Q4(b)(i). Rate constants of A ⇌ B + C from a temperature jump
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
- 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
- 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
- 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
- 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)/(τ)=13.0×10⁻⁶ s=3.33×10⁵s⁻¹
- 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
- 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
- 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

All parts of Question 4
- 4(a) The polarographic wave at a dropping mercury electrode10 marks
- 4(b)(i) Rate constants of A ⇌ B + C from a temperature jump15 marks
- 4(b)(ii) Rate coefficients of two parallel first-order reactions10 marks
- 4(c) Pressure dependence of the order of a surface-catalysed decomposition15 marks