VKC ScienceChemistry for written exams Telegram

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 · 2026 · Paper I

Q4(b)(i). Rate constants of A ⇌ B + C from a temperature jump

  • 15 marks
  • Answered
Question

The equilibrium A⇌ B+C at 25°C is subjected to a sudden temperature increase that slightly increases the concentrations of B and C. The measured relaxation time is 3.0 µs. The equilibrium constant for the system is 2.0×10⁻¹⁶ mol dm⁻³ at 25°C, and the equilibrium concentrations of B and C at 25°C are both 2.0×10⁻⁴ mol dm⁻³. Calculate the rate constants involved in the system.

Answer in brief

kᵣ (B + C → A) =8.3×10⁸ dm³ mol⁻¹ s⁻¹; k_f (A → B + C) =1.7×10⁻⁷ s⁻¹

Key points

  • Net rate of change of A: (d[A])/(dt)=-k_f[A]+kᵣ[B][C]
  • Let x be the deviation from the new equilibrium: [A]=[A]_(eq)+x, [B]=[B]_(eq)-x, [C]=[C]_(eq)-x
  • 4pt] (dx)/(dt)=-k_f[A]_(eq)+kᵣ[B]_(eq)[C]_(eq)-k_f+kᵣ([B]_(eq)+[C]_(eq)) x+kᵣ x²
  • At the new equilibrium k_f[A]_(eq)=kᵣ[B]_(eq)[C]_(eq), so the first two terms cancel. The jump is small, so the term in x² is neglected (linearisation): (dx)/(dt)=-k_f+kᵣ([B]_(eq)+[C]_(eq)) x

The full model answer

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

Reading of the data: the data imply [A]_(eq)=[B][C]/K=2×10⁸ mol dm⁻³, which is unphysical; but [A] is not needed, since only K=k_f/kᵣ and [B]_(eq)+[C]_(eq) enter 1/τ.

Temperature jump: a capacitor discharge or laser pulse raises T by 5 to 10 K in about 1 µs. The rate constants take their new values at once, but the concentrations are still those of the old equilibrium. The system then relaxes to the new equilibrium, and a relaxation time of 3.0 µs is easily followed.

Derivation of the relaxation time

  1. Net rate of change of A: (d[A])/(dt)=-k_f[A]+kᵣ[B][C]
  2. Let x be the deviation from the new equilibrium: [A]=[A]_(eq)+x, [B]=[B]_(eq)-x, [C]=[C]_(eq)-x
  3. Substituting, with d[A]/dt=dx/dt: 4pt] (dx)/(dt)=-k_f[A]_(eq)+kᵣ[B]_(eq)[C]_(eq)-k_f+kᵣ([B]_(eq)+[C]_(eq)) x+kᵣ x²
  4. At the new equilibrium k_f[A]_(eq)=kᵣ[B]_(eq)[C]_(eq), so the first two terms cancel. The jump is small, so the term in x² is neglected (linearisation): (dx)/(dt)=-k_f+kᵣ([B]_(eq)+[C]_(eq)) x
  5. This is first order in x, so x=x₀ e^(-t/τ) with (1)/(τ)=k_f+kᵣ([B]_(eq)+[C]_(eq))

Calculation

Given: τ=3.0×10⁻⁶ s; K=k_f/kᵣ=2.0×10⁻¹⁶ mol dm⁻³; [B]_(eq)=[C]_(eq)=2.0×10⁻⁴ mol dm⁻³.

  1. (1)/(τ)=13.0×10⁻⁶ s=3.33×10⁵s⁻¹
  2. With k_f=K kᵣ: (1)/(τ)=kᵣ(K+[B]_(eq)+[C]_(eq)), and K+[B]_(eq)+[C]_(eq)=2.0×10⁻¹⁶+4.0×10⁻⁴=4.0×10⁻⁴ mol dm⁻³
  3. kᵣ=1τ (K+[B]_(eq)+[C]_(eq))=1(3.0×10⁻⁶)(4.0×10⁻⁴), kᵣ=8.3×10⁸ dm³ mol⁻¹ s⁻¹
  4. k_f=K kᵣ=(2.0×10⁻¹⁶)(8.33×10⁸), k_f=1.7×10⁻⁷ s⁻¹

Reading the answer: the units agree: k_f in s⁻¹ divided by kᵣ in dm³ mol⁻¹ s⁻¹ gives K in mol dm⁻³, as printed. k_f supplies only 5×10⁻¹³ of 1/τ, so the relaxation is set almost wholly by the recombination of B and C, and k_f follows only through K. A value of kᵣ near 10⁹ dm³ mol⁻¹ s⁻¹ marks a reaction close to the diffusion-controlled limit. A, by contrast, dissociates slowly: t_(1/2)=2/k_f=4.2×10⁶ s, about 48 days.

Application. the same analysis for H₂O ⇌ H⁺ + OH⁻ (relaxation time 37 µs at 298 K) gives kᵣ=1.4×10¹¹ dm³ mol⁻¹ s⁻¹ for H⁺ + OH⁻, one of the fastest reactions in solution. Relaxation methods (temperature and pressure jump) are the standard way to measure such fast reactions.

Practising this answer? Write it in a Mains test and get it checked by VKC. Civil Services Mains test series · Telegram

Model answer to Q4(b)(i), 2026 Paper I: solution page 23
Solution page 23 of 55