AQA A-Level Chemistry Paper 1, June 2025: Question 4

12 marks · Hard difficulty · State/Explain/Numerical

Calculate the feasibility temperature of magnesium nitrate decomposition, determine the N-O bond enthalpy from reaction data, identify a catalyst type, and explain equilibrium shifts with temperature.

Practise this question

Question

An AQA A-Level Chemistry exam question consisting of five parts. Part 4.1 asks for standard conditions of enthalpy and entropy. Part 4.2 provides a table of thermodynamic data (enthalpy of formation and entropy) for magnesium nitrate decomposition and asks to calculate the temperature at which the reaction becomes feasible. Part 4.3 provides a table of mean bond enthalpies and asks to calculate the mean bond enthalpy of the N-O bond in nitrogen dioxide. Part 4.4 is a multiple-choice question about the type of catalyst used. Part 4.5 asks to explain the effect of temperature on the equilibrium mixture's mole fraction of nitrogen dioxide.
Question text

04 Nitrogen dioxide gas is formed when magnesium nitrate is heated.

2Mg(NO3)2(s) → 2MgO(s) + 4NO2(g) + O2(g)

Table 4 shows some thermodynamic data.

Table 4

Substance Mg(NO3)2(s) MgO(s) NO2(g) O2(g)

Δ Ho / kJ mol–1 –790 –602 +34 0

f

So / J K–1 mol–1 +164 +27 +240 +205

04.1 Give the standard conditions indicated by the symbol o in Δ Ho and So

f

[1 mark]

04.2 Use the equation and the data in Table 4 to calculate the temperature at which this

reaction becomes feasible.

[5 marks]

Temperature K

Nitrogen dioxide can also be produced in a reaction between ammonia and oxygen.

4 NH (g) + 7 O (g) ⇌ 4 NO (g) + 6 H O(g) ΔHo = –1132 kJ mol–1

*14* 3 2 2 2

Table 5 shows some bond enthalpy values.

Table 5

N–H O=O O–H

Mean bond enthalpy / kJ mol–1 388 496 463

04.3 Assume that the bonds between N and O in NO2 are the same.

Calculate the mean bond enthalpy, in kJ mol–1, for a bond between N and O in NO

[3 marks]

Bond enthalpy kJ mol–1

04.4 An alloy of platinum and rhodium is used as a catalyst in the reaction.

What type of catalyst is this?

[1 mark]

Tick ( ) one box.

Autocatalyst

Heterogeneous catalyst

Homogeneous catalyst 16

04.5 Some ammonia and oxygen are reacted and the mixture is left to reach equilibrium at

a fixed temperature.

Explain why increasing the temperature of this equilibrium mixture decreases the

*15mole fraction of NO*2 in the equilibrium mixture.

[2 marks]

Mark scheme

Show the mark scheme The mark scheme for question 4. Part 4.1 accepts standard pressure of 100 kPa and a stated temperature of 298 K. Part 4.2 shows the calculation steps: Delta H equals 512 kJ/mol, Delta S equals 891 J/K/mol (or 0.891 kJ/K/mol), and T equals Delta H divided by Delta S, giving 575 K. Part 4.3 shows the bond enthalpy calculation: -1132 equals reactants' bonds minus products' bonds, leading to 8 N-O bonds equaling 3704 kJ/mol, so one N-O bond is 463 kJ/mol. Part 4.4 identifies the catalyst as heterogeneous. Part 4.5 awards marks for stating the forward reaction is exothermic and that the equilibrium shifts to oppose the temperature increase.

Question Answers Additional comments/Guidelines Mark

allow 1 bar

04.1 Standard pressure/100 kPa and a stated temperature/298 K

do not accept 1 atm (1 x AO1)

M1 ΔHϴ = (2 × –602) + (4 × 34) – (2 × –790) = 512 kJ mol–1

M2 ΔSϴ = (27 × 2) + (4 × 240) + 205 – ( 2 × 164) =

891 J K 1 mol–1

04.2 ϴ –1 –1

M3 ΔS = 0.891 kJ K mol M3: ∆S divided by 1000 (1 x AO1,

4 x AO2)

M3: Allow ∆H = 512 000 Jmol-1

∆H

M4 T = ∆S

M5 T = = 575 K

0.891

M1 Allow 8x =8128-5556+1132

M1 –1132 = (12 × 388) + (7 × 496) – (8 × N–O) – (12 × 463)

M2 (8 N–O) = 3704 kJ mol–1 3

04.3

(3 x AO2)

M3 = M2 ÷ 8 to give a positive value

M3 BE N–O = 463 kJ mol–1

– A-LEVEL CHEMISTRY – –

04.4 Heterogeneous

(1 x AO1)

M1 (forward) reaction is exothermic M1 reverse reaction is endothermic

04.5 M2 equilibrium moves/shifts to lower the temperature OR Ignore favours (1 x AO1,

equilibrium moves/shifts to oppose the increase in temperature M2: Equilibrium moves/shifts to the 1 x AO2)

left/backwards is insufficient

How to answer it

Thermodynamics, Bond Enthalpies, and Equilibrium

What this question tests

This exam question assesses your ability to define standard thermodynamic conditions, perform multi-step calculations using Gibbs Free Energy ( ΔG = ΔH - TΔS ) and mean bond enthalpies, identify catalyst types, and apply Le Chatelier's principle to explain equilibrium shifts quantitatively and qualitatively.

Part 04.1

Standard Conditions Definition

Define the standard conditions indicated by the symbol ⊖

Correct Answer

  • Standard pressure of 100 kPa (or 1 bar )
  • AND a stated temperature (usually 298 K / 25 °C )

Common Pitfalls

  • Do not write 1 atm for pressure; AQA specifically penalises this. Use 100 kPa .
  • Forgetting to state both pressure and temperature. You must have both to secure the mark.
Mark scheme breakdown: 1 mark for stating both standard pressure (100 kPa / 1 bar) and a stated temperature (298 K).
Part 04.2

Feasibility Temperature Calculation

Calculate the temperature at which the decomposition of magnesium nitrate becomes feasible

2Mg(NO₃)₂(s) → 2MgO(s) + 4NO₂(g) + O₂(g)

Step-by-Step Calculation

  1. Calculate Enthalpy Change (ΔH°):
    ΔH° = ΣΔ_f H°(products) - ΣΔ_f H°(reactants)
    ΔH° = [(2 × -602) + (4 × +34) + 0] - [2 × -790]
    ΔH° = [-1204 + 136] - [-1580] = +512 kJ mol⁻¹
  2. Calculate Entropy Change (ΔS°):
    ΔS° = ΣS°(products) - ΣS°(reactants)
    ΔS° = [(2 × 27) + (4 × 240) + 205] - [2 × 164]
    ΔS° = [54 + 960 + 205] - 328 = +891 J K⁻¹ mol⁻¹
  3. Convert Entropy Units to kJ K⁻¹ mol⁻¹:
    ΔS° = 891 / 1000 = 0.891 kJ K⁻¹ mol⁻¹ (Crucial Step!)
  4. Set up the Feasibility Condition:
    A reaction becomes feasible when ΔG ≤ 0 . At the exact point of feasibility, let ΔG = 0 :
    0 = ΔH - TΔS  ⇒  T = ΔH / ΔS
  5. Calculate Temperature (T):
    T = 512 / 0.891 = 574.6 K (rounds to 575 K )

Exam Technique

Always write down the formula T = ΔH / ΔS . Even if you make an arithmetic error earlier, you can still secure method marks (M4 and M5) for using your calculated values correctly.

The Unit Trap!

Entropy values are given in J K⁻¹ mol⁻¹, while enthalpy values are in kJ mol⁻¹. You must divide ΔS by 1000 before using it in the Gibbs equation, or multiply ΔH by 1000.

Mark scheme breakdown: M1: ΔH calculation (512); M2: ΔS calculation (891); M3: Division of ΔS by 1000 (0.891); M4: Rearrangement of formula; M5: Final temperature (575 K).
Part 04.3

Mean Bond Enthalpy Calculation

Calculate the mean bond enthalpy of the N-O bond in NO₂

4NH₃(g) + 7O₂(g) ⇌ 4NO₂(g) + 6H₂O(g)      ΔH° = -1132 kJ mol⁻¹

Step-by-Step Calculation

  1. Count the bonds broken (Reactants):
    • 4 moles of NH₃: 4 × 3 = 12 N-H bonds
    • 7 moles of O₂: 7 × 1 = 7 O=O bonds
    Σ(bonds broken) = (12 × 388) + (7 × 496) = 4656 + 3472 = 8128 kJ mol⁻¹
  2. Count the bonds formed (Products):
    • 4 moles of NO₂: 4 × 2 = 8 N-O bonds (let this be 8x )
    • 6 moles of H₂O: 6 × 2 = 12 O-H bonds
    Σ(bonds formed) = 8x + (12 × 463) = 8x + 5556 kJ mol⁻¹
  3. Use the Bond Enthalpy Equation:
    ΔH = Σ(bonds broken) - Σ(bonds formed)
    -1132 = 8128 - (8x + 5556)
  4. Solve for x (N-O bond enthalpy):
    -1132 = 2572 - 8x
    8x = 2572 + 1132 = 3704
    x = 3704 / 8 = 463 kJ mol⁻¹

Stoichiometry Traps

Students often forget that water (H₂O) has two O-H bonds, meaning 6 moles of H₂O contains 12 moles of O-H bonds. Similarly, NO₂ contains two N-O bonds, giving 8 moles of N-O bonds total.

Sign Check

Bond enthalpies are always positive values because bond breaking is an endothermic process. If you get a negative value, re-check your algebraic rearrangement!

Mark scheme breakdown: M1: Correctly setting up the equation with all stoichiometry; M2: Simplifying to find total value for 8 N-O bonds (3704); M3: Dividing by 8 to get final positive value (463 kJ mol⁻¹).
Part 04.4

Catalyst Classification

Identify the type of catalyst used when a solid platinum-rhodium alloy catalyses gaseous reactants

Correct Answer

Heterogeneous catalyst

Key Knowledge

A heterogeneous catalyst is in a different physical state (phase) than the reactants. Here, the catalyst is a solid alloy, while the reactants (ammonia and oxygen) are gases.

Mark scheme breakdown: 1 mark for identifying "Heterogeneous".
Part 04.5

Le Chatelier's Principle & Equilibrium

Explain why increasing the temperature decreases the mole fraction of NO₂ in the equilibrium mixture

Model Answer

  1. The forward reaction is exothermic (or the reverse reaction is endothermic).
  2. According to Le Chatelier's principle, the equilibrium shifts/moves to oppose the increase in temperature (by shifting in the endothermic/reverse direction to absorb heat).

Examiner Tip: Use Active Verbs

The examiner requires you to state that the equilibrium shifts/moves to oppose the change. Simply writing "the reverse reaction is favoured" or "it shifts left" without stating why (to oppose the temperature increase) will lose you the second mark.

Mark scheme breakdown: M1: Stating forward reaction is exothermic (or reverse is endothermic); M2: Explaining that the equilibrium shifts to oppose the temperature increase.

Topics

Physical Chemistry · 3.1.4 Energetics · 3.1.5 Kinetics · 3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc · 3.1.8 Thermodynamics

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.