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 $Hg$.

The total pressure of the vessel after $75$ minute, if the volume of liquid $S$ is supposed to be negligible is:

(Given : Vapour pressure of $S\left(l\right)$ at temperature $T=32.5$mm $Hg$)

• A. $P_t=379.55$ mm $Hg$
• B. $P_t=387.64$ mm $Hg$
• C. $P_t=468.23$ mm $Hg$
• D. $P_t=489.44$ mm $Hg$

Answer 1

The correct option is A $P_t=379.55$ mm $Hg$

After long time, total pressure is 617 mm Hg.
$5P=617-32.5$
$2P=255.4$ .....(1)
After 30 min, $255+4x=317-32.5$
$x=7.375$......(2)
The expression for the rate of the reaction is $k=\frac{2.303}{t}loga(a-x)=\frac{2.303}{30}log(255\times \left(247.62\right))$......(3)
After 75 min, $k=\frac{2.303}{t}loga(a-x)=\frac{2.303}{\left(75\right)}log(255\times (255-y))$......(4)
From (3) and (4), $y=23$
Total pressure $=255+4y+32.5=255+4\left(23\right)+32.5=379.55mmHg$