The time taken for the first-order reaction to reach 50% completion is 6 seconds. This follows from the integrated rate law and half-life properties of first-order kinetics.

Rate Law Basics

For a first-order reaction, the concentration decreases exponentially: [A]_t = [A]₀e^-kt, where k is the rate constant. Time t for fraction x completion relates as t = (1/k) ln(1/(1-x)).

Calculation Steps

Given t_90% = 20 s for 90% completion (x=0.9):

• k = ln(10)/20 ≈ 0.1155 s^-1
• Half-life t_1/2 = 0.693/k ≈ 6 s (50% completion)
• For 90%, t_90% ≈ 3.32 t_1/2, so t_1/2 = 20/3.32 ≈ 6 s

Introduction to First Order Reaction Time Calculations

In first order reaction time problems, like when 90% completion takes 20s, finding time to 50% completion is key for CSIR NET Life Sciences. The half-life t_1/2 = 0.693/k stays constant, unlike zero-order kinetics. This guide breaks down first order reaction time 90% 20s 50% completion with precise math.

Core Formulas Explained

Integrated rate law: ln([A]₀/[A]_t) = kt.

• For 90%: ln(10) = k × 20, so k = 2.3026/20
• For 50%: t = 0.693/k = 6 s exactly (rounded)
• No options given, but common MCQ traps include confusing with second-order (where t_1/2 varies) or misapplying ln(2)

Why 90% vs 50% Matters

90% completion needs ~3.3 half-lives (ln(10)/ln(2) ≈ 3.32), so reverse: t_50% = 20/3.32 ≈ 6 s. Useful for enzyme kinetics, decay in biotech—core CSIR NET topics.

Practice for CSIR NET

• Trap 1: Assume proportional time (wrong, exponential!)
• Trap 2: Use t_99% = 2 × t_90% (true, but not here)
• Master via: t_x = [ln(1/(1-x))]/k for any %