AQA A-Level Chemistry AS Paper 1, June 2025: Question 7

11 marks · Medium difficulty · State/Explain/Numerical

Draw a concentration-time graph for HI, explain the effect of temperature on yield using Le Chatelier's principle, and calculate the equilibrium constant Kc and its units for the reaction between carbon dioxide and hydrogen.

Practise this question

Question

Question 7 covers chemical equilibria. It starts with the reaction H2(g) + I2(g) ⇌ 2HI(g) with ΔH = +52 kJ mol⁻¹. Figure 1 shows a grid plotting concentration against time, depicting the concentration of I2 starting at a high initial value and decreasing asymptotically to a constant level. Sub-question 07.1 asks to draw a line on Figure 1 showing how the concentration of HI changes over time (2 marks). Sub-question 07.2 asks to state and explain the effect of increasing temperature on the equilibrium yield of HI (3 marks). Next, the reaction CO2(g) + 3H2(g) ⇌ CH3OH(g) + H2O(g) is given where 2.50 mol CO2 and 6.00 mol H2 are placed in a 4.60 dm³ container, leaving 1.30 mol CO2 at equilibrium. Sub-question 07.3 asks for the Kc expression, the calculated value of Kc, and its units (6 marks).
Question text

07 This question is about chemical equilibria.

A mixture of hydrogen and iodine is allowed to reach equilibrium in a sealed container.

H (g) + I (g) ⇌ 2 HI(g) ΔH = +52 kJ mol–1

Figure 1 shows how the concentration of iodine changes with time.

Figure 1

07.1 Draw a line on Figure 1 to show how the concentration of hydrogen iodide changes

with time.

[2 marks]

07.2 State and explain the effect of increasing the temperature on the equilibrium yield of

hydrogen iodide.

[3 marks]

Effect on yield

Explanation

Carbon dioxide can react with hydrogen to form methanol.

*16* CO2(g) + 3H2(g) ⇌ CH3OH(g) + H2O(g)

2.50 mol of CO2 are mixed with 6.00 mol of H2 in a sealed container with a

volume of 4.60 dm3

The mixture is left to reach equilibrium at a fixed temperature.

At equilibrium, 1.30 mol of CO2 remain.

07.3 Give an expression for the equilibrium constant, Kc, for this reaction.

Calculate a value for the equilibrium constant at this fixed temperature.

Give the units of the equilibrium constant.

[6 marks]

Expression for Kc

Calculation

Kc Units

Section B

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CORRECT METHOD WRONG METHODS

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Mark scheme

Show the mark scheme The mark scheme provides answers for question 7. For 07.1: M1 is for a curve starting at the origin (0,0) and reaching a final horizontal plateau at twice the change in iodine concentration; M2 is for the gradient becoming zero at the same time iodine reaches equilibrium. For 07.2: M1 states yield would increase; M2 states the equilibrium shifts in the endothermic direction (to the right); M3 states this opposes the temperature increase / absorbs heat. For 07.3: M1 gives Kc = [CH3OH][H2O] / ([CO2][H2]³); M2 calculates mol CO2 reacted = 1.20 mol; M3 gives equilibrium moles: H2 = 2.40 mol, CH3OH = 1.20 mol, H2O = 1.20 mol; M4 converts to concentrations by dividing moles by 4.60; M5 calculates Kc = 1.69 to 1.72; M6 gives units as mol⁻² dm⁶.

Question Marking guidance Additional Comments/Guidelines Mark

07.1

(2 x AO2)

M1 = starts at zero and correct final concentration

– AS CHEMISTRY – 7404/1 –

M2 = gradient becomes zero at correct time

M1 Yield would increase

M2 The equilibrium shifts in the endothermic direction (to the right) ignore reference to forward reaction being favoured 3

07.2

– AS CHEMISTRY – 7404/1 – (3 x AO2)

M3 To oppose the temperature increase / to oppose the temperature

change / to decrease the temperature Mark each point independently

[CH3OH][H2O] Do not accept round brackets in expression for Kc

M1: Expression for Kc Kc = 3

[CO2][H2]

M2: Amount of carbon dioxide reacted

n(CO2) = 1.20 mol

M3: Equilibrium amounts of hydrogen, methanol

and water

M3: Equilibrium amounts of hydrogen, methanol and water

n(H2) = 2.40 mol

n(H2) = 6.00 – (3 × M2) (mol) n(CH3OH) = 1.20 mol

n(CH3OH) = M2 (mol) n(H2O) = 1.20 mol

n(H2O) = M2 (mol)

M4: Converts their amounts to concentrations M4: Converts their amounts to concentrations

07.3 1.20

M3 n(CH3OH) [CH OH] = = 0.261 mol dm–3 (6 x AO2)

[CH OH] = 3

34.60

4.60

M3 n(H O) 1.20 –3

[H O] = 2 [H2O] = = 0.261 mol dm

24.60

4.60

M3 n(H ) 2.40 –3

[H ] = 2 [H2] = = 0.522 mol dm

24.60

4.60

1.30 1.30 –3

[CO ] = = 0.283 mol dm–3 [CO2] = = 0.283 mol dm

26 2 4.60

4.60

M5: Calculation of Kc

M5: calculation of Kc using M1 and M4 – AS CHEMISTRY – 7404/1 –

0.261 × 0.261

Kc = 3 = 1.69 / 1.7(0) / 1.7(2)

0.283 × (0.522)

M6: Units

M6: units using their M1 mol–2 dm6

Alternative approach

07.3

(cont)

M4: shows division by 4.60 in expression and

cancels down

M5: calculation of Kc using M1 and M4

M6: units using their M1

How to answer it

Dynamic Equilibria, Le Chatelier's Principle & Kc

WHAT THIS QUESTION TESTS

This question assesses key competencies from Physical Chemistry (Section 3.1.6: Chemical Equilibria, Le Chatelier's principle and Kc):

  • Translating stoichiometric ratios into concentration-time curves on a graph.
  • Understanding dynamic equilibrium characteristics (plateau timing and starting points).
  • Applying Le Chatelier’s Principle systematically to predict changes in yield with temperature.
  • Writing homogeneous Kc expressions using strict square-bracket notation.
  • Setting up an ICE table (Initial, Change, Equilibrium) with unequal reacting ratios (1:3).
  • Converting equilibrium amounts into concentrations using system volume and calculating Kc with correct units.

Question 07.1

Concentration–Time Graph for Hydrogen Iodide [2 Marks]

✅ Model Answer & Visual Description

Draw a curve for [HI] that:

  1. Starts at the origin (0, 0): since only H₂ and I₂ are initially introduced.
  2. Plateaus at exactly twice the decrease in [I₂]:
    • Initial [I₂] = 16 grid units; Equilibrium [I₂] = 6 grid units (drop of 10 grid units).
    • Stoichiometry is 1 I₂ : 2 HI, so [HI] must rise by 2 × 10 = 20 grid units.
  3. Reaches a plateau (gradient = 0) at the exact same time that the [I₂] curve levels off.

💡 Key Knowledge

  • Equation: H₂(g) + I₂(g) ⇌ 2HI(g)
  • For every 1 mole of I₂ consumed, 2 moles of HI are formed (1:2 molar ratio).
  • Dynamic equilibrium is reached when macroscopic concentrations remain constant. Therefore, the concentrations of all reactants and products become horizontal at the exact same point in time.

🧠 Exam Technique

  • M1: Origin check + final height check. Count small grid squares carefully! The drop in reactant concentration multiplied by the stoichiometric coefficient gives the product's rise.
  • M2: Line up a vertical ruler with the point where [I₂] becomes completely flat. Your HI curve must become horizontal at that precise vertical line.

❌ Common Errors

  • Ignoring stoichiometry: Mirroring the curve so that HI levels off at the same height I₂ dropped (10 units instead of 20 units).
  • Timing mismatch: Having the HI curve plateau earlier or later than the I₂ curve. All species reach equilibrium concurrently.
Mark Scheme Breakdown:
• M1: Starts at zero and reaches the correct final concentration (+1 mark)
• M2: Gradient becomes zero at the correct equilibrium time (+1 mark)

Question 07.2

Effect of Increasing Temperature on Yield [3 Marks]

✅ Model Answer

Effect on yield: Yield would increase (or increases) [M1]

Explanation:

  • The equilibrium shifts in the endothermic direction (to the right / towards products) [M2]
  • In order to oppose the increase in temperature / to lower the temperature by absorbing heat [M3]

💡 Key Knowledge

  • Given: ΔH = +52 kJ mol⁻¹ . A positive ΔH means the forward reaction is endothermic.
  • Le Chatelier's Principle: If an external condition is changed, the position of equilibrium shifts in the direction that acts to oppose that change.

🧠 Top-Scorer Structure (3 Steps)

  1. State the direct effect: "Yield increases."
  2. State which reaction is endothermic & which way the shift occurs: "The forward reaction is endothermic, so the equilibrium shifts to the right."
  3. State the opposition: "To oppose the increase in temperature."

❌ Examiner Pitfalls & Lost Marks

  • Missing the word "equilibrium": Simply stating "it favours the forward reaction" or "the reaction shifts right" without specifying the position of equilibrium shifts can cost M2.
  • Confusing Rate with Yield: Stating that molecules have more kinetic energy explains rate, not equilibrium yield. Stick strictly to enthalpy arguments.
Mark Scheme Breakdown:
• M1: Yield would increase [1 mark]
• M2: The equilibrium shifts in the endothermic direction (or to the right) [1 mark]
• M3: To oppose the temperature increase / change (or to decrease the temperature) [1 mark]
Note: Each mark is awarded independently.

Question 07.3

Equilibrium Constant (Kc) Calculation & Units [6 Marks]

💡 Reaction Context

CO₂(g) + 3H₂(g) ⇌ CH₃OH(g) + H₂O(g)

Initial: 2.50 mol CO₂, 6.00 mol H₂ in 4.60 dm³. At equilibrium: 1.30 mol CO₂ remains.

📐 Step-by-Step Calculation

1 Expression for Kc [M1]

Kc = [CH₃OH][H₂O] [CO₂][H₂]³

⚠️ Must use square brackets [ ] . Round brackets ( ) score 0.

2 ICE Table (Moles) [M2, M3]

Species CO₂(g) H₂(g) CH₃OH(g) H₂O(g)
Initial 2.50 mol 6.00 mol 0 mol 0 mol
Change -1.20 mol -3 × (1.20) = -3.60 mol +1.20 mol +1.20 mol
Equilibrium 1.30 mol (given) 2.40 mol 1.20 mol 1.20 mol

• Moles CO₂ reacted = 2.50 - 1.30 = 1.20 mol [M2]

• Moles H₂ at equilibrium = 6.00 - (3 × 1.20) = 2.40 mol; CH₃OH = 1.20 mol; H₂O = 1.20 mol [M3]

3 Convert Amounts to Concentrations (Volume = 4.60 dm³) [M4]

  • [CO₂] = 1.30 / 4.60 = 0.283 mol dm⁻³
  • [H₂] = 2.40 / 4.60 = 0.522 mol dm⁻³
  • [CH₃OH] = 1.20 / 4.60 = 0.261 mol dm⁻³
  • [H₂O] = 1.20 / 4.60 = 0.261 mol dm⁻³

4 Calculate Value of Kc [M5]

Kc = (0.261 × 0.261) / (0.283 × (0.522)³)

Kc = 0.06812 / (0.283 × 0.14224) = 0.06812 / 0.04025 = 1.69 (accepts 1.7, 1.70, 1.72 depending on rounding)

5 Determine Units [M6]

Units = (mol dm⁻³)² / (mol dm⁻³)⁴ = (mol dm⁻³)⁻² = mol⁻² dm⁶

❌ Common Calculation Traps

  • Forgetting the 1:3 ratio: Subtracting 1.20 mol instead of 3 × 1.20 = 3.60 mol from H₂.
  • Forgetting to divide by volume: Because the sum of powers in numerator (1+1=2) does not equal denominator (1+3=4), volume terms do not cancel! You must divide by 4.60 dm³.
  • Omitting the power of 3: Forgetting to cube the concentration of H₂ in the denominator.
  • Wrong unit signs: Writing mol² dm⁻⁶ instead of mol⁻² dm⁶.

🧠 Exam Technique: Quick Unit Check

Let [M] = mol dm⁻³:

Units = [M]² / [M]⁴ = [M]⁻²

Expand (mol dm⁻³)⁻²:
• mol⁻²
• (dm⁻³)⁻² = dm⁺⁶
Result: mol⁻² dm⁶

Mark Scheme Breakdown (6 Marks):
• M1: Correct expression for Kc with square brackets.
• M2: Calculation of moles CO₂ reacted = 1.20 mol.
• M3: Correct equilibrium moles: H₂ = 2.40, CH₃OH = 1.20, H₂O = 1.20.
• M4: Dividing equilibrium moles by 4.60 to obtain concentrations.
• M5: Correct numerical answer for Kc = 1.69 (or 1.7 / 1.70).
• M6: Correct units: mol⁻² dm⁶.

Topics

Physical Chemistry · 3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc

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