Question

For a first order reaction, the time required for $99\%$ completion is $x$ times the time required for the completion of $90\%$ of the reaction. Find the value of $\frac{x}{2}\times 100$.

Answer 1

For the first order reaction, $t=\frac{2.303}{k}log\frac{a}{a-x}$
$1^{st}\text{case}:$
$a=100\%,x=99\%,a-x=1\%$
$t_{99\%}=\frac{2.303}{k}log\frac{100}{1}=\frac{2.303}{k}log{10}^2=\frac{4.606}{k}...\left(i\right)$

$2^{nd}\text{case}$:
$a=100\%,x=90\%,a-x=10\%$
$t_{90\%}=\frac{2.303}{k}log\frac{100}{10}=\frac{2.303}{k}log10=\frac{2.303}{k}...\left(ii\right)$
Dividing equation $\left(ii\right)$ by $\left(i\right)$,
$x=\frac{t_{99\%}}{t_{90\%}}=\frac{4.606}{2.303}=2$
$\therefore \frac{x}{2}\times 100=\frac{2}{2}\times 100=100$