Question

The decomposition of a compound P, at temperature T according to the equation
$2P_{\left(g\right)}\to 4Q_{\left(g\right)}+R_{\left(g\right)}+S_{\left(l\right)}$ is the first order reaction. After 30 minutes from the start of decomposition in a closed vessel, the total pressure developed is found to be 317 mm Hg and after a long period of time the total pressure observed to be 617 mm Hg. Calculate the total pressure of the vessel after 75 minute, if volume of liquid S is supposed to be negligible. Also calculate the time fraction $t_{7/8}$?
Given : Vapour pressure of S(l) at temperature T = 32.5 mm Hg.

• A. $P_t=37.955mmHg,t_{7/8}=39.996min$
• B. $P_t=379.55mmHg,t_{7/8}=399.96min$
• C. $P_t=179.55mmHg,t_{7/8}=199.96min$
• D. None of these

Answer 1

The correct option is B $P_t=379.55mmHg,t_{7/8}=399.96min$

$2P_{\left(g\right)}\to 4Q_{\left(g\right)}+R_{\left(g\right)}+S_{\left(l\right)}$
$t=0$ $P_0$
$t=30min.$ $P_0-P$2PP/2
$t=\infty$- $2P_0$ $P_0/2$
so $P_0-P+2P+P/2=317-32.5$
i.e. $P_0+1.5P=284.5$.........(i)
& $2.5P_0=617-32.5=584.5$
so$P_0=233.8$
$P=33.8$
$k\times 30=\ln \frac{233.8}{200}\Rightarrow k=0.0052$
At$t=75min$
$0.0052\times 75=\ln \frac{233.8}{P^{\circ }-P}$
$P^{\circ }-P=158.23\Rightarrow P=75.57$
$P_T=32.5+P_0+1.5P=347.155+32.5$
$P_T=379.65mmHg$
(ii) $0.0052\times t=\ln 8$
$t=399.89min.$