Q15 · CSIR NET Chemistry, December 2011
Paper: CSIR NET December 2011 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Rate Laws Order · Marks: 2 · Difficulty: Easy
The half life of a zero order reaction $(A \rightarrow P)$ is given by $(k=$ rate constant $)$ :
(a)$t_{1/2}=\dfrac{[A]_{0}}{2k}$
(b)$t_{1/2}=\dfrac{2.303}{k}$
(c)$t_{1/2}=\dfrac{[A]_{0}}{k}$
(d)$t_{1/2}=\dfrac{1}{k[A]_{0}}$
Answer
Answer: A ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
Zero order: [A] = [A]₀ − kt, so [A]₀/2 = [A]₀ − kt₁/₂ ⇒ t₁/₂ = [A]₀/2k.
Study loop for Chemical Kinetics