lattice energy and hydration enthalpy become more exothermic when:
    1. ionic charge increases;
    2. ionic radius decreases.
20 marks
04

Topic 12B — Extended Explanation

Solubility, Hydration and Enthalpy of Solution

The solubilities of Group 2 hydroxides generally increase down the group, whereas the solubilities of Group 2 sulfates generally decrease down the group.

Extended-response question: compare the relative changes in lattice energy and hydration enthalpy. Simply stating that both become less exothermic will not fully explain either trend.
  1. Define:
    1. standard enthalpy change of hydration of an ion;
    2. standard enthalpy change of solution of an ionic solid.
  2. Explain why the magnitude of the hydration enthalpy of M2+ decreases from Mg2+ to Ba2+.
  3. Explain why the magnitude of the lattice energy of an ionic compound generally decreases as the radius of its cation increases.
  4. Explain fully why Group 2 hydroxides generally become more soluble down the group.
  5. Explain fully why Group 2 sulfates generally become less soluble down the group.
  6. Explain why an enthalpy change of solution value alone cannot always be used to predict whether an ionic compound will be soluble.
  7. A student states:
    “An ionic solid with an endothermic enthalpy change of solution must be insoluble.”
    Evaluate this statement using the concepts of entropy of the system, entropy of the surroundings and total entropy.
18 marks
05

Topic 12A — Entropy

Feasibility and Thermal Decomposition

Calcium carbonate decomposes when heated according to the equation:

CaCO3(s) ⟶ CaO(s) + CO2(g)

ΔrH° = +178.3 kJ mol−1
Substance S° / J K−1 mol−1
CaCO3(s) 92.9
CaO(s) 39.8
CO2(g) 213.7
  1. Calculate ΔSsystem for the reaction.
  2. Explain the sign of ΔSsystem in terms of the dispersal of matter and energy.
  3. Calculate the temperature at which ΔStotal = 0. Assume that ΔH and ΔSsystem do not change with temperature.
  4. State the temperature range over which the forward reaction is thermodynamically feasible under standard conditions.
  5. Calculate:
    1. ΔSsurroundings at 1200 K;
    2. ΔStotal at 1200 K.
  6. Explain why a positive value of ΔStotal does not mean that calcium carbonate decomposes rapidly.
  7. Distinguish between thermodynamic stability and kinetic stability with reference to this reaction.
  8. Predict and explain how increasing the partial pressure of carbon dioxide affects the decomposition equilibrium.
18 marks
06

Topic 13 — Chemical Equilibria

Kp, Compression and a New Equilibrium

Dinitrogen tetroxide and nitrogen dioxide establish the following equilibrium in a sealed container at 350 K.

N2O4(g) ⇌ 2NO2(g)

At equilibrium, the total pressure is 2.40 atm. The mole fraction of NO2 is 0.600 and the mole fraction of N2O4 is 0.400.

  1. Write the expression for Kp, including its units.
  2. Calculate the equilibrium partial pressure of each gas.
  3. Calculate Kp at 350 K.
  4. The volume of the container is suddenly halved at constant temperature. Calculate the partial pressure of each gas immediately after compression, before the equilibrium position changes.
  5. Predict the direction in which the equilibrium moves after compression. Explain your answer in terms of the numbers of moles of gas.
  6. Let the increase in the partial pressure of N2O4 as equilibrium is re-established be x atm. Show that the new equilibrium partial pressures are:
    p(N2O4) = 1.92 + x
    p(NO2) = 2.88 − 2x
  7. Use Kp = 2.16 atm to show that x satisfies:
    4x2 − 13.68x + 4.1472 = 0
    Hence calculate the new equilibrium partial pressure of each gas.
  8. Explain why compression changes the equilibrium composition but does not change Kp.
  9. The value of Kp increases when temperature is increased. Deduce whether the forward reaction is endothermic or exothermic, and explain your answer.
  10. Explain the effect of a catalyst on:
    1. the time taken to reach equilibrium;
    2. the equilibrium composition;
    3. the value of Kp.
  11. Use:
    ΔStotal = R ln K
    to calculate ΔStotal at 350 K, using the numerical value of Kp and R = 8.31 J K−1 mol−1.
  12. Comment on the extent of the forward reaction at 350 K using the magnitude of Kp.
23 marks

Question 1

  1. Comparing Experiments 1 and 2, [A] doubles while [B] remains constant. The rate increases by a factor of four.
    2m = 4
    m = 2
    The reaction is second order with respect to A.
  2. Comparing Experiments 2 and 3, [B] increases by a factor of three while [A] remains constant. The rate also increases by a factor of three.
    3n = 3
    n = 1
    The reaction is first order with respect to B.
  3. Rate = k[A]2[B]
    Overall order = 3
  4. Using Experiment 1:
    2.40 × 10−4 = k(0.100)2(0.100)

    k = 0.240 dm6 mol−2 s−1
    k = 0.240 dm6 mol−2 s−1.
  5. Rate = 0.240(0.300)2(0.200)
    Rate = 4.32 × 10−3 mol dm−3 s−1
  6. The slow step is the rate-determining step:
    Rate = k2[I][A]
    The fast equilibrium gives an intermediate concentration proportional to [A][B]:
    [I] ∝ [A][B]
    Substitution gives:
    Rate ∝ [A][B][A]
    Rate ∝ [A]2[B]
    This agrees with the experimental rate equation. Adding the two steps and cancelling intermediate I gives:
    2A + B ⟶ products
    The mechanism is consistent with both the rate equation and the overall equation.
  7. The overall equation shows only the net chemical change. It does not show the individual elementary steps or identify the rate-determining step. The orders depend on the mechanism and must normally be determined experimentally.

Question 2

  1. The reaction is first order with respect to each reactant.
    Overall order = 1 + 1 = 2
    Rate has units mol dm−3 s−1 and the concentration product has units mol2 dm−6.
    Units of k = mol dm−3 s−1 / mol2 dm−6
    = dm3 mol−1 s−1
    1. Ea is the activation energy. It is the minimum energy required for a collision to lead to reaction.
    2. A is the Arrhenius pre-exponential factor. It is related to collision frequency and the probability that collisions have a suitable orientation.
  3. ln(k2/k1) = ln[(1.10 × 10−3) / (1.25 × 10−4)]
    = ln(8.80) = 2.175
    1/318 − 1/298 = −2.1105 × 10−4 K−1
    2.175 = −Ea/8.31 (−2.1105 × 10−4)
    Ea = 8.56 × 104 J mol−1
    = 85.6 kJ mol−1
    Activation energy = 85.6 kJ mol−1.
  4. At constant concentrations:
    Rate ∝ k
    Rate factor = (1.10 × 10−3) / (1.25 × 10−4) = 8.80
    The rate increases by a factor of 8.80.
  5. Comparing:
    ln k = (−Ea/R)(1/T) + ln A
    with y = mx + c:
    Gradient = −Ea/R
    Intercept = ln A
  6. Ethene and hydrogen molecules are adsorbed onto active sites on the nickel surface. Adsorption brings the reactant molecules close together and in a suitable orientation. Electron density may be transferred between the metal surface and the adsorbed reactants. The H–H bond and the π component of the C=C bond are weakened. The catalyst provides an alternative pathway with a lower activation energy. Hydrogen atoms react with the adsorbed ethene to form ethane. Ethane is then desorbed from the surface. Desorption releases the product and leaves the active sites available for further reactant molecules. The nickel is not consumed overall.
  7. A catalyst lowers the activation energy for both the forward and reverse reactions. Both reaction rates increase, so equilibrium is reached more quickly. However, the forward and reverse rates are increased without changing the relative thermodynamic stability of the reactants and products. The equilibrium composition and Kp are unchanged. At this level, the equilibrium constant changes only when temperature changes.
Exam warning: Do not state that the catalyst gives particles more energy. It provides a different pathway with a lower activation energy.

Question 3

  1. Lattice energy is the standard enthalpy change when one mole of an ionic solid is formed from its gaseous ions under standard conditions.
    Mg2+(g) + 2Cl−(g) ⟶ MgCl2(s)
  2. ΔfH° = ΔatH°(Mg) + IE1 + IE2 + D(Cl2) + 2EA(Cl) + LE
    −642 = 148 + 738 + 1451 + 243 + 2(−349) + LE
    −642 = 1882 + LE
    LE = −2524 kJ mol−1
    Experimental lattice energy = −2524 kJ mol−1.
  3. Difference = 2524 − 2326 = 198 kJ mol−1
    The experimental value is 198 kJ mol−1 more exothermic.
  4. The theoretical value assumes spherical ions with purely electrostatic ionic attractions. The more exothermic experimental value shows that the real attractions are stronger than predicted by the purely ionic model. MgCl2 has some covalent character.
  5. Mg2+ has a higher charge and a smaller radius than Na+. It therefore has a higher charge density and greater polarising power. Mg2+ distorts the electron cloud of Cl− more strongly. Electron density is drawn towards Mg2+, giving partial sharing of electron density and increased covalent character. The purely ionic model is less accurate for MgCl2, so its theoretical and experimental lattice energies differ by a greater amount.
  6. Separating the lattice requires:
    +2524 kJ mol−1
    Hydration releases:
    −1920 + 2(−364) = −2648 kJ mol−1
    Therefore:
    ΔsolH° = +2524 − 2648 = −124 kJ mol−1
    ΔsolH° = −124 kJ mol−1.
    1. Increasing ionic charge increases the electrostatic attraction between oppositely charged ions. It also increases the ion–dipole attraction between an ion and water molecules. Lattice formation and hydration both become more exothermic.
    2. Decreasing ionic radius reduces the distance between centres of charge. The electrostatic attraction becomes stronger. Smaller ions have more exothermic hydration enthalpies and generally form lattices with more exothermic lattice energies.

Question 4

    1. The standard enthalpy change of hydration of an ion is the enthalpy change when one mole of gaseous ions becomes hydrated to form aqueous ions under standard conditions.
    2. The standard enthalpy change of solution is the enthalpy change when one mole of an ionic solid dissolves in sufficient water to form an infinitely dilute solution under standard conditions.
  2. The ionic charge remains +2, but ionic radius increases from Mg2+ to Ba2+. Charge density decreases and water molecules cannot approach the centre of charge as closely. Ion–dipole attractions become weaker, so hydration enthalpy becomes less exothermic down the group.
  3. Increasing cation radius increases the distance between the centres of oppositely charged ions. The electrostatic attractions become weaker. Less energy is released during lattice formation, so lattice energy becomes less exothermic.
  4. Down Group 2, both the lattice energy and cation hydration enthalpy become less exothermic. For the hydroxides, OH− is relatively small. Increasing the size of M2+ causes a significant reduction in the magnitude of the lattice energy. The energy required to separate the lattice therefore decreases more rapidly than the hydration enthalpy becomes less exothermic. The enthalpy change of solution becomes more favourable. Group 2 hydroxides generally become more soluble down the group.
  5. SO42− is a large ion. The distance between the ions is already relatively large, so increasing the radius of M2+ has a smaller proportional effect on lattice energy. However, the hydration enthalpy of M2+ becomes significantly less exothermic down the group as charge density decreases. The reduction in favourable hydration enthalpy is greater than the reduction in the energy needed to separate the lattice. The enthalpy change of solution therefore becomes less favourable. Group 2 sulfates generally become less soluble down the group.
  6. Solubility depends on the total entropy change, not only the enthalpy change.
    ΔStotal = ΔSsystem + ΔSsurroundings
    Dissolving usually increases the dispersal of ions, which may give a positive ΔSsystem. However, hydration can also order water molecules. Enthalpy change of solution alone does not include all factors controlling the feasibility and extent of dissolution.
  7. The statement is incorrect. For an endothermic dissolution:
    ΔH > 0
    ΔSsurroundings = −ΔH/T < 0
    However, separating an ordered ionic lattice into dispersed aqueous ions may produce a positive ΔSsystem. If the positive entropy change of the system is greater than the magnitude of the negative entropy change of the surroundings:
    ΔStotal > 0
    Dissolution can be thermodynamically feasible even when the enthalpy change of solution is endothermic.
Exam technique: A correct trend explanation must compare which changes more: lattice energy or hydration enthalpy.

Question 5

  1. ΔSsystem = ΣS°(products) − ΣS°(reactants)
    ΔSsystem = (39.8 + 213.7) − 92.9 = +160.6 J K−1 mol−1
  2. A gas is produced from solid reactants. Gas particles have much greater freedom of movement and a much larger number of possible arrangements than particles in a solid. Energy is also distributed among a greater number of accessible arrangements. The dispersal of matter and energy increases, so ΔSsystem is positive.
  3. At the feasibility boundary:
    ΔStotal = 0
    Therefore:
    T = ΔH / ΔSsystem
    T = 178300 / 160.6 = 1.11 × 103 K
    Boundary temperature = approximately 1110 K.
  4. The forward reaction is feasible above approximately 1110 K under the stated standard conditions.
    1. ΔSsurroundings = −178300/1200 = −148.6 J K−1 mol−1
    2. ΔStotal = 160.6 − 148.6 = +12.0 J K−1 mol−1
  6. A positive ΔStotal indicates thermodynamic feasibility. It gives no direct information about the reaction rate. The reaction may still have a large activation energy. A feasible reaction may therefore occur very slowly.
  7. Thermodynamic stability concerns whether a change is feasible according to the sign of ΔStotal. Kinetic stability concerns whether the reaction rate is sufficiently slow for the substance to persist. Above the boundary temperature, decomposition may be thermodynamically feasible. However, calcium carbonate can remain present temporarily if the activation-energy barrier makes the reaction slow. A substance can be thermodynamically unstable but kinetically stable.
  8. Increasing the partial pressure of CO2 increases the amount of a product. The reverse reaction is favoured until equilibrium is re-established. The equilibrium moves towards CaCO3, reducing the extent of decomposition. The equilibrium constant remains unchanged if temperature is unchanged.

Question 6

  1. Kp = p(NO2)2 / p(N2O4)
    Units = atm.
  2. p(NO2) = 0.600 × 2.40 = 1.44 atm
    p(N2O4) = 0.400 × 2.40 = 0.960 atm
  3. Kp = (1.44)2 / 0.960 = 2.16 atm
  4. Halving the volume immediately doubles both partial pressures.
    p(NO2) = 2.88 atm
    p(N2O4) = 1.92 atm
  5. The left-hand side contains one mole of gas for every two moles of gas on the right-hand side. Compression moves the equilibrium to the left, towards the side with fewer moles of gas.
  6. The reverse reaction is:
    2NO2 &longrightarrow; N2O4
    If N2O4 increases by x atm, NO2 must decrease by 2x atm.
    p(N2O4) = 1.92 + x
    p(NO2) = 2.88 − 2x
  7. 2.16 = (2.88 − 2x)2 / (1.92 + x)
    (2.88 − 2x)2 = 2.16(1.92 + x)
    8.2944 − 11.52x + 4x2 = 4.1472 + 2.16x
    4x2 − 13.68x + 4.1472 = 0
    The physically possible root is:
    x = 0.336 atm
    Therefore:
    p(N2O4) = 1.92 + 0.336 = 2.26 atm
    p(NO2) = 2.88 − 2(0.336) = 2.21 atm
    New equilibrium partial pressures: N2O4 = 2.26 atm and NO2 = 2.21 atm.
  8. Compression initially changes the partial pressures, so the mixture is no longer at equilibrium. The forward and reverse rates are temporarily unequal, causing the composition to change. Kp remains unchanged because the temperature remains at 350 K.
  9. Increasing temperature increases Kp, so the equilibrium contains proportionally more NO2 at the higher temperature. Heating favours the forward reaction. The forward reaction is endothermic.
  10. A catalyst lowers the activation energy for both forward and reverse reactions.
    • Equilibrium is reached more quickly.
    • The equilibrium composition is unchanged.
    • Kp is unchanged.
  11. ΔStotal = 8.31 ln(2.16)
    ΔStotal = 8.31 × 0.770 = +6.40 J K−1 mol−1
  12. Kp is greater than 1, so products are favoured relative to reactants. However, Kp is not extremely large. The forward reaction occurs to a significant extent, but appreciable quantities of both gases remain at equilibrium.
Scroll to Top