Question

$N_2{O}_5$ decomposes to $N{O}_2$ and $O_2$ and follows first order kinetics. After $50$ minutes, the pressure inside the vessel increases from $50$ mm Hg of $87.5$ mm Hg. The pressure of the gaseous mixture after $100$ minute at constant temperature will be __________.

• A. $116.25$ mmHg
• B. $106.25$ mmHg
• C. $136.25$ mmHg
• D. $175.0$ mmHg

Answer 1

Answer: (B)

$106.25$ mmHg

$N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2$

$R=k[N_2{O}_5{]}^1$

$\begin{array}{cccccc}1.& P_o& O& O& & \\ 2.& P_o-p& 2p& \frac{1}{2}p& \to & 50min\\ 3.& P_o-p^1& 2p^1& \frac{1}{2}{p}^1& \to & 100min\end{array}$

$P_o=50$

$\therefore P_{50min}=P_o+\frac{3}{2}P=87.5$

$P=\frac{2}{3}(87.5-50)$

$P=\frac{37.5}{3}\times 2=25$

First order $E{q}^n:t=\frac{2.303}{k}\log \left[\frac{\left(N_2{O}_5\right)}{(N_2{O}_5{)}_t}\right]$

At 50 min : $t=\frac{2.303}{k}\log 2$

$k=\frac{2.303}{50}\times 0.3010$

At 100min : $100=\frac{2.303\times 50}{2.303\times 0.3010}\log \left[\frac{(N_2{O}_5{)}_o}{(N_2{O}_5{)}_{100}}\right]$

$2\times 0.3010={\log }_{10}\left[\frac{50}{x}\right]$

$4=\frac{50}{x}$

$x=\frac{50}{4}$

$\therefore P_o-p^1=12.5$

$p^1=50-12.5$

$p^1=37.5$

$\therefore$ Total Pressure $=P^o+\frac{3}{2}\times 37.5=106.25$ mm of Hg

Hence, the correct option is $\text{B}$