A pharmaceutical company is studying the degradation of a drug compound X in aqueous solution. The degradation follows the mechanism given below:
Step I : X + H⁺ → Y (slow)
Step II : Y + H₂O → Z + H⁺ (fast)
In a separate kinetics experiment, the rate of decomposition of X was monitored at two temperatures. At 300 K, the rate constant k₁ = 2.0 × 10⁻³ s⁻¹. At 320 K, the rate constant k₂ = 8.0 × 10⁻³ s⁻¹.
A pharmaceutical company is studying the degradation of a drug compound X in aqueous solution. The degradation follows the mechanism given below:
Step I : X + H⁺ → Y (slow)
Step II : Y + H₂O → Z + H⁺ (fast)
In a separate kinetics experiment, the rate of decomposition of X was monitored at two temperatures. At 300 K, the rate constant k₁ = 2.0 × 10⁻³ s⁻¹. At 320 K, the rate constant k₂ = 8.0 × 10⁻³ s⁻¹.
(a) Write the rate law expression for the overall reaction and identify the order with respect to each reactant. Also state the role of H⁺ in this mechanism. [2]
(b) Calculate the energy of activation (Eₐ) for the degradation of drug X.
[Given: 2.303R = 19.15 J K⁻¹ mol⁻¹, log 2 = 0.301] [2]
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The rate-determining step (slow step) governs the rate law.
Step I (slow): X + H⁺ → Y
∴ Rate = k[X][H⁺]
Order with respect to X = 1 (first order)
Order with respect to H⁺ = 1 (first order)
Overall order of reaction = 1 + 1 = 2 (second order)
Role of H⁺: H⁺ is a catalyst — it is consumed in Step I (slow) but regenerated in Step II (fast), so its concentration remains unchanged at the end of the reaction. It increases the rate without being permanently consumed.
(½ mark: correct rate law expression; ½ mark: orders stated; ½ mark: overall order; ½ mark: role of H⁺ as catalyst correctly justified)
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(b) Calculation of Energy of Activation (Eₐ): [2 marks]
Given:
k₁ = 2.0 × 10⁻³ s⁻¹ at T₁ = 300 K
k₂ = 8.0 × 10⁻³ s⁻¹ at T₂ = 320 K
2.303R = 19.15 J K⁻¹ mol⁻¹, log 2 = 0.301
Using the Arrhenius equation:
log(k₂/k₁) = (Eₐ / 2.303R) × (1/T₁ − 1/T₂)
Substituting:
log(8.0 × 10⁻³ / 2.0 × 10⁻³) = (Eₐ / 19.15) × (1/300 − 1/320)
log 4 = (Eₐ / 19.15) × (1/300 − 1/320)
log 4 = log 2² = 2 × 0.301 = 0.602
(1/300 − 1/320) = (320 − 300)/(300 × 320) = 20/96000 = 2.083 × 10⁻⁴ K⁻¹
∴ 0.602 = (Eₐ / 19.15) × 2.083 × 10⁻⁴
Eₐ = (0.602 × 19.15) / (2.083 × 10⁻⁴)
Eₐ = 11.529 / (2.083 × 10⁻⁴)
∴ Eₐ = 5.535 × 10⁴ J mol⁻¹ ≈ 55.35 kJ mol⁻¹
(½ mark: correct formula written; ½ mark: correct substitution including correct log ratio and temperature term; ½ mark: correct simplification; ½ mark: correct final answer with unit)