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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2019 · Paper I

Q5(e). Surface reactions: first order at low pressure, zero order at high

  • 10 marks
  • Answered
Question

Unimolecular surface catalysed gas phase reactions follow first order kinetics at low pressures and zero order kinetics at high pressures. Why?

Answer in brief

The rate is proportional to the fraction of surface covered, given by the Langmuir isotherm: rate =kθ=(kKp)/(1+Kp). At low p the coverage is proportional to p (first order); at high p the surface is saturated, θ→1 (zero order).

Key points

  • the gas A is adsorbed rapidly and reversibly on surface sites S; the adsorbed molecule decomposes slowly and the products leave at once: A(g) + S ⇌ A-S (fast) A-S → products + S (slow).
  • adsorption ∝ p(1-θ), desorption ∝θ; at equilibrium θ=(Kp)/(1+Kp) (monolayer, equivalent sites, no interaction between adsorbed molecules).
  • Only adsorbed molecules react, so rate=kθ=(kKp)/(1+Kp)
  • θ≈ Kp and rate =kKp: first order. The surface is nearly bare; doubling p doubles the number of adsorbed molecules.

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

  1. Mechanism: the gas A is adsorbed rapidly and reversibly on surface sites S; the adsorbed molecule decomposes slowly and the products leave at once: A(g) + S ⇌ A-S (fast) A-S → products + S (slow).
  2. Langmuir isotherm: adsorption ∝ p(1-θ), desorption ∝θ; at equilibrium θ=(Kp)/(1+Kp) (monolayer, equivalent sites, no interaction between adsorbed molecules).
  3. Only adsorbed molecules react, so rate=kθ=(kKp)/(1+Kp)
  4. Low pressure (Kp≪1): θ≈ Kp and rate =kKp: first order. The surface is nearly bare; doubling p doubles the number of adsorbed molecules.
  5. High pressure (Kp≫1): θ→1 and rate =k: zero order. Every site is occupied; extra gas cannot find room, so the rate no longer depends on p.
  6. In between: the order is (dln(rate))/(dln p)=(1)/(1+Kp), falling smoothly from 1 to 0 (fractional order). PH₃ on tungsten shows both limits; HI on Pt (weak adsorption) stays first order, while NH₃ on W at high pressure is zero order.

Application. in industrial catalysis the feed pressure is chosen to keep the surface close to saturation, so the rate is maximal and insensitive to small pressure changes, as in the decomposition of ammonia on tungsten and the oxidation of CO on platinum in catalytic converters.

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Model answer to Q5(e), 2019 Paper I: solution page 26
Solution page 26 of 40
Read the whole 2019 solution2019 question paper

All parts of Question 5