$\mathrm{N}_2 \mathrm{O}_5$ decomposes to $\mathrm{NO}_2$ and $\mathrm{O}_2$ and follows first order…
$\mathrm{N}_2 \mathrm{O}_5$ decomposes to $\mathrm{NO}_2$ and $\mathrm{O}_2$ and follows first order kinetics. After $50$ minutes, the pressure inside the vessel increases from $50 \mathrm{~mm} \mathrm{~Hg}$ to $87.5 \mathrm{~mm} \mathrm{~Hg}$. The pressure of the gaseous mixture after $100$ minutes at constant temperature will be ______.
$136.25 \mathrm{~mm} \mathrm{~Hg}$
$106.25 \mathrm{~mm} \mathrm{~Hg}$
$175.0 \mathrm{~mm~Hg}$
$116.25 \mathrm{~mm} \mathrm{~Hg}$
Solution
$\mathrm{N}_2 \mathrm{O}_5 \rightarrow 2 \mathrm{NO}_2+\frac{1}{2} \mathrm{O}_2$
$\begin{array}{llll}\text { At } \mathrm{t}=0 & 50 & 0 & 0 \\ \text { At } \mathrm{t}=50 \mathrm{~min} & 50-\mathrm{p}_1 & 2 \mathrm{p}_1 & \frac{\mathrm{p}_1}{2}\end{array}$
Total pressure at $50$ minutes
$\begin{aligned}
&=50-\mathrm{p}_1+2 \mathrm{p}_1+\frac{\mathrm{p}_1}{2}=87.5 \\
&50+\frac{3 \mathrm{p}_1}{2}=87.5 \\
&\frac{3 \mathrm{p}_1}{2}=37.5 \\
&\therefore \quad \mathrm{p}_1=\frac{37.5 \times 2}{3}=25
\end{aligned}$
At $\mathrm{t}=100 \mathrm{~min} \quad 50-\mathrm{p}_2 \quad 2 \mathrm{p}_2 \quad \frac{\mathrm{p}_2}{2}$
$50$ minutes is half life period
For $100$ minutes i.e. for $2$ half lives $50-p_2=12.5$
$\therefore \mathrm{p}_2=37.5 \mathrm{~mm} \text { of } \mathrm{Hg}$
Total pressure at $100$ minutes
$=50-\mathrm{p}_2+2 \mathrm{p}_2+\frac{\mathrm{p}_2}{2}$
$=50+\frac{3 \mathrm{p}_2}{2}=50+\frac{3}{2} \times 37.5$
$=50+56.25$
$=106.25 \mathrm{~mm}$ of $\mathrm{Hg}$