Half-life t1 / 2 for a third order reaction 3C → products, where C₀ is the initial concentration of C, will be

Q60 · CSIR-NET Chemistry, June 2016

Paper: CSIR-NET June 2016 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Rate Laws Order · Marks: 2 · Difficulty: Medium

Half-life $\mathrm{t}_{1 / 2}$ for a third order reaction 3C → products, where $\mathrm{C}_{0}$ is the initial concentration of C, will be

(a)$\frac{3}{2 \mathrm{kC}_{0}^{2}}$

(b)$\frac{3}{\mathrm{kC}_{0}}$

(c)$\frac{3}{2 \mathrm{kC}_{0}}$

(d)$\frac{2}{3 \mathrm{kC}_{0}^{2}}$

Answer

Answer: A ✓ checked by 4AB · confidence medium

The source book printed no answer; this one was worked out and checked — see the explanation.

Explanation

For a third-order decay 1/C² − 1/C₀² = 2kt (with −dC/dt = kC³); at C = C₀/2: 3/C₀² = 2kt₁/₂ ⇒ t₁/₂ = 3/(2kC₀²).

Study loop for Chemical Kinetics