Q6(b). Rate constants of H⁺ + OH⁻ ⇌ H₂O from the relaxation time

Given: τ=36×10⁻⁶ s; K_w=[H⁺][OH⁻]=1.0×10⁻¹⁴ (mol dm⁻³)²; [H₂O]=55.5 mol dm⁻³.

Deriving the relaxation time

1. Rate equation: (d[H₂O])/(dt)=k₁[H⁺][OH⁻]-k₋₁[H₂O] (forward second order, reverse first order in water).
2. After the pulse, let x be the small deviation from the new equilibrium: [H⁺]=a+x, [OH⁻]=a+x, [H₂O]=w-x, where a and w are the equilibrium values.
3. Since [H₂O]=w-x, (d[H₂O])/(dt)=-(dx)/(dt), so -(dx)/(dt)=k₁(a+x)²-k₋₁(w-x)
4. Use the equilibrium condition k₁a²=k₋₁w and drop x² (small jump): -(dx)/(dt)=(2k₁a+k₋₁)x
5. Hence x=x₀e^(-t/τ) with (1)/(τ)=k₋₁+k₁([H⁺]+[OH⁻])

Numbers

1. Concentrations: in pure water [H⁺]=[OH⁻]=√(K_w)=1.0×10⁻⁷ mol dm⁻³, so the sum is 2.0×10⁻⁷ mol dm⁻³.
2. Link between k₋₁ and k₁: k₋₁=(k₁K_w)/([H₂O]). Then k₋₁ is only 9×10⁻¹⁰ of 2k₁a and can be dropped in 1/τ: k₁=(1)/(τ([H⁺]+[OH⁻]))=1(36×10⁻⁶)(2.0×10⁻⁷), k₁=1.4×10¹¹dm³ mol⁻¹ s⁻¹
3. Dissociation: k₋₁=(1.39×10¹¹)(1.0×10⁻¹⁴)55.5, k₋₁=2.5×10⁻⁵s⁻¹
4. If the water concentration is absorbed into k₋₁ (rate =k₋₁', zero order), k₋₁'=k₁K_w=1.4×10⁻³ mol dm⁻³ s⁻¹; the physics is the same.
5. Meaning: k₁ is about the largest second-order rate constant known: H⁺ and OH⁻ react at almost every encounter, their approach speeded by proton hopping through the hydrogen-bonded water (Grotthuss). A single water molecule ionises only once in about 1/k₋₁≈4×10⁴ s (about 11 hours).

Application. relaxation methods (temperature jump, pressure jump, microwave pulse) measure reactions too fast for mixing, with half-lives down to nanoseconds; they gave the rates of proton transfer, metal–ligand substitution and enzyme–substrate binding.