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Indian Forest Service (Mains)

Chemistry optional

  • 2 papers
  • 200 marks each
  • 3 h per paper

Paper I covers physical and inorganic chemistry.

Indian Forest Service (Mains), Chemistry optional · 2016 · Paper I

Q1(b)(i). Activation energy from rate constants at two temperatures

Question

Assuming that the activation energy and pre-exponential factor are temperature independent, obtain the expression relating the activation energy (Eₐ) and the rate parameters k(T₁) and k(T₂) for the same reaction at temperatures T₁ and T₂.

Answer in brief

ln(k(T₂))/(k(T₁))=(Eₐ)/(R)((1)/(T₁)-(1)/(T₂))=(Eₐ(T₂-T₁))/(RT₁T₂)

Key points

  • Arrhenius equation: k=A e^(-Eₐ/RT); taking logarithms, ln k=ln A-(Eₐ)/(RT)
  • Write it at the two temperatures, with the same A and Eₐ (both temperature independent): ln k(T₂)=ln A-(Eₐ)/(RT₂) ln k(T₁)=ln A-(Eₐ)/(RT₁)
  • Subtract the second from the first; ln A cancels: ln(k(T₂))/(k(T₁))=(Eₐ)/(R)((1)/(T₁)-(1)/(T₂))=(Eₐ (T₂-T₁))/(R T₁T₂)
  • In common logarithms: log(k₂)/(k₁)=(Eₐ)/(2.303R)((T₂-T₁)/(T₁T₂)); rearranged, Eₐ=(R ln(k₂/k₁))/(1/T₁-1/T₂).

The full model answer

As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.

Model answer to Q1(b)(i), 2016 Paper I: solution pages 3 and 4
Solution pages 3 and 4 of 30
Read the whole 2016 solution2016 question paper

Topic notes: Arrhenius equation, collision theory and transition-state theory, with every Civil Services, IFoS and Geo-Scientist PYQ →

All parts of Question 1