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Class 11 Chemistry Chapter 6: Equilibrium case study question with answers

One case-based question of the kind CBSE-pattern school exams ask: a passage and a table, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks
A sulphuric acid plant oxidises sulphur dioxide by the contact process: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). The plant chemist notes the values of Kp for this equilibrium at three temperatures (values based on NCERT). At 298 K the equilibrium lies almost entirely on the side of SO₃, yet without a catalyst hardly any SO₃ forms in the converter. The plant therefore passes the gases over divanadium pentoxide (V₂O₅) at a raised temperature.
Temperature / KKp
2984.0 × 10²⁴
5002.5 × 10¹⁰
7003.0 × 10⁴
Part 1 · 1 mark
From the table, the forward reaction is
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Answer: (b) exothermic, as Kp falls with temperature

K of an exothermic reaction decreases as temperature rises; Kp falls from 4.0 × 10²⁴ to 3.0 × 10⁴ between 298 K and 700 K. The size of K alone does not give the sign of ΔH.

Part 2 · 1 mark
V₂O₅ is used because
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Answer: (c) it makes the slow oxidation of SO₂ fast without changing Kp

A large K means the products are favoured, but says nothing about rate. The oxidation of SO₂ is very slow; the catalyst increases the rate but does not change K or the equilibrium composition.

Part 3 · 2 marks
In the converter at 700 K the partial pressure of O₂ at equilibrium is 0.12 bar. Calculate the ratio p(SO₃)/p(SO₂) at equilibrium.
Answer and marking scheme

Kp = p(SO₃)²/[p(SO₂)² p(O₂)] (Kp in bar⁻¹, as Δn = −1, with pressures in bar), so [p(SO₃)/p(SO₂)]² = Kp × p(O₂) = 3.0 × 10⁴ × 0.12 = 3.6 × 10³, and p(SO₃)/p(SO₂) = 60.

  • ratio² = Kp × p(O₂) from the Kp expression (1)
  • ratio = 60 (1)
Or
State, with reasons, the effect on the amount of SO₃ at equilibrium of (i) raising the pressure by compressing the mixture, and (ii) adding argon at constant volume.
Answer and marking scheme

(i) Three moles of gas give two; compression favours the side with fewer moles of gas, so the amount of SO₃ increases. (ii) At constant volume argon changes neither the concentrations nor the partial pressures of the reacting gases, so the amount of SO₃ is unchanged.

  • (i) increases: fewer moles of gas on the product side (1)
  • (ii) no change: partial pressures unchanged at constant volume (1)
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