The reaction given below, involving the gases is observed to be first order with rate constant $7.48 \times 10^{-3} {sec}^{-1}$.

Question

The reaction given below, involving the gases is observed to be first order with rate constant $7.48 \times 10^{-3} \mathrm{sec}^{-1}$. Calculate the time required for the total pressure (in a system containing $A$ at an initial pressure of $0.1 \mathrm{~atm}$ ) to rise to $0.145 \mathrm{~atm}$ and also, find the total pressure after $100 \mathrm{sec}$.

$2 A_{(g)} \rightarrow 4 B_{(g)}+C_{(g)}$

(A) $46.5 \mathrm{sec}, 0.12 \mathrm{~atm}$

(B) $47.7 \mathrm{sec}, 0.18 \mathrm{~atm}$

(C) $50.2 \mathrm{sec}, 0.16 \mathrm{~atm}$

(D) None of these

Answer 1

$\quad 2 A_{(g)} \rightarrow 4 B_{(g)}+C_{(g)}$

Initial : $P_0 \quad 0 \quad 0$

At time : $t \quad P_0-P^{\prime} \quad 2 P^{\prime} \quad P^{\prime} / 2$

$\begin{array}{l}P_{\text {total }}=P_0-P^{\prime}+2 P^{\prime}+P^{\prime} / 2=P_0+\frac{3 P^{\prime}}{2}\\P^{\prime}=\frac{2}{3}\left(P_{\text {total }}-P_0\right) \\P^{\prime}=\frac{2}{3}(0.145-0.1)=0.03 \mathrm{~atm} \\K=\frac{2.303}{t} \log \frac{P_0}{P_0-P^{\prime}} \\t=\frac{2.303}{7.48 \times 10^{-3}} \log \frac{0.1}{0.07}=47.7 \mathrm{sec}\end{array}$

Also, $k=\frac{2.303}{t} \log \frac{P_0}{P_0-P^{\prime}}$

$\begin{aligned}k= & \frac{2.303}{100} \log \left(\frac{0.1}{0.1-P^{\prime}}\right) \\\Rightarrow & 0.1-P^{\prime}=0.047 \text { or } P^{\prime}=0.053 \\& P_{\text {total }}=0.1+\frac{3}{2}(0.053) \approx 0.18 \mathrm{~atm}\end{aligned}$

So, The correct option is (B).