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AQA A-LEVEL CHEMISTRY PAPER 2 2026/2027 |
CHEMISTRY | 100 VERIFIED Q&A | DETAILED
RATIONALES | PASS GUARANTEED – A+ GRADED
SECTION 1: THERMODYNAMICS & ENERGETICS - Questions 1-15
Q1: Enthalpy of Formation Definition
The student is reviewing thermodynamics. The standard enthalpy of formation of a compound is the
enthalpy change when:
A. One mole of the compound is formed from its elements under standard conditions
B. One mole of the compound is burned in oxygen
C. One mole of bonds is broken
D. One mole of gas dissolves in water
Correct Answer: A
Rationale: Standard enthalpy of formation (Δ𝐻𝑓∘) is the enthalpy change when one mole of a
compound is formed from its constituent elements in their standard states under standard
conditions (298 K, 100 kPa). [100% CORRECT]
Q2: Enthalpy of Combustion
The student is reviewing combustion. The standard enthalpy of combustion of ethanol is −1367 kJ
mol⁻¹. This means:
A. 1367 kJ is released when one mole of ethanol is completely burned in oxygen
B. 1367 kJ is absorbed when one mole of ethanol is burned
C. 1367 kJ is released when one mole of ethanol is formed
D. 1367 kJ is absorbed when one mole of ethanol is formed
Correct Answer: A
Rationale: Negative sign indicates exothermic; 1367 kJ released per mole burned completely in
oxygen under standard conditions. [100% CORRECT]
Q3: Calorimetry Calculation
The student is reviewing calorimetry. 50.0 g of water is heated from 20.0°C to 30.0°C. Specific heat
capacity = 4.18 J g⁻¹ K⁻¹. Heat absorbed is:
A. 2090 J
B. 418 J
C. 209 J
D. 4180 J
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Correct Answer: A
Rationale: 𝑞 = 𝑚𝑐Δ𝑇 = 50.0 × 4.18 × 10.0 = 2090 J. [100% CORRECT]
Q4: Hess’s Law Application
The student is reviewing Hess’s Law. Given:
C(s) + O₂(g) → CO₂(g) ΔH = −394 kJ mol⁻¹
CO(g) + ½O₂(g) → CO₂(g) ΔH = −283 kJ mol⁻¹
Calculate ΔH for C(s) + ½O₂(g) → CO(g).
A. −111 kJ mol⁻¹
B. +111 kJ mol⁻¹
C. −677 kJ mol⁻¹
D. +677 kJ mol⁻¹
Correct Answer: A
Rationale: Reverse second equation: CO₂ → CO + ½O₂ ΔH = +283. Add to first: C + O₂ → CO₂ −394.
Sum: C + ½O₂ → CO ΔH = −394 + 283 = −111 kJ mol⁻¹. [100% CORRECT]
Q5: Mean Bond Enthalpy
The student is reviewing bond enthalpies. Mean bond enthalpy is:
A. The average energy required to break one mole of a covalent bond in gaseous molecules
B. The energy released when one mole of bonds forms
C. The energy required to break one mole of ionic bonds
D. The energy required to melt one mole of a solid
Correct Answer: A
Rationale: Mean bond enthalpy is the average energy needed to break one mole of a given covalent
bond in gaseous molecules, averaged over different compounds. [100% CORRECT]
Q6: Bond Enthalpy Calculation
The student is reviewing calculations. Use bond enthalpies (kJ mol⁻¹): H–H = 436, Cl–Cl = 242, H–Cl =
431. ΔH for H₂ + Cl₂ → 2HCl is:
A. −184 kJ mol⁻¹
B. +184 kJ mol⁻¹
C. −247 kJ mol⁻¹
D. +247 kJ mol⁻¹
Correct Answer: A
Rationale: Bonds broken: H–H + Cl–Cl = 436 + 242 = 678. Bonds formed: 2 × H–Cl = 862. ΔH = 678 −
862 = −184 kJ mol⁻¹. [100% CORRECT]
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Q7: Entropy Definition
The student is reviewing entropy. Entropy is a measure of:
A. Disorder or randomness of a system
B. Enthalpy change
C. Activation energy
D. Rate of reaction
Correct Answer: A
Rationale: Entropy (S) quantifies disorder; gases have higher entropy than liquids, which have higher
entropy than solids. [100% CORRECT]
Q8: Entropy Change Sign
The student is reviewing entropy changes. For the reaction CaCO₃(s) → CaO(s) + CO₂(g), ΔS is:
A. Positive
B. Negative
C. Zero
D. Cannot be determined
Correct Answer: A
Rationale: A gas is produced from a solid, increasing disorder; ΔS is positive. [100% CORRECT]
Q9: Gibbs Free Energy Equation
The student is reviewing spontaneity. The Gibbs free energy equation is:
A. ΔG = ΔH − TΔS
B. ΔG = ΔH + TΔS
C. ΔG = TΔS − ΔH
D. ΔG = ΔH × TΔS
Correct Answer: A
Rationale: ΔG = ΔH − TΔS. Reaction feasible when ΔG < 0. [100% CORRECT]
Q10: Feasibility Conditions
The student is reviewing feasibility. A reaction is always feasible when:
A. ΔH is negative and ΔS is positive
B. ΔH is positive and ΔS is negative
C. ΔH is positive and ΔS is positive
D. ΔH is negative and ΔS is negative
Correct Answer: A
Rationale: ΔG = ΔH − TΔS. If ΔH < 0 and ΔS > 0, ΔG is always negative. [100% CORRECT]
AQA A-LEVEL CHEMISTRY PAPER 2 2026/2027 |
CHEMISTRY | 100 VERIFIED Q&A | DETAILED
RATIONALES | PASS GUARANTEED – A+ GRADED
SECTION 1: THERMODYNAMICS & ENERGETICS - Questions 1-15
Q1: Enthalpy of Formation Definition
The student is reviewing thermodynamics. The standard enthalpy of formation of a compound is the
enthalpy change when:
A. One mole of the compound is formed from its elements under standard conditions
B. One mole of the compound is burned in oxygen
C. One mole of bonds is broken
D. One mole of gas dissolves in water
Correct Answer: A
Rationale: Standard enthalpy of formation (Δ𝐻𝑓∘) is the enthalpy change when one mole of a
compound is formed from its constituent elements in their standard states under standard
conditions (298 K, 100 kPa). [100% CORRECT]
Q2: Enthalpy of Combustion
The student is reviewing combustion. The standard enthalpy of combustion of ethanol is −1367 kJ
mol⁻¹. This means:
A. 1367 kJ is released when one mole of ethanol is completely burned in oxygen
B. 1367 kJ is absorbed when one mole of ethanol is burned
C. 1367 kJ is released when one mole of ethanol is formed
D. 1367 kJ is absorbed when one mole of ethanol is formed
Correct Answer: A
Rationale: Negative sign indicates exothermic; 1367 kJ released per mole burned completely in
oxygen under standard conditions. [100% CORRECT]
Q3: Calorimetry Calculation
The student is reviewing calorimetry. 50.0 g of water is heated from 20.0°C to 30.0°C. Specific heat
capacity = 4.18 J g⁻¹ K⁻¹. Heat absorbed is:
A. 2090 J
B. 418 J
C. 209 J
D. 4180 J
,2
Correct Answer: A
Rationale: 𝑞 = 𝑚𝑐Δ𝑇 = 50.0 × 4.18 × 10.0 = 2090 J. [100% CORRECT]
Q4: Hess’s Law Application
The student is reviewing Hess’s Law. Given:
C(s) + O₂(g) → CO₂(g) ΔH = −394 kJ mol⁻¹
CO(g) + ½O₂(g) → CO₂(g) ΔH = −283 kJ mol⁻¹
Calculate ΔH for C(s) + ½O₂(g) → CO(g).
A. −111 kJ mol⁻¹
B. +111 kJ mol⁻¹
C. −677 kJ mol⁻¹
D. +677 kJ mol⁻¹
Correct Answer: A
Rationale: Reverse second equation: CO₂ → CO + ½O₂ ΔH = +283. Add to first: C + O₂ → CO₂ −394.
Sum: C + ½O₂ → CO ΔH = −394 + 283 = −111 kJ mol⁻¹. [100% CORRECT]
Q5: Mean Bond Enthalpy
The student is reviewing bond enthalpies. Mean bond enthalpy is:
A. The average energy required to break one mole of a covalent bond in gaseous molecules
B. The energy released when one mole of bonds forms
C. The energy required to break one mole of ionic bonds
D. The energy required to melt one mole of a solid
Correct Answer: A
Rationale: Mean bond enthalpy is the average energy needed to break one mole of a given covalent
bond in gaseous molecules, averaged over different compounds. [100% CORRECT]
Q6: Bond Enthalpy Calculation
The student is reviewing calculations. Use bond enthalpies (kJ mol⁻¹): H–H = 436, Cl–Cl = 242, H–Cl =
431. ΔH for H₂ + Cl₂ → 2HCl is:
A. −184 kJ mol⁻¹
B. +184 kJ mol⁻¹
C. −247 kJ mol⁻¹
D. +247 kJ mol⁻¹
Correct Answer: A
Rationale: Bonds broken: H–H + Cl–Cl = 436 + 242 = 678. Bonds formed: 2 × H–Cl = 862. ΔH = 678 −
862 = −184 kJ mol⁻¹. [100% CORRECT]
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Q7: Entropy Definition
The student is reviewing entropy. Entropy is a measure of:
A. Disorder or randomness of a system
B. Enthalpy change
C. Activation energy
D. Rate of reaction
Correct Answer: A
Rationale: Entropy (S) quantifies disorder; gases have higher entropy than liquids, which have higher
entropy than solids. [100% CORRECT]
Q8: Entropy Change Sign
The student is reviewing entropy changes. For the reaction CaCO₃(s) → CaO(s) + CO₂(g), ΔS is:
A. Positive
B. Negative
C. Zero
D. Cannot be determined
Correct Answer: A
Rationale: A gas is produced from a solid, increasing disorder; ΔS is positive. [100% CORRECT]
Q9: Gibbs Free Energy Equation
The student is reviewing spontaneity. The Gibbs free energy equation is:
A. ΔG = ΔH − TΔS
B. ΔG = ΔH + TΔS
C. ΔG = TΔS − ΔH
D. ΔG = ΔH × TΔS
Correct Answer: A
Rationale: ΔG = ΔH − TΔS. Reaction feasible when ΔG < 0. [100% CORRECT]
Q10: Feasibility Conditions
The student is reviewing feasibility. A reaction is always feasible when:
A. ΔH is negative and ΔS is positive
B. ΔH is positive and ΔS is negative
C. ΔH is positive and ΔS is positive
D. ΔH is negative and ΔS is negative
Correct Answer: A
Rationale: ΔG = ΔH − TΔS. If ΔH < 0 and ΔS > 0, ΔG is always negative. [100% CORRECT]
