Question

Rate of the chemical reaction: $nA\to \text{Products}$, is doubled when the concentration of $A$ is increased four times. If the half time of the reaction at any temperature is $16min$, time required for $75\%$ of the reaction to complete is:

• A. $24.0min$
• B. $27.3min$
• C. $48.0min$
• D. $49.4min$

Answer 1

The correct option is B $27.3min$

$\text{Rate}\left(r\right)=k[A{]}^x\text{where,}x=\text{order}$
$r_1=k[A{]}^x....(1)$
$2r_1=k[4A{]}^x....(2)$
$\text{Dividing}\left(1\right)\text{by}\left(2\right)\text{,}$
$\left(\frac{r_1}{2r_1}\right)={\left(\frac{A}{4A}\right)}^x$
$\frac{1}{2}={\left(\frac{1}{4}\right)}^x\therefore x=\frac{1}{2}$

Let initial concentration be $A_0$.

We know that,
$t_{1/2}\propto \frac{1}{[A_0{]}^{x-1}}{t}_{1/2}\propto \frac{1}{[A_0{]}^{\frac{1}{2}-1}}{t}_{1/2}\propto [A_0{]}^{1/2}\therefore t_{1/2}=k\sqrt{\left[A_0\right]}$

$t_{1/2\left(I\right)}=k\sqrt{\left[A_0\right]}.....\left(1\right)$

$t_{1/2\left(II\right)}=k\sqrt{\frac{\left[A_0\right]}{2}}.....\left(2\right)$

$\text{Dividing}\left(1\right)\text{by}\left(2\right)\text{,}$
$\frac{16}{t_{1/2\left(II\right)}}=\sqrt{2}{t}_{1/2\left(II\right)}=\frac{16}{\sqrt{2}}=11.3$

$\therefore t_{75}=t_{1/2\left(I\right)}+t_{1/2\left(II\right)}=16+11.3=27.3min$