Question

The reaction
$2N_2{O}_5\left(g\right)\to 4N{O}_2\left(g\right)+O_2\left(g\right)$
Follows first order kinetics. The pressure of a vessel containing only $N_2{O}_5$ was found to increase from 50 mm Hg to 87.5 mm Hg in 30 min. The pressure exerted by the gases after 60 min. will be :(Assume temperature remains constant)

Answer 1

$2N_2{O}_5\left(g\right)⟶4N{O}_2\left(g\right)+O_2\left(g\right)$

$t=0$ $P_0$ $0$ $0$

$t=t$ $P_0-P$ $2P$ $P/2$

Total pressure after time $t=P_0-P+2P+P/2$

Total= $P_0+3P/2$

Given, $P_0=50mm$ $Hg$ & $P_{total}=87.5mm$ $Hg$ at $t=30$ min

$\Rightarrow 87.5=50+3P/2$

$\Rightarrow 3P/2=37.5\Rightarrow P=\frac{37.5\times 2}{3}=25mm$ $Hg$

Since the reaction follows first order kinetics,

so, $Kt=ln\frac{P_0}{P_0-P}$

$\Rightarrow Kt=ln\left(\frac{50}{25}\right)$ (at $t=30$ min)

$\Rightarrow K=\frac{1}{30}ln2$ $-\left(i\right)$

After $60$ min,

$\Rightarrow K=\frac{1}{60}ln\left(\frac{50}{50-P^{\prime }}\right)$ $-\left(ii\right)$

$\Rightarrow \frac{1}{30}ln2=\frac{1}{60}ln\left(\frac{50}{50-P^{\prime }}\right)$ (From $\left(i\right)$)

$\Rightarrow 2l{n}^2=ln(50/50-P^{\prime })$

$\Rightarrow ln{2}^2=ln(50/50-P^{\prime })$

$\Rightarrow 4=\frac{50}{50-P^{\prime }}\Rightarrow 200-4P^{\prime }=50$

$\Rightarrow 4P^{\prime }=150$

$\Rightarrow P^{\prime }=37.5mm$ $Hg$

So, total pressure at $t=60$ min=$P_0+3P^{\prime }/2$

$=50+\frac{3\times 37.5}{2}=106.25mm$ $Hg$