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

(a) 3/(2kC₀²). Explanation: Third order: 1/C² − 1/C₀² = 2kt; at C = C₀/2, t½ = 3/(2kC₀²) (a).

Q14 · 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 $t_{1/2}$ for a third order rection $3\mathrm{C}\rightarrow$ products, where $C_0$ is the initial concentration of C, will be

(a)$\dfrac{3}{2kC_0^{2}}$

(b)$\dfrac{3}{kC_0}$

(c)$\dfrac{3}{2kC_0}$

(d)$\dfrac{2}{3kC_0^{2}}$

Answer

Answer: A ✓ checked by 4AB · confidence high

Explanation

Third order: 1/C² − 1/C₀² = 2kt; at C = C₀/2, t½ = 3/(2kC₀²) (a).

Study loop for Chemical Kinetics