Question

In a first-order reaction A → B, if k is rate constant and inital
concentration of the reactant A is 0.5 M, then the half-life is
(a) log2/k
(b) log2/k√0.5
(c) ln 2/k
(d) 0.693/0.5k

Answer 1

Answer:

It requires counting to decide the answer

Answer 2

Option (a)

$t_{\frac{1}{2} }=\frac{ln2}{k}$

Explanation:

First order reaction: The rate of reaction of a first order reaction depends only the concentration of the reactant.

A→B

$t=\frac{1}{k} ln\frac{[A]_0}{[A]}$

[A]₀ = initial concentration, [A] = final concentration, k= rate constant.

Half life: The half life of a reaction is denoted a time in that the concentration of the reactant reduced to its 50% of the initial concentration.

$t_{\frac{1}{2} }=\frac{1}{k} ln\frac{[A]_0}{[\frac{A}{2}]}$

Here [A]₀ = 0.5 M  ,

$[\frac{A}{2} ]= \frac{0.5}{2} M$= 0.25 M

Therefore

$t_{\frac{1}{2} }=\frac{1}{k} ln\frac{0.5}{0.25}$

$\Rightarrow t_{\frac{1}{2} }=\frac{1}{k} ln2$

$\Rightarrow t_{\frac{1}{2} }=\frac{ln2}{k}$