a. The pKa of NH4+ is 9.25 at 25°C. The rate constant at 25°C for the reaction of NH4+ and OH− to form aqueous NH3 is 4.0 × 1010 dm3 mol−1 s−1. Calculate the rate constant for proton transfer to NH3. What relaxation time would be observed if a temperature jump were applied to a solution of 0.15 mol dm−3 NH3(aq) at 25°C?
b. The equilibrium A 5B + C at 25°C is subjected to a temperature jump 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 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 for steps (1) and (2).
The Correct Answer and Explanation is:
Let’s solve each part step-by-step.
a. Reaction of NH₄⁺ with OH⁻ and Relaxation Time
Given:
- pKa of NH₄⁺ = 9.25 → Ka = 10^(-9.25) ≈ 5.62 × 10⁻¹⁰
- Rate constant for forward reaction (NH₄⁺ + OH⁻ → NH₃ + H₂O):
k₁ = 4.0 × 10¹⁰ dm³ mol⁻¹ s⁻¹ - NH₄⁺ ⇌ NH₃ + H⁺ is the relevant acid-base equilibrium
- Equilibrium constant K = [NH₃][H⁺]/[NH₄⁺] = Ka
So, we know:
K=k1k−1⇒k−1=k1KK = \frac{k_1}{k_{-1}} \Rightarrow k_{-1} = \frac{k_1}{K}
Substitute: k−1=4.0×10105.62×10−10≈7.12×1019 s−1k_{-1} = \frac{4.0 \times 10^{10}}{5.62 \times 10^{-10}} ≈ 7.12 \times 10^{19} \ \text{s}^{-1}
Now, calculate the relaxation time (τ) for a simple acid-base reaction: τ=1k1[OH−]+k−1\tau = \frac{1}{k_1[\text{OH}^-] + k_{-1}}
Assuming [OH⁻] is from 0.15 mol dm⁻³ NH₃:
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻⇒Kb=KwKa=10−145.62×10−10≈1.78×10−5\text{NH₃ + H₂O ⇌ NH₄⁺ + OH⁻} \Rightarrow K_b = \frac{K_w}{K_a} = \frac{10^{-14}}{5.62 \times 10^{-10}} \approx 1.78 \times 10^{-5}
For NH₃ (0.15 M): [OH−]≈Kb⋅[NH₃]=1.78×10−5⋅0.15≈1.63×10−3 mol dm−3[\text{OH}^-] ≈ \sqrt{K_b \cdot [\text{NH₃}]} = \sqrt{1.78 \times 10^{-5} \cdot 0.15} ≈ 1.63 \times 10^{-3} \ \text{mol dm}^{-3} τ=1(4.0×1010)(1.63×10−3)+7.12×1019≈17.12×1019 s\tau = \frac{1}{(4.0 \times 10^{10})(1.63 \times 10^{-3}) + 7.12 \times 10^{19}} ≈ \frac{1}{7.12 \times 10^{19}} \ \text{s} τ≈1.4×10−20 s\tau ≈ 1.4 \times 10^{-20} \ \text{s}
Answer (a):
- k₋₁ = 7.12 × 10¹⁹ s⁻¹
- τ ≈ 1.4 × 10⁻²⁰ s
b. Equilibrium A ⇌ 5B + C (relaxation time method)
Given:
- Relaxation time, τ = 3.0 μs = 3.0 × 10⁻⁶ s
- Equilibrium constant, K = [B]^5[C]/[A] = 2.0 × 10⁻¹⁶
- [B] = [C] = 2.0 × 10⁻⁴ mol dm⁻³
We use the relation: τ=1kf+kr\tau = \frac{1}{k_f + k_r}
Where:
- Forward: A → 5B + C → rate = k_f [A]
- Reverse: 5B + C → A → rate = k_r [B]^5[C]
From K: K=kfkr=2.0×10−16⇒kf=K⋅krK = \frac{k_f}{k_r} = 2.0 \times 10^{-16} \Rightarrow k_f = K \cdot k_r
Also: τ=1kf+kr=1Kkr+kr=1kr(1+K)\tau = \frac{1}{k_f + k_r} = \frac{1}{K k_r + k_r} = \frac{1}{k_r(1 + K)} kr=1τ(1+K)≈1(3.0×10−6)(1+2.0×10−16)≈3.33×105 s−1k_r = \frac{1}{\tau (1 + K)} ≈ \frac{1}{(3.0 \times 10^{-6})(1 + 2.0 \times 10^{-16})} ≈ 3.33 \times 10^{5} \ \text{s}^{-1} kf=K⋅kr=(2.0×10−16)(3.33×105)≈6.67×10−11 s−1k_f = K \cdot k_r = (2.0 \times 10^{-16})(3.33 \times 10^5) ≈ 6.67 \times 10^{-11} \ \text{s}^{-1}
Final Answers:
(a)
- Proton transfer rate constant (k₋₁): 7.12 × 10¹⁹ s⁻¹
- Relaxation time (τ): 1.4 × 10⁻²⁰ s
(b)
- k_f = 6.67 × 10⁻¹¹ s⁻¹
- k_r = 3.33 × 10⁵ s⁻¹
Explanation
In acid-base equilibrium systems, understanding forward and reverse rate constants provides insights into dynamic behavior. For part (a), the ammonium ion (NH₄⁺) reacts rapidly with hydroxide (OH⁻) to form ammonia (NH₃). The forward rate constant, k₁, is known. Using the acid dissociation constant (Ka), the reverse rate constant (k₋₁) is found via the relation K=k1/k−1K = k₁/k₋₁. A large k₋₁ (~10¹⁹ s⁻¹) indicates proton transfer is extremely fast, approaching diffusion control. The relaxation time τ, representing how quickly equilibrium re-establishes after a perturbation (like a temperature jump), is dominated by this large k₋₁ value, making τ extraordinarily short (∼10⁻²⁰ s), implying virtually instantaneous relaxation.
In part (b), we use the temperature-jump relaxation method to analyze a more complex equilibrium: A ⇌ 5B + C. Given the relaxation time (τ = 3 μs), equilibrium constant, and concentrations, we derive rate constants for both directions. The reverse rate constant (k_r) dominates due to the extremely small equilibrium constant (K = 2.0 × 10⁻¹⁶), reflecting a reaction that heavily favors reactants. The forward rate constant (k_f) is extremely small, consistent with the low equilibrium product concentrations. The combination of a fast reverse step and slow forward step produces the observed relaxation behavior. This method illustrates how kinetic data can reveal the detailed mechanism and speed of chemical equilibria, beyond what equilibrium constants alone can show.
