OCR A-Level Chemistry AS Breadth in chemistry (01), June 2025: Question 12
1 mark · Easy difficulty · Multiple Choice
Identify the effect of increasing pressure on the rate of reaction and equilibrium yield of HI for the reaction H2(g) + I2(g) ⇌ 2HI(g).
Practise this questionQuestion
Question text
12 Hydrogen and iodine react in the reaction below.
H2(g) + I2(g) 2HI(g)
Which row is correct when pressure is increased?
Rate of reaction Equilibrium yield of HI
A Increases Increases
B Increases No effect
C Increases Decreases
D Decreases No effect
Your answer [1]
Mark scheme
Show the mark scheme
12 B 1
How to answer it
Effect of Pressure on Reaction Rate and Equilibrium Position
📋 What this question tests
This multiple-choice question assesses your ability to decouple and evaluate the dual effects of changing pressure on:
- Reaction kinetics: How increased pressure influences the frequency of successful collisions between gas particles.
- Dynamic equilibria (Le Chatelier's Principle): How changing system pressure affects the position of equilibrium by comparing moles of gaseous reactants and products.
Analysis of Question 12
Reaction: H₂(g) + I₂(g) ⇌ 2HI(g)
✅ Correct Answer
B: Increases | No effect
- Rate of reaction: Increases
- Equilibrium yield of HI: No effect
💡 Key Knowledge
- Rate: Increasing pressure compresses gas into a smaller volume. Particles are closer together (higher concentration), leading to more frequent collisions per second. Rate always increases when pressure rises.
- Equilibrium Position: Increasing pressure shifts equilibrium toward the side with fewer moles of gas.
- Count moles of gas:
• Left side: 1 mol H₂ + 1 mol I₂ = 2 moles of gas
• Right side: 2 mol HI = 2 moles of gas - Since moles of gas are equal on both sides ( 2 = 2 ), a pressure change has no effect on the equilibrium position or yield.
📐 Step-by-Step Decision Process
- Assess Column 1 (Rate of reaction):
Increasing pressure increases collision frequency.
→ Rate increases (eliminates row D). - Assess Column 2 (Equilibrium yield):
Compare gaseous stoichiometry:
Left: 1 + 1 = 2 mol(g)
Right: 2 mol(g)
→ Moles of gas are identical on both sides.
→ No shift in equilibrium position.
→ Yield has no effect (eliminates rows A and C). - Select Row: Only row B matches both criteria.
🧠 Exam Technique & Strategy
- Split the table: Solve one column at a time to immediately rule out distractor options.
- Never confuse rate and equilibrium: Rate is about how fast collisions occur; equilibrium yield is about where the final balance lies.
- Remember: Catalysts and pressure (when Δn = 0) increase the rate of both forward and reverse reactions by the same factor, leaving the equilibrium position unaltered.
❌ Common Errors & Pitfalls
- Confusing the effect on rate: Assuming that because the equilibrium position doesn't shift, the rate must also have "no effect". Rate depends on collision frequency, which always rises with pressure.
- Assuming higher pressure always favours products: Students frequently forget to count the stoichiometric coefficients of gaseous species.
- Misreading states: Always verify state symbols. Only gaseous moles (g) count when applying Le Chatelier's principle to pressure changes (though here all species are gases).
Topics
Module 3: Periodic table and energy · 3.2 Physical chemistry
Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.