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AQA A Level Chemistry 7405/2 Paper 2 Actual
Exam June 2026 | Complete Exam-Style
Questions | 100% Verified – Detailed Rationales –
Pass Guaranteed – A+ Graded
SECTION 1: PHYSICAL CHEMISTRY — THERMODYNAMICS AND KINETICS
Question 1 of 50
A student performs a calorimetry experiment by mixing 50.0 cm³ of 1.00 mol dm⁻³ hydrochloric
acid with 50.0 cm³ of 1.00 mol dm⁻³ sodium hydroxide in an insulated polystyrene cup. The
temperature of the mixture rises from 19.5 °C to 26.2 °C. Assuming the specific heat capacity of
the solution is 4.18 J g⁻¹ K⁻¹ and the density is 1.00 g cm⁻³, calculate the enthalpy change of
neutralisation per mole of water formed.
A. −28 kJ mol⁻¹ B. −112 kJ mol⁻¹ C. −56 kJ mol⁻¹ ✓ CORRECT D. −5.6 kJ mol⁻¹
Correct Answer: C Rationale: The total heat evolved is q = mcΔT = (100 g)(4.18 J g⁻¹ K⁻¹)(6.7
K) = 2.80 kJ, and since 0.0500 mol of water is formed from 0.0500 mol of each reagent, ΔH =
−2.80 kJ / 0.0500 mol = −56 kJ mol⁻¹. Option A arises from using only 50 g for the mass rather
than the combined 100 g, which halves the energy calculated. In practical calorimetry, always
use the total mass of the resulting solution and remember that exothermic enthalpy changes carry
a negative sign.
Question 2 of 50
For a particular chemical reaction, the standard enthalpy change ΔH° is +88 kJ mol⁻¹ and the
standard entropy change ΔS° is +120 J K⁻¹ mol⁻¹. Determine the minimum temperature, in
kelvin, at which this reaction becomes thermodynamically feasible under standard conditions.
A. 0.73 K B. 733 K ✓ CORRECT C. 1467 K D. 293 K
Correct Answer: B Rationale: A reaction becomes feasible when ΔG° ≤ 0, so setting ΔG° = ΔH°
− TΔS° = 0 gives T = ΔH°/ΔS° = 88000 J mol⁻¹ / 120 J K⁻¹ mol⁻¹ = 733 K. Option A results from
failing to convert ΔH° from kJ to J, giving a value three orders of magnitude too small. This
calculation is essential when designing industrial processes such as the Contact Process, where
temperature must be optimised between thermodynamic feasibility and kinetic rate.
Question 3 of 50
The table below shows initial rate data for the reaction A + B → C at a constant temperature.
Table
,2
Experiment [A] / mol dm⁻³ [B] / mol dm⁻³ Initial rate / mol dm⁻³ s⁻¹
1 0.10 0.10 0.20
2 0.20 0.10 0.40
3 0.10 0.20 0.80
Determine the rate equation for this reaction.
A. rate = k[A][B]² ✓ CORRECT B. rate = k[A]²[B] C. rate = k[A][B] D. rate = k[B]²
Correct Answer: A Rationale: Doubling [A] while holding [B] constant doubles the rate,
indicating first order with respect to A, whereas doubling [B] while holding [A] constant
quadruples the rate, indicating second order with respect to B, so the overall rate equation is rate
= k[A][B]². Option B incorrectly swaps the orders, which might arise from reading the columns
in reverse rather than comparing experiments systematically. When deducing orders from initial
rates data, always isolate one variable at a time across experiments.
Question 4 of 50
A kinetic study of a gas-phase reaction produces an Arrhenius plot of ln k versus 1/T. The
straight line obtained has a gradient of −5200 K. Calculate the activation energy for this reaction,
given that R = 8.31 J K⁻¹ mol⁻¹.
A. 625 kJ mol⁻¹ B. −43.2 kJ mol⁻¹ C. 6.25 kJ mol⁻¹ D. 43.2 kJ mol⁻¹ ✓ CORRECT
Correct Answer: D Rationale: From the Arrhenius equation in linearised form, ln k = ln A −
Ea/RT, the gradient equals −Ea/R, so Ea = −(gradient)(R) = 5200 × 8.31 = 43 212 J mol⁻¹ ≈ 43.2
kJ mol⁻¹. Option B incorrectly retains the negative sign from the gradient, producing a physically
impossible negative activation energy. Understanding this graphical method allows chemists to
predict how temperature changes will affect reaction rates in industrial catalytic processes.
Question 5 of 50
In an equilibrium mixture for the Haber process at 450 °C, the partial pressures are measured as
N₂ = 2.0 atm, H₂ = 6.0 atm, and NH₃ = 4.0 atm. Calculate the value of Kp for the reaction N₂(g)
+ 3H₂(g) ⇌ 2NH₃(g), including appropriate units.
A. 0.056 atm⁻² B. 27 atm² C. 0.037 atm⁻² ✓ CORRECT D. 0.333
Correct Answer: C Rationale: Substituting into Kp = (P_NH₃)² / (P_N₂ × P_H₂³) gives (4.0)² /
(2.0 × 6.0³) = = 0.037 atm⁻². Option B represents the value of the inverse expression
, 3
raised incorrectly, a common error when reversing the numerator and denominator. When writing
equilibrium expressions, always raise each partial pressure to the power of its stoichiometric
coefficient and check that the units balance to the correct dimension.
Question 6 of 50
Use the enthalpy of combustion data below to calculate the standard enthalpy of formation of
liquid ethanol, C₂H₅OH(l).
C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l) ΔH°c = −1367 kJ mol⁻¹ C(s) + O₂(g) → CO₂(g) ΔH°c
= −394 kJ mol⁻¹ H₂(g) + ½O₂(g) → H₂O(l) ΔH°c = −286 kJ mol⁻¹
A. −2019 kJ mol⁻¹ B. −279 kJ mol⁻¹ ✓ CORRECT C. +279 kJ mol⁻¹ D. −1367 kJ mol⁻¹
Correct Answer: B Rationale: Applying Hess’s law, ΔH°f = ΣΔH°c(reactants) − ΣΔH°c(products)
= [2(−394) + 3(−286)] − [−1367] = −1646 + 1367 = −279 kJ mol⁻¹. Option A arises from
summing all combustion values without reversing the sign for the target product, which violates
the fundamental cycle direction in Hess’s law calculations. Constructing a clear Born-Haber style
cycle prevents sign errors when combining thermochemical equations.
Question 7 of 50
Use the mean bond enthalpies provided to estimate the enthalpy change for the hydrogenation of
ethene: C₂H₄(g) + H₂(g) → C₂H₆(g).
Mean bond enthalpies: C=C = 612 kJ mol⁻¹, C−C = 347 kJ mol⁻¹, C−H = 413 kJ mol⁻¹, H−H =
436 kJ mol⁻¹.
A. −125 kJ mol⁻¹ ✓ CORRECT B. +125 kJ mol⁻¹ C. −1048 kJ mol⁻¹ D. −347 kJ mol⁻¹
Correct Answer: A Rationale: Bonds broken total 612 + 436 = 1048 kJ mol⁻¹, while bonds
formed total 2(413) + 347 = 1173 kJ mol⁻¹, giving ΔH = 1048 − 1173 = −125 kJ mol⁻¹. Option B
simply reverses the subtraction order, which would incorrectly suggest an endothermic process
for a reaction that is clearly exothermic. Bond enthalpy calculations are valuable in industrial
hydrogenation processes for estimating energy requirements without direct calorimetry.
Question 8 of 50
The Maxwell-Boltzmann distribution curves for a gas at two temperatures, T₁ and T₂, are plotted
on the same axes, where T₂ is greater than T₁. Which statement correctly describes the change
observed as the temperature increases from T₁ to T₂?
A. The peak of the curve increases in height and shifts to higher molecular speeds. B. The total
area under the curve decreases because fewer molecules are present. C. All molecules move with
a speed greater than the most probable speed at T₂. D. A greater proportion of molecules
possesses kinetic energy exceeding the activation energy. ✓ CORRECT
AQA A Level Chemistry 7405/2 Paper 2 Actual
Exam June 2026 | Complete Exam-Style
Questions | 100% Verified – Detailed Rationales –
Pass Guaranteed – A+ Graded
SECTION 1: PHYSICAL CHEMISTRY — THERMODYNAMICS AND KINETICS
Question 1 of 50
A student performs a calorimetry experiment by mixing 50.0 cm³ of 1.00 mol dm⁻³ hydrochloric
acid with 50.0 cm³ of 1.00 mol dm⁻³ sodium hydroxide in an insulated polystyrene cup. The
temperature of the mixture rises from 19.5 °C to 26.2 °C. Assuming the specific heat capacity of
the solution is 4.18 J g⁻¹ K⁻¹ and the density is 1.00 g cm⁻³, calculate the enthalpy change of
neutralisation per mole of water formed.
A. −28 kJ mol⁻¹ B. −112 kJ mol⁻¹ C. −56 kJ mol⁻¹ ✓ CORRECT D. −5.6 kJ mol⁻¹
Correct Answer: C Rationale: The total heat evolved is q = mcΔT = (100 g)(4.18 J g⁻¹ K⁻¹)(6.7
K) = 2.80 kJ, and since 0.0500 mol of water is formed from 0.0500 mol of each reagent, ΔH =
−2.80 kJ / 0.0500 mol = −56 kJ mol⁻¹. Option A arises from using only 50 g for the mass rather
than the combined 100 g, which halves the energy calculated. In practical calorimetry, always
use the total mass of the resulting solution and remember that exothermic enthalpy changes carry
a negative sign.
Question 2 of 50
For a particular chemical reaction, the standard enthalpy change ΔH° is +88 kJ mol⁻¹ and the
standard entropy change ΔS° is +120 J K⁻¹ mol⁻¹. Determine the minimum temperature, in
kelvin, at which this reaction becomes thermodynamically feasible under standard conditions.
A. 0.73 K B. 733 K ✓ CORRECT C. 1467 K D. 293 K
Correct Answer: B Rationale: A reaction becomes feasible when ΔG° ≤ 0, so setting ΔG° = ΔH°
− TΔS° = 0 gives T = ΔH°/ΔS° = 88000 J mol⁻¹ / 120 J K⁻¹ mol⁻¹ = 733 K. Option A results from
failing to convert ΔH° from kJ to J, giving a value three orders of magnitude too small. This
calculation is essential when designing industrial processes such as the Contact Process, where
temperature must be optimised between thermodynamic feasibility and kinetic rate.
Question 3 of 50
The table below shows initial rate data for the reaction A + B → C at a constant temperature.
Table
,2
Experiment [A] / mol dm⁻³ [B] / mol dm⁻³ Initial rate / mol dm⁻³ s⁻¹
1 0.10 0.10 0.20
2 0.20 0.10 0.40
3 0.10 0.20 0.80
Determine the rate equation for this reaction.
A. rate = k[A][B]² ✓ CORRECT B. rate = k[A]²[B] C. rate = k[A][B] D. rate = k[B]²
Correct Answer: A Rationale: Doubling [A] while holding [B] constant doubles the rate,
indicating first order with respect to A, whereas doubling [B] while holding [A] constant
quadruples the rate, indicating second order with respect to B, so the overall rate equation is rate
= k[A][B]². Option B incorrectly swaps the orders, which might arise from reading the columns
in reverse rather than comparing experiments systematically. When deducing orders from initial
rates data, always isolate one variable at a time across experiments.
Question 4 of 50
A kinetic study of a gas-phase reaction produces an Arrhenius plot of ln k versus 1/T. The
straight line obtained has a gradient of −5200 K. Calculate the activation energy for this reaction,
given that R = 8.31 J K⁻¹ mol⁻¹.
A. 625 kJ mol⁻¹ B. −43.2 kJ mol⁻¹ C. 6.25 kJ mol⁻¹ D. 43.2 kJ mol⁻¹ ✓ CORRECT
Correct Answer: D Rationale: From the Arrhenius equation in linearised form, ln k = ln A −
Ea/RT, the gradient equals −Ea/R, so Ea = −(gradient)(R) = 5200 × 8.31 = 43 212 J mol⁻¹ ≈ 43.2
kJ mol⁻¹. Option B incorrectly retains the negative sign from the gradient, producing a physically
impossible negative activation energy. Understanding this graphical method allows chemists to
predict how temperature changes will affect reaction rates in industrial catalytic processes.
Question 5 of 50
In an equilibrium mixture for the Haber process at 450 °C, the partial pressures are measured as
N₂ = 2.0 atm, H₂ = 6.0 atm, and NH₃ = 4.0 atm. Calculate the value of Kp for the reaction N₂(g)
+ 3H₂(g) ⇌ 2NH₃(g), including appropriate units.
A. 0.056 atm⁻² B. 27 atm² C. 0.037 atm⁻² ✓ CORRECT D. 0.333
Correct Answer: C Rationale: Substituting into Kp = (P_NH₃)² / (P_N₂ × P_H₂³) gives (4.0)² /
(2.0 × 6.0³) = = 0.037 atm⁻². Option B represents the value of the inverse expression
, 3
raised incorrectly, a common error when reversing the numerator and denominator. When writing
equilibrium expressions, always raise each partial pressure to the power of its stoichiometric
coefficient and check that the units balance to the correct dimension.
Question 6 of 50
Use the enthalpy of combustion data below to calculate the standard enthalpy of formation of
liquid ethanol, C₂H₅OH(l).
C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l) ΔH°c = −1367 kJ mol⁻¹ C(s) + O₂(g) → CO₂(g) ΔH°c
= −394 kJ mol⁻¹ H₂(g) + ½O₂(g) → H₂O(l) ΔH°c = −286 kJ mol⁻¹
A. −2019 kJ mol⁻¹ B. −279 kJ mol⁻¹ ✓ CORRECT C. +279 kJ mol⁻¹ D. −1367 kJ mol⁻¹
Correct Answer: B Rationale: Applying Hess’s law, ΔH°f = ΣΔH°c(reactants) − ΣΔH°c(products)
= [2(−394) + 3(−286)] − [−1367] = −1646 + 1367 = −279 kJ mol⁻¹. Option A arises from
summing all combustion values without reversing the sign for the target product, which violates
the fundamental cycle direction in Hess’s law calculations. Constructing a clear Born-Haber style
cycle prevents sign errors when combining thermochemical equations.
Question 7 of 50
Use the mean bond enthalpies provided to estimate the enthalpy change for the hydrogenation of
ethene: C₂H₄(g) + H₂(g) → C₂H₆(g).
Mean bond enthalpies: C=C = 612 kJ mol⁻¹, C−C = 347 kJ mol⁻¹, C−H = 413 kJ mol⁻¹, H−H =
436 kJ mol⁻¹.
A. −125 kJ mol⁻¹ ✓ CORRECT B. +125 kJ mol⁻¹ C. −1048 kJ mol⁻¹ D. −347 kJ mol⁻¹
Correct Answer: A Rationale: Bonds broken total 612 + 436 = 1048 kJ mol⁻¹, while bonds
formed total 2(413) + 347 = 1173 kJ mol⁻¹, giving ΔH = 1048 − 1173 = −125 kJ mol⁻¹. Option B
simply reverses the subtraction order, which would incorrectly suggest an endothermic process
for a reaction that is clearly exothermic. Bond enthalpy calculations are valuable in industrial
hydrogenation processes for estimating energy requirements without direct calorimetry.
Question 8 of 50
The Maxwell-Boltzmann distribution curves for a gas at two temperatures, T₁ and T₂, are plotted
on the same axes, where T₂ is greater than T₁. Which statement correctly describes the change
observed as the temperature increases from T₁ to T₂?
A. The peak of the curve increases in height and shifts to higher molecular speeds. B. The total
area under the curve decreases because fewer molecules are present. C. All molecules move with
a speed greater than the most probable speed at T₂. D. A greater proportion of molecules
possesses kinetic energy exceeding the activation energy. ✓ CORRECT
