AQA A-Level Chemistry Paper 1, 2022: Question 1

12 marks · Medium difficulty · State/Explain/Numerical

State the features of dynamic equilibrium, deduce the relationship between gaseous moles from pressure effects, and perform calculations involving mole fractions, partial pressures, and the equilibrium constant Kp.

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Question

Question 1 contains six parts on chemical equilibria: 01.1 asks for two features of a reaction in dynamic equilibrium (2 marks); 01.2 asks what can be deduced about a chemical equation if increasing pressure decreases product yield (1 mark); 01.3 provides the reaction CO(g) + 2H2(g) ⇌ CH3OH(g) with initial moles and equilibrium conditions, asking to calculate the moles and partial pressure of CO (3 marks); 01.4 asks for the Kp expression (1 mark); 01.5 gives a table of equilibrium values at temperature T to calculate the partial pressure of H2 (3 marks); 01.6 asks to calculate Kp and its units for the reverse reaction (2 marks).
Question text

01 This question is about equilibria.

01.1 Give two features of a reaction in dynamic equilibrium.

[2 marks]

Feature 1

Feature 2

01.2 A gas-phase reaction is at equilibrium.

When the pressure is increased the yield of product decreases.

State what can be deduced about the chemical equation for this equilibrium.

[1 mark]

01.3 Carbon monoxide and hydrogen react to form methanol.

CO(g) + 2H2(g) ⇌ CH3OH(g)

0.430 mol of carbon monoxide is mixed with 0.860 mol of hydrogen.

At equilibrium, the total pressure in the flask is 250 kPa and the mixture contains

0.110 mol of methanol.

Calculate the amount, in moles, of carbon monoxide present at equilibrium.

Calculate the partial pressure, in kPa, of carbon monoxide in this equilibrium mixture.

[3 marks]

Amount of carbon monoxide mol

Partial pressure kPa

01.4 Give an expression for the equilibrium constant (Kp) for this reaction.

CO(g) + 2H2(g) ⇌ CH3OH(g)

[1 mark]

Kp

01.5 A different mixture of carbon monoxide and hydrogen is left to reach equilibrium at a

temperature T.

Some data for this equilibrium are shown in Table 1.

Table 1

Partial pressure of CO 125 kPa

Partial pressure of CH3OH 5.45 kPa

K 1.15 x 10–6 kPa–2

p

CO(g) + 2H2(g) ⇌ CH3OH(g)

Calculate the partial pressure, in kPa, of hydrogen in this equilibrium mixture.

[3 marks]

Partial pressure kPa

01.6 Use the Kp value from Table 1 to calculate a value for Kp for the following reaction at

temperature T.

CH3OH(g) ⇌ CO(g) + 2H2(g)

Give the units for Kp

[2 marks]

Kp

Units

Mark scheme

Show the mark scheme Mark scheme for Question 01 detailing marks for each sub-question: 01.1 awards 1 mark for forward and reverse reactions proceeding at equal rates, and 1 mark for constant concentrations; 01.2 awards 1 mark for more moles of gaseous products than reactants; 01.3 awards 3 marks for finding 0.32 mol of CO, total moles 1.07, and partial pressure 74.8 kPa; 01.4 gives Kp = p(CH3OH) / (p(H2)^2 * p(CO)); 01.5 awards 3 marks for rearranging to find p(H2) = 194.7 kPa (or 195 kPa); 01.6 awards 1 mark for calculating 8.7 x 10^5 and 1 mark for units of kPa^2.

Question Answers Additional Comments/Guidelines Mark

allow answers in either order

forward and reverse reactions proceed at equal rates 1

01.1

concentrations (of reactants and products) remain constant do not accept equal concentrations 1

or do not accept concentrations are the same AO1

concentrations (of reactants and products) stay the same ignore closed system

more moles of (gaseous) products (than (gaseous) reactants) allow molecules 1

01.2 or do not accept atoms AO3

more moles on the RHS (than LHS)

– A-LEVEL CHEMISTRY – –

M1 (at equilibrium) n(CO) = 0.32 (mol) 1

M2 total number of moles (at equilibrium) = 1.07 (mol) 1

or mole fraction (CO) = 0.299

01.3 0.320 × 250 M1 × 250

M3 p(CO) = = 74.8 (kPa) M3 = 1

1.07 M2 11

AO2

allow 75 (kPa)

an answer of 67.8 (kPa) = 2 marks max

p(CH3OH)

01.4 Kp = 2 do not accept square brackets 1

p(H2) p(CO) AO1

– A-LEVEL CHEMISTRY – –

2 p(CH3OH) 5.45 rearrangement with or without numbers 1

M1 p(H2) = or –6

Kp × p(CO) 1.15 × 10 × 125

M2 p(H ) = √37 913 or p(H )2 = 37 913 1

M3 p(H2) = 194.7 (kPa) M3 = √M2 1

allow 195 (kPa) AO2

01.5

if rearrangement incorrect in M1 allow M3 only

if p(H2) is not squared in Question 01.4 allow

p(CH3OH)

p(H2) = for M1 and 37913 for M2

Kp × p(CO)

(max 2)

= –6 = 8.7(0) × 10 allow 869 565 1

01.6 1.15 × 10

kPa2 1

AO2

How to answer it

Gaseous Equilibria, Mole Fractions & Kₚ Calculations

What this question tests

  • Fundamental Equilibrium Concepts: Defining dynamic equilibrium and deducing stoichiometric imbalances from Le Chatelier's principle.
  • Quantitative Gas Equilibria: Finding equilibrium quantities via ICE tables, calculating mole fractions, and deriving partial pressures.
  • Equilibrium Constant (Kₚ): Formulating gas expressions correctly (using partial pressures, not square brackets).
  • Algebraic Manipulation: Rearranging Kₚ expressions to solve for an unknown squared partial pressure term.
  • Reverse Reaction Properties: Inverting equilibrium constant values ( Kₚ' = 1 / Kₚ ) and recalculating appropriate units.

Part 01.1: Features of Dynamic Equilibrium 2 Marks

✅ Mark Scheme Answers

  • Feature 1: Forward and reverse reactions proceed at equal rates.
  • Feature 2: Concentrations of reactants and products remain constant (or stay the same).

❌ Common Errors & Pitfalls

  • "Concentrations are equal": Wrong! Rates are equal; concentrations only remain constant.
  • "Closed system": While dynamic equilibrium can only be achieved in a closed system, this is a condition to establish it, not a feature of the equilibrium state itself (mark scheme directs examiners to ignore this).
Mark Breakdown: 1 mark per valid feature [AO1]. Acceptable in any order.

Part 01.2: Deducing the Chemical Equation 1 Mark

✅ Mark Scheme Answer

There are more moles of (gaseous) products than (gaseous) reactants (or "more moles on the RHS than LHS").

🧠 Exam Technique & Logic

By Le Chatelier's principle, increasing total pressure causes the equilibrium position to shift toward the side with fewer moles of gas to reduce pressure.

Since increasing pressure decreased product yield, the equilibrium shifted left (backward). Therefore, the left (reactants) must have fewer gas moles, meaning the products side has more gas moles.

Mark Breakdown: 1 mark [AO3]. "Molecules" is permitted, but do not accept "atoms". Must specify moles or molecules.

Part 01.3: Equilibrium Moles & Partial Pressure 3 Marks

CO(g) + 2H₂(g) ⇌ CH₃OH(g)

📐 Step-by-Step Calculation

  1. Find equilibrium moles of CO:
    Mole ratio is 1 CO : 1 CH₃OH. Since 0.110 mol of CH₃OH is formed, 0.110 mol of CO is consumed.
    n(CO) = 0.430 - 0.110 = 0.320 mol (M1)
  2. Determine total moles at equilibrium:
    • n(CO) = 0.320 mol
    • n(H₂) = 0.860 - (2 × 0.110) = 0.860 - 0.220 = 0.640 mol (remember the 1 : 2 stoichiometry!)
    • n(CH₃OH) = 0.110 mol
    Total moles = 0.320 + 0.640 + 0.110 = 1.07 mol (M2)
    Alternatively: Mole fraction of CO = 0.320 / 1.07 = 0.299
  3. Calculate partial pressure of CO:
    p(CO) = mole fraction × total pressure
    p(CO) = (0.320 / 1.07) × 250 kPa = 74.8 kPa (or 75 kPa) (M3)

✅ Final Answers

  • Amount of CO = 0.320 mol
  • Partial pressure = 74.8 kPa (allow 75 kPa)

❌ Common Calculation Traps

  • Forgetting 1:2 ratio for H₂: Subtracting 0.110 instead of 0.220 gives an incorrect total of 1.18 mol and a final answer of 67.8 kPa (capped at 2 marks max).
  • Omitting product in total moles: Total moles must include the 0.110 mol of CH₃OH produced.
Mark Breakdown: M1 = 0.320 mol CO | M2 = 1.07 total moles (or mole fraction 0.299) | M3 = 74.8 kPa [AO2].

Part 01.4: Expression for Kₚ 1 Mark

✅ Correct Expression

Kₚ = p(CH₃OH) / [p(CO) × p(H₂)²]

Can also be written as: Kₚ = p(CH₃OH) / [p(H₂)² · p(CO)]

❌ The #1 Critical Penalty

NEVER use square brackets [ ]!

Square brackets [ ] denote concentration in mol dm⁻³ (for Kc). Writing [CH₃OH] / [CO][H₂]² scores 0 marks. You must use p(CH₃OH) or pCH₃OH .

Mark Breakdown: 1 mark [AO1]. Strictly penalises square brackets.

Part 01.5: Calculating Unknown Partial Pressure 3 Marks

📐 Step-by-Step Calculation

  1. Rearrange expression to make p(H₂)² the subject:
    p(H₂)² = p(CH₃OH) / [Kₚ × p(CO)]
    Substitute values: p(H₂)² = 5.45 / (1.15 × 10⁻⁶ × 125) (M1)
  2. Evaluate p(H₂)²:
    Denominator: 1.15 × 10⁻⁶ × 125 = 1.4375 × 10⁻⁴
    p(H₂)² = 5.45 / 1.4375 × 10⁻⁴ = 37 913 (M2)
  3. Square root to find p(H₂):
    p(H₂) = √(37 913) = 194.7 kPa (or 195 kPa) (M3)

✅ Final Answer

Partial pressure = 194.7 kPa (allow 195 kPa)

🧠 Top Tip: Calculator Input

When dividing by a product in standard form, always use brackets around the denominator: 5.45 ÷ (1.15×10⁻⁶ × 125) . Leaving out brackets leads to arithmetic errors.

Mark Breakdown: M1 = Correct rearrangement | M2 = p(H₂)² = 37 913 (or √37913) | M3 = 194.7 (or 195) kPa [AO2].

Part 01.6: Reverse Reaction Kₚ & Units 2 Marks

CH₃OH(g) ⇌ CO(g) + 2H₂(g)

📐 Calculating Reversed Kₚ

Reversing the equation inverts the equilibrium constant expression:

Kₚ(rev) = 1 / Kₚ(fwd)

Kₚ(rev) = 1 / (1.15 × 10⁻⁶) = 8.70 × 10⁵

(Also allow 8.7 × 10⁵ or 869 565)

💡 Deriving the Units

Write out the unit dimensions for the reverse reaction expression:

Units = [p(CO) × p(H₂)²] / p(CH₃OH)

Units = (kPa × kPa²) / kPa = kPa²

Notice: Inverting the value also inverts the original units ( 1 / kPa⁻² = kPa² ).

✅ Mark Scheme Answers

  • Kₚ = 8.70 × 10⁵ (or 8.7 × 10⁵ / 869 565) (1 mark)
  • Units = kPa² (1 mark)

❌ Common Misconceptions

  • Changing the sign instead of inverting: writing -1.15 × 10⁻⁶ instead of taking the reciprocal 1 / Kₚ .
  • Keeping the forward units ( kPa⁻² ) rather than flipping them to kPa² .
Mark Breakdown: 1 mark for calculated value | 1 mark for correct unit (kPa²) [AO2].

Topics

Physical Chemistry · 3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc · 3.1.10 Equilibrium Constant Kp · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.