Question

$75\%$ of a first order reaction was completed in $32$ minutes. When was $50\%$ of the reaction completed?

• A. $4\text{min}$
• B. $8\text{min}$
• C. $16\text{min}$
• D. $24\text{min}$

Answer 1

The correct option is C $16\text{min}$

Since it is a first order reaction, for $75\%$ completion,

$k=\frac{2.303}{32}log\left(\frac{1}{0.25}\right)....\left(i\right)$
Now, let $t$ is the time taken for $50\%$ completion, then

$\Rightarrow k=\frac{2.303}{t}log\left(\frac{1}{0.5}\right)....\left(ii\right)$
Equating (i) and (ii)
$\Rightarrow \frac{2.303}{32}log4=\frac{2.303}{t}log2$
$\Rightarrow t=32\times \frac{log2}{log4}$
$\Rightarrow t=32\times \frac{log\left(2\right)}{log(2{)}^2}$
$\Rightarrow t=32\times \frac{log\left(2\right)}{2log\left(2\right)}$
$\Rightarrow t=16\text{min}$