Edexcel A-Level Chemistry AS Paper 2, November 2021: Question 2

7 marks · Medium difficulty · Calculations

Complete a Hess's Law cycle, calculate the enthalpy change of thermal decomposition of PCl5 using given formation and vaporisation enthalpy data, and explain the effect of increasing temperature on an equilibrium position.

Practise this question

Question

Exam question about the thermal decomposition of phosphorus(v) chloride into phosphorus(iii) chloride and chlorine. Part (a)(i) asks to complete a Hess's Law cycle with missing species and label arrows. Part (a)(ii) provides a table of enthalpy of formation and vaporisation data to calculate the enthalpy change of the reaction. Part (b) presents a reversible equilibrium equation and asks to explain the effect of increasing temperature on the position of equilibrium.
Question text

2 Phosphorus(V) chloride, PCl5, can be thermally decomposed to

phosphorus(III) chloride, PCl3, and chlorine, Cl2. The equation for this reaction is

PCl5(g) → PCl3(g) + Cl2(g)

The enthalpy change for this reaction cannot be measured directly.

(a) (i) Complete the Hess’s Law cycle to include the enthalpy change of formation of

both phosphorus chlorides.

Include the labels of the missing enthalpy changes.

ΔvH is the enthalpy change for the vaporisation of the substance from the

state shown to the gaseous state.

(3)

ΔrH

PCl5(g) PCl3(g) + Cl2(g)

ΔvH [PCl5(s)] ΔvH [PCl3(l)]

PCl5(s) PCl3(l) + Cl2(g)

(ii) Calculate the enthalpy change for the thermal decomposition of PCl5(g) to

PCl3(g) and Cl2(g), using the data given in the table.

Include a sign and units in your answer.

(2)

Enthalpy change / kJ mol–1

ΔfH [PCl5(s)] –443.5

ΔfH [PCl3(l)] –319.7

ΔvH [PCl5(s)] +64.9

ΔvH [PCl3(l)] +30.5

*P67084A0428*

(b) Another source gave a different value for the enthalpy change of this reaction.

PCl (g) PCl (g) + Cl (g) Δ H = +87.9 kJ mol–1

53 2 r

Explain the effect, if any, of increasing the temperature on the position of the

equilibrium at constant volume.

(2)

(Total for Question 2 = 7 marks)

Mark scheme

Show the mark scheme Mark scheme showing the expected answers for completing the Hess's Law cycle with elements P(s) and 2.5Cl2(g), the calculation steps giving +89.4 kJ mol-1, and the explanation that increasing temperature shifts the equilibrium to the right because the forward reaction is endothermic.

Question

Answer Additional Guidance Mark

Number

2(a)(i) An answer that makes reference to the following points: Penalise missing states only once (M1) (3)

• correct elements in the lower box (P(s), Cl2(g)) (1) States are required

Allow P4(s)

Ignore balancing numbers for M1

• correct moles of each element, P(s) Allow ¼P4(s)

and Ignore state symbols (if given)

2½Cl2(g) (1)

• arrows correctly labelled (∆fH [PCl5], ∆fH [PCl3]) (1) Ignore state symbols (if given) on

arrows

ExaExample of cycle

Question

Answer Additional Guidance Mark

Number

2(a)(ii) Example of calculation (2)

• use of ∑(∆fH[products] - ∑(∆fH[reactants] (1) (-319.7 + 30.5) - (-443.5 + 64.9)

Allow correct sums (−289.2 and −378.6) but must be negative

• correct answer with sign and units (1) = +89.4 kJ mol-1

Sign and units must be shown

Allow TE from M1(for omission of ∆ H data (+123.8 kJ mol-1)

v

Correct answer with no working scores (2)

Question

Answer Additional Guidance Mark

Number

2(b) An explanation that makes reference to the following (2)

points:

• (increasing the temperature) will move the

equilibrium position to the right/ in forward direction Allow more products will form

(1)

• because the (forward) reaction is endothermic (1)

M2 conditional on M1

(Total for Question 2 = 7 marks)

How to answer it

Thermochemistry & Le Chatelier's Principle Study Guide

📌 What this question tests

This question assesses core AS Level physical chemistry concepts: constructing and interpreting Hess's Law enthalpy cycles using standard enthalpy changes of formation and vaporisation, performing enthalpy calculations using thermochemical data, and applying Le Chatelier's Principle to predict the effect of temperature changes on chemical equilibria.

Part (a)(i): Completing the Hess's Law Cycle

Building standard enthalpy cycles with physical states

✅ Correct Answer

  • Lower Box Elements: P(s) + 2½Cl₂(g) (or equivalent balanced species like ½P₄(s) + 2½Cl₂(g) ).
  • Arrow 1 Label: Δ_fH [PCl₅(s)] (pointing from elements up to PCl₅(s) ).
  • Arrow 2 Label: Δ_fH [PCl₃(l)] (pointing from elements up to PCl₃(l) + Cl₂(g) ).

💡 Key Knowledge

  • Enthalpy of formation (Δ_fH): The enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions.
  • Balancing Ratios: Because PCl₅ contains 1 phosphorus and 5 chlorines, the elements must be P(s) + 2½Cl₂(g) to match the stoichiometry of the cycle.

🧠 Exam Technique

Always double-check state symbols when dealing with vaporisation steps. The cycle bridges solid reactants/products through their standard elemental states at the base.

❌ Common Errors

  • Omitting state symbols in the lower box ( (s) and (g) are strictly required).
  • Incorrect balancing numbers (e.g., using Cl instead of Cl₂ ).
🎯 Mark Breakdown: (1) Correct elements in lower box | (1) Correct moles/stoichiometry | (1) Correct arrow labels for formation enthalpies.

Part (a)(ii): Calculating Reaction Enthalpy

Applying Hess's Law to find Δ_r H

📐 Step-by-Step Calculation

  1. Identify the formula:
    Δ_r H = Σ(Δ_fH [products]) − Σ(Δ_fH [reactants])
  2. Calculate Products Total:
    Path goes through PCl₃(l) + Δ_vH [PCl₃(l)] .
    Sum = (−319.7) + (+30.5) = −289.2 kJ mol⁻¹
  3. Calculate Reactants Total:
    Path goes through PCl₅(s) + Δ_vH [PCl₅(s)] .
    Sum = (−443.5) + (+64.9) = −378.6 kJ mol⁻¹
  4. Subtract Reactants from Products:
    (−289.2) − (−378.6) = +89.4 kJ mol⁻¹

✅ Final Answer

+89.4 kJ mol⁻¹

Note: A correct final answer with sign and units scores all 2 marks instantly, even without working shown.

❌ Common Calculation Traps

  • Sign Errors: Forgetting that subtracting a negative number is equivalent to addition ( − (−378.6) ).
  • Forgetting Units: Omitting kJ mol⁻¹ or leaving off the + sign for endothermic values.
🎯 Mark Breakdown: (1) Correct use of products minus reactants expression | (1) Correct final value with correct sign and units.

Part (b): Effect of Temperature on Equilibrium

Applying Le Chatelier's Principle

✅ Correct Answer

  • Direction: The position of equilibrium shifts to the right (or towards the forward reaction / more products form).
  • Reason: Because the forward reaction is endothermic ( Δ_r H is positive ).

💡 Key Knowledge

  • Le Chatelier's Principle: If a factor connected to an equilibrium is changed, the position of equilibrium shifts to counteract that change.
  • Temperature rule: Increasing temperature favours the endothermic direction to absorb excess heat. Decreasing temperature favours the exothermic direction.

🧠 Exam Technique

Structure your answer in two clear parts: 1. State the shift (right/products), then 2. Give the justification linked to the sign of enthalpy ( Δ_r H is positive / endothermic ). Remember that mark 2 is strictly conditional on getting mark 1 correct!

🎯 Mark Breakdown: (1) Stating equilibrium shifts to the right / forward direction | (1) Explaining that the forward reaction is endothermic. (Note: M2 is conditional on M1).

Topics

Physical Chemistry · Topic 8: Energetics I · Topic 10: Equilibrium I

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.