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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 · 2017 · Paper I

Q6(a)(ii). Activation energy from two equal-conversion times

  • 15 marks
  • Answered
Question

Define activation energy, Eₐ of a chemical reaction. A gas-phase reaction completes 25% in 30 min at 27°C and in 10 min at 37°C. Determine the Eₐ in SI unit.

Answer in brief

Eₐ=8.49×10⁴ J mol⁻¹ (84.9 kJ mol⁻¹): the same fraction is converted three times faster at 37°C, so k₃₁₀/k₃₀₀=3.

Key points

  • Definition. The activation energy is the minimum energy, above the average energy of the reactants, that colliding molecules must have to react; it is the height of the barrier on the reaction profile. Operationally, Eₐ=RT²…
  • Rate constant from the times. For a reaction of any order n started from the same initial concentrations, the time to reach a given fraction of conversion is inversely proportional to k (first order: kt=ln(1)/(1-0.25), the…
  • Arrhenius equation for two temperatures (T₁=300 K, T₂=310 K): ln(k₂)/(k₁)=(Eₐ)/(R)((1)/(T₁)-(1)/(T₂))
  • 3=1.0986; (1)/(300)-(1)/(310)=1.0753×10⁻⁴ K⁻¹ Eₐ=(8.314)(1.0986)1.0753×10⁻⁴ Eₐ=8.49×10⁴J mol⁻¹=84.9 kJ mol⁻¹

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. Definition. The activation energy is the minimum energy, above the average energy of the reactants, that colliding molecules must have to react; it is the height of the barrier on the reaction profile. Operationally, Eₐ=RT² (dln k)/(dT) (from k=Ae^(-Eₐ/RT)).
  2. Rate constant from the times. For a reaction of any order n started from the same initial concentrations, the time to reach a given fraction of conversion is inversely proportional to k (first order: kt=ln(1)/(1-0.25), the same at both temperatures). So k₃₁₀k₃₀₀=t₃₀₀t₃₁₀=(30)/(10)=3
  3. Arrhenius equation for two temperatures (T₁=300 K, T₂=310 K): ln(k₂)/(k₁)=(Eₐ)/(R)((1)/(T₁)-(1)/(T₂))
  4. Substitute: 3=1.0986; (1)/(300)-(1)/(310)=1.0753×10⁻⁴ K⁻¹ Eₐ=(8.314)(1.0986)1.0753×10⁻⁴ Eₐ=8.49×10⁴J mol⁻¹=84.9 kJ mol⁻¹
  5. Check with first-order constants: k₃₀₀=ln(4/3)/30=9.59×10⁻³ min⁻¹ and k₃₁₀=2.88×10⁻² min⁻¹; their ratio is 3, as used. (With 300.15 and 310.15 K, Eₐ=85.0 kJ mol⁻¹.)
  6. Meaning: a 10 K rise triples the rate. Eₐ near 85 kJ mol⁻¹ is typical of many gas-phase decompositions; the rule of thumb "rate doubles for 10 K near room temperature" corresponds to about 53 kJ mol⁻¹.

Application. the same two-temperature method is used to predict shelf life of drugs and foods from accelerated tests at higher temperature.

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Model answer to Q6(a)(ii), 2017 Paper I: solution page 25
Solution page 25 of 38
Read the whole 2017 solution2017 question paper

All parts of Question 6