Question

$2P\left(g\right)\to 4Q\left(g\right)+R\left(g\right)+S\left(l\right)$
The decomposition of compound P, at temperature T follows first order kinetics.
After 30 minutes, the total pressure developed in the closed vessel is found to be $310mmHg$ and the total pressure observed at the end of reaction is $610mmHg$.
Calculate the total pressure of the vessel after 75 minutes, if volume of liquid (S) is supposed to be negligible.
Given: Vapour pressure of $S\left(l\right)$ at temperature, $T=30mmHg$
Take:
$e^{0.4}=1.5$

• A. $P_t=37.9mmHg$
• B. $P_t=377.5mmHg$
• C. $P_t=179.55mmHg$
• D. None of these

Answer 1

The correct option is B

$P_t=377.5mmHg$

Since the reaction mixture contains a liquid compound, its vapour pressure will also contributes to the total pressure in the vessel.
$2P_{\left(g\right)}\to 4Q_{\left(g\right)}+R_{\left(g\right)}+S_{\left(1\right)}$
$t=0P_{0}00$
$t=30min(P_0-P)2P\frac{P}{2}$

$t=\infty 02P_0\frac{P_0}{2}$

So, at $t=30min$
$\text{Total pressure}=P_0-P+2P+\frac{P}{2}+\text{Vapour pressure of (S)}$
$310=P_0+1.5P+30$
i.e.
$P_0+1.5P=\mathrm{280.........}eqn\left(i\right)$

At end of reaction, $t=\infty$
$\text{Total pressure}={2P}_0+\frac{P_0}{2}+\text{Vapour pressure of (S)}$
$610=2.5P_0+30$
So,
$P_0=232mmHg$
Substituting $P_0$ in equation (1), we get,
$P=32mmHg$

Given reaction is first order kinetics so,
At 30 min,
$k=\frac{1}{t}ln\left(\frac{P_0}{P_0-P}\right)\text{(In terms of pressure})$

$k\times 30=ln\left(\frac{232}{200}\right)\Rightarrow k\approx 0.005$

At, $t=75min$
$0.0052\times 75=ln\left(\frac{232}{P_0-P}\right)$
$0.4=ln\left(\frac{232}{P_0-P}\right)$
$1.5=\frac{232}{P_0-P}$
$P_0-P=155\Rightarrow P=77mmHg$
$P_T=30+P_0+1.5P$
$P_T=377.5mmHg$