Q.2 The deactivation rate constant of an enzyme is 0.346 h –1 . Assuming that the deactivation process
follows first order kinetics, the half life of the enzyme in minutes is _____
The half-life of the enzyme is 120 minutes. This value is calculated using the first-order kinetics formula for enzyme deactivation.
Calculation Steps
For first-order deactivation kinetics, the half-life t1/2 is given by t1/2 = ln(2)/k, where k = 0.346 h-1.
Since ln(2) ≈ 0.693, first compute t1/2 in hours: t1/2 = 0.693 / 0.346 ≈ 2.003 hours.
Convert to minutes by multiplying by 60: 2.003 × 60 ≈ 120 minutes.
First-Order Kinetics Explained
Enzyme deactivation often follows first-order kinetics, where the rate of inactivation is proportional to the active enzyme concentration.
The integrated rate law is [E] = [E]0e-kt, and at half-life, [E] = [E]0/2, leading to t1/2 = ln(2)/k independent of initial concentration.
This contrasts with zero-order (constant rate, t1/2 = [E]0/(2k)) or second-order kinetics, but the problem specifies first-order.
Introduction to Enzyme Deactivation Half Life
Enzyme deactivation follows first order kinetics in many biological processes, where the enzyme deactivation half life depends solely on the deactivation rate constant. For a rate constant of 0.346 h⁻¹, the half life of the enzyme in minutes is a key calculation for biochemistry students preparing for exams like CSIR NET. This guide breaks down the formula t1/2 = 0.693/k, derivation, and precise computation.
Step-by-Step Half Life Calculation
Identify the formula for first order enzyme deactivation: t1/2 = ln(2)/k ≈ 0.693/k.
Plug in k = 0.346 h⁻¹: t1/2 = 0.693 / 0.346 = 2.003 hours.
Convert hours to minutes: 2.003 × 60 = 120 minutes— the exact answer for this enzyme deactivation rate constant problem.
This matches numerical examples where similar k values (e.g., 0.346 min⁻¹) yield t1/2 = 2 min, scaled by units.
Why First Order Kinetics Applies to Enzymes
First order means deactivation rate −d[E]/dt = k[E], exponential decay.
Common in thermal or pH-induced enzyme inactivation.
Half life is constant, unlike zero-order (linear decay).
| Order | Half Life Formula | Dependency on [E]₀ |
|---|---|---|
| Zero | [E]₀/(2k) | Directly proportional |
| First | 0.693/k | Independent |
| Second | 1/(k[E]₀) | Inversely proportional |
CSIR NET Exam Tips
Practice enzyme deactivation first order kinetics problems by verifying units (h⁻¹ to minutes) and using ln(2) = 0.693. Avoid errors like forgetting unit conversion, which changes 2 hours to 120 minutes. For competitive exams, derive from [E] = [E]0/2 at t1/2.
