Q.60 In an enzyme catalyzed first-order reaction, the substrate conversion follows an exponential
pattern such that 80 % of the substrate is converted in 10 minutes. The first-order rate
constant (in min–1) of the reaction, rounded off to THREE decimal places, is __________.
The first-order rate constant for this enzyme-catalyzed reaction is 0.161 min⁻¹.
Problem Analysis
In first-order kinetics, substrate conversion follows [S] = [S]₀ e^(-kt), where [S]/[S]₀ = 0.20 after 80% conversion in t = 10 min. Rearranging gives k = -ln(0.20)/10. This exponential decay pattern holds for enzyme reactions at low substrate concentrations ([S] << Kₘ).
Step-by-Step Solution
Take natural log: ln([S]/[S]₀) = -kt, so ln(0.20) = -k(10). ln(0.20) ≈ -1.60944, thus k = 1.60944/10 = 0.160944 min⁻¹. Rounded to three decimal places: 0.161 min⁻¹.
In enzyme-catalyzed first-order reactions, substrate conversion follows an exponential pattern governed by the rate equation [S] = [S]₀ e^(-kt). This is crucial for CSIR NET Life Sciences preparation, where such problems test understanding of enzyme kinetics under low substrate conditions ([S] << Kₘ). For 80% conversion in 10 minutes, 20% substrate remains, enabling direct calculation of the first-order rate constant in min⁻¹.
Core Formula and Derivation
The integrated rate law is ln([S]₀/[S]_t) = kt. Here, [S]_t/[S]₀ = 0.20 and t = 10 min, so k = -ln(0.20)/10 ≈ 0.161 min⁻¹ (rounded to three decimals). No options are provided in the numerical answer-type question, but common errors include using log₁₀ instead of ln (yielding ~0.070 min⁻¹) or half-life approximations.
Common Mistakes Explained
-
Using common log: k = 2.303 log(1/0.20)/10 ≈ 0.070 min⁻¹ (wrong base).
-
Half-life confusion: t₁/₂ = 0.693/k assumes 50% conversion, not 80%.
-
Zero-order assumption: Linear decay, inapplicable here.
Correct approach always uses natural log for exponential first-order decay.
CSIR NET Application
This matches CSIR NET enzyme kinetics syllabus, emphasizing first-order behavior in Michaelis-Menten kinetics at low [S]. Practice similar: 50% conversion uses t₁/₂ = 0.693/k; verify units stay min⁻¹.
