OCR A-Level Chemistry AS Breadth in chemistry (01), November 2021: Question 22

✅ Correct Answer Requirements

• Products line 3H₂(g) + CO(g) drawn positioned above the reactants line (reflecting positive ΔH).
• ΔH labelled clearly with an upwards arrow pointing from the reactant level to the product level.
• Curves showing activation energies: Two curves or a single curve with two peaks. The peak for the catalysed reaction ( E_c ) must be lower than the uncatalysed reaction ( E_a ). Both arrows must originate from the reactants line.

💡 Key Knowledge

• Endothermic reactions: Products have higher energy than reactants, giving a positive ΔH value ( +206 kJ mol⁻¹ ).
• Catalysts: Provide an alternative reaction pathway with a lower activation energy ( E_c < E_a ).

🧠 Exam Technique

• Draw your energy levels clearly using a ruler if sketching on paper. Make sure the product line is visibly higher than the reactant line.
• Ensure activation energy arrows ( E_a and E_c ) start precisely from the reactant energy level and reach the apex of their respective curves.

❌ Common Errors

• Drawing products lower than reactants for an endothermic reaction.
• Starting the activation energy arrow from the bottom axis or the product line instead of the reactant level.
• Failing to show two distinct peaks or curves when comparing catalysed and uncatalysed pathways.

✅ Correct Answer Requirements

• Pressure reasoning: State that the right-hand side has more gaseous moles (4 moles) compared to the left-hand side (2 moles). Conclude that a low pressure (or decreasing pressure) shifts equilibrium to the right.
• Temperature reasoning: State that the forward reaction is endothermic ( ΔH is positive ). Conclude that a high temperature (or increasing temperature) shifts equilibrium to the right.

💡 Key Knowledge

• Le Chatelier's Principle: If a factor changing the conditions of a system in equilibrium is altered, the system shifts to counteract that change.
• Count total moles of gas on each side of the balanced equation: CH₄ + H₂O (2 moles) ⇌ 3H₂ + CO (4 moles).

🧠 Exam Technique

• Structure your answer into two distinct paragraphs or clear bullet points for Pressure and Temperature.
• Always explicitly state both the condition change (e.g., "low pressure") and the reasoning (moles of gas / endothermic nature).

❌ Common Errors

• Confusing the number of moles by ignoring stoichiometric coefficients (e.g., forgetting the 3 in front of H₂ ).
• Stating that high pressure favours the side with more moles, getting the shift backwards.

📐 Step-by-Step Calculation

Equation: CH₄(g) + H₂O(g) ⇌ 3H₂(g) + CO(g) with ΔH = +206 kJ mol⁻¹

1. Calculate energy required to break bonds (Reactants):
   Bonds broken: 4 × (C–H) and 2 × (O–H)
   = (4 × 413) + (2 × 464) = 1652 + 928 = 2580 kJ
2. Set up the enthalpy change equation:
   ΔH = Energy input (bonds broken) - Energy output (bonds formed)
   Bonds formed in products: 1 × (C≡O) and 3 × (H–H)
   +206 = 2580 - (1077 + 3 × (H–H))
3. Rearrange to solve for the H–H bond enthalpy:
   3 × (H–H) = 2580 - 1077 - 206 = 1297 kJ
   H–H bond enthalpy = 1297 / 3 = 432.33... kJ mol⁻¹

💡 Key Knowledge & Guidance

• Final Answer: +432 or 432.3 (Accept 432 or 433).
• Sign convention: Examiners ignore sign on the final answer line as specified in the mark scheme, but standard bond enthalpies are positive values.
• Error Carried Forward (ECF): If an arithmetic error occurs in step 1, subsequent steps following the correct method still pick up intermediate marks.

❌ Common Calculation Traps

• Forgetting to multiply the C–H bond value by 4 and O–H by 2.
• Subtracting ΔH instead of accounting for it properly in the rearrangement.
• Dividing by the wrong number of moles of H–H bonds (must divide by 3).

🧠 Exam Technique

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• Always show your working clearly line-by-line. Even if arithmetic slips occur, method marks are heavily rewarded.
• Double-check stoichiometry from the balanced equation before multiplying bond values.