QJEEAdv18Ch30b9: JEE Advanced 2018 Chemistry - Class 12 JEEAdv18Ch30b9 Solution: For a first order reaction at constant volume and the total…

Total pressure of gaseous mixture at beginning $(T=0)=\mathrm{P}_0$

Total pressure of gaseous mixture after time $t$ $(\mathrm{T}=t)=\mathrm{P}_t$

Initial concentration of $A=[A]_0$

Temperature of the reaction mixture (T) $=300 \mathrm{~K}$

For the first order reaction,


Total pressure after time $(\mathrm{T}=t)=\mathrm{P}_0-\mathrm{P}+2 \mathrm{P} +\mathrm{P}$

$\begin{aligned}\mathrm{P}_t=\mathrm{P}_0+2 \mathrm{P} \\\\\mathrm{P} =\frac{\mathrm{P}_t-\mathrm{P}_0}{2}\end{aligned}$

According to integrated rate law for the first order reaction:

$\begin{aligned}&t =\frac{2.303}{k} \log \frac{\mathrm{P}_0}{\mathrm{P}_t} ~~~~...(i)\\\\&t =\frac{2.303}{k} \log \frac{\mathrm{P}_0}{\mathrm{P}_0-\mathrm{P}} \\\\&t =\frac{2.303}{k} \log \frac{\mathrm{P}_0}{\mathrm{P}_0-\frac{\left(\mathrm{P}_t-\mathrm{P}_0\right)}{2}} \\\\&t =\frac{2.303}{k} \log \frac{2 \mathrm{P}_0}{2 \mathrm{P}_0-\mathrm{P}_t+\mathrm{P}_0} \\\\&t =\frac{2.303}{k} \log \frac{2 \mathrm{P}_0}{3 \mathrm{P}_0-\mathrm{P}_t} \\\\&t =\frac{1}{k} \ln \frac{2 \mathrm{P}_0}{3 \mathrm{P}_0-\mathrm{P}_t} \\\\&\ln \left(3 \mathrm{P}_0-\mathrm{P}_t\right) =\ln \left(2 \mathrm{P}_0-\mathrm{P}_t\right)-k t ~~~~...(ii)\end{aligned}$

Comparing equation (ii) with equation of straight option line

$\begin{aligned}& \ln \left(3 \mathrm{P}_0-\mathrm{P}_t\right) =\ln 2 \mathrm{P}_0-k t \\\\&y =c+m x\end{aligned}$

Plot of $\ln \left(3 \mathrm{P}_0-\mathrm{P}_t\right)$ verses time $(t)$ is represented as:


Option (A) is correct.

(ii) For the first order reaction,


$\mathrm{P}_t=\frac{\mathrm{P}_0}{3}+\frac{4 \mathrm{P}_0}{3}+\frac{2 \mathrm{P}_0}{3}=\frac{7 \mathrm{P}_0}{3}$

Substituting the value of $P_t$ in $e q$ (i)

$\begin{aligned}t_{1 / 3} & =\frac{2.303}{k} \log \frac{\mathrm{P}_0}{\frac{7 \mathrm{P}_0}{3}} \\\\t_{1 / 3} & =\frac{2.303}{k} \log \frac{3}{7}\end{aligned}$

$\mathrm{P}_{t_3}$ is independent of concentration of (A).

Option (B) is not correct.

(iii) At any time $t$ , rate constant $(k)$ is independent of concentration of [A].

For example:

At $t_{y_3}$ :

$k=\frac{2.303}{t_{1 / 3}} \times \log \frac{3}{7}$

Hence, Option (D) is correct.