Question

A first order reaction, calculate the ratio between the time taken to complete one half of the reaction and the time taken to complete one third of the reaction

Answer 1

Answer: 1.71

Explanation:

Rate law expression for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = time taken for decay process

a = initial amount of the reactant

a - x = amount left after decay process

The time taken for half of the reaction is designated as $t_{\frac{1}{2}$ and the concentration left after $t_{\frac{1}{2}$ is half of initial concentration.

Thus a-x= a/2

$t_\frac{1}{2}=\frac{2.303}{k}\log\frac{a}{a/2}$

    $=\frac{2.303}{k}\log{2}$

   $=\frac{0.693}{k}$     (1)

The time taken for one third of the reaction is designated as $t_{\frac{1}{3}$ and the concentration left after  $t_{\frac{1}{3}$ is 2/3 of initial concentration.

Thus  a-x= 2a/3

$t=\frac{2.303}{k}\log\frac{a}{2a/3}$

   $=\frac{2.303}{k}\log\frac{3}{2}$

   $=\frac{0.405}{k}$     (2)

Thus dividing (1) by (2)

  $\frac{\frac{0.693}{k}}{\frac{0.405}{k}}=1.71$