Question

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

• A. 24.0 min
• B. 27.3 min
• C. 48 min
• D. 49.4 min

Answer 1

The correct option is B 27.3 min

$Rate=k[A{]}^x$

$r_1=k[A{]}^x.....(1)$
$2r_1=k[4A{]}^x.....(2)$

Dividing (1) by (2)

$\left(\frac{r_1}{2r_1}\right)=(\frac{A}{4A}{)}^x$

$\frac{1}{2}=(\frac{1}{4}{)}^x;x=\frac{1}{2}$

We know that $t_{\frac{1}{2}}\propto \frac{1}{a^{n-1}}$

$t_{\frac{1}{2}}\propto \frac{1}{a^{\frac{1}{2}-1}}$

$t_{\frac{1}{2}}\propto a^{\frac{1}{2}}$

$t_{\frac{1}{2}}=k\sqrt{a}$

For $1^{st}{t}_{\frac{1}{2}}=k\sqrt{a}$....(1)
For $2^{nd}{t}_{\frac{1}{2}}=k\sqrt{\frac{a}{2}}$....(2)

Dividing (1) by (2)

$\frac{16}{t_{\frac{1}{2\left(II\right)}}}=\sqrt{2}$
$t_{\frac{1}{2\left(II\right)}}=\frac{16}{\sqrt{2}}=11.3$

For 75% completion, we need first and second $t_{\frac{1}{2}}=16+11.3=27.3min.$

Hence, option B is correct.