Question

For a ${\text{I}}^{\text{st}}$ order reaction $nA\to B$, whose concentration Vs time curve is as shown below.

If half life for the reaction is $24\text{min}$, the value of $n$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

$\begin{array}{cccc}& nA& \to & b\\ \text{initially}& a_o& & 0\\ \text{at t time}& a_o-nx& & x\end{array}$

Given half life $\left(t_{1/2}\right)=24\text{min}$
So $48\text{min}$ is the second half life of the reaction.
$nx=\frac{75}{100}{a}_0$ or $\frac{3}{4}{a}_0$

According to the diagram,
$a_0-nx=x$
$\Rightarrow a_0=(n+1)x$
$\Rightarrow a_0-\frac{3}{4}{a}_0=x$
$\Rightarrow 4a_0-3a_0=4x$
$\Rightarrow x=\frac{a_0}{4}$
$\therefore a_0=(n+1)x$
$\Rightarrow a_0=(n+1)\times \frac{a_0}{4}$
$\Rightarrow (n+1)=4$
$\Rightarrow n=3$