Question

The decomposition of a compound A, 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 160.25 mm of Hg and after a long period of time the total pressure observed to be 220.25 mm of Hg. Calculate the total pressure of the vessel just after 2 hours in mm of Hg, if volume of liquid S is supposed to be
negligible.
Given : Vapour pressure of S (l) at temperature T = 20.25 mm of Hg.

Answer 1

$2P\left(g\right)\to 4Q\left(g\right)+R\left(g\right)+S\left(l\right)$
$\begin{array}{cccc}p& 0& 0& mmHg\left(initialpressure\right(let\left)\right)\\ p-2x& 4x& x& mmHgat30min\\ p-2y& 4y& y& mmHgat2hours\\ 0& 2p& \frac{p}{2}& mmHgatlongtime\end{array}$
At 30 min total pressure due to gases = 160.25-20.25= 140 mm of Hg (v.p.=20.25 mm of Hg)
$\Rightarrow p-2x+4x+x=p+3x=140$---(1)
As the total pressure after long time is 220.25 mm of Hg
Means the total pressure due to gases = 220.25-20.25= 200 mm of Hg
Hence 2p + $\frac{p}{2}$ = 200
so p = $80mmofHg$
by using equation (1)
So, $80+3x=140\Rightarrow x=20mmofHg$
So, pressure of P(g) at 30 min $=p-2x=80-2\times 20=40mmofHg,$ which is half of initial pressure, hence 30 min is half life.
So, just after 2 hours (i.e.4 half lives),
pressure of $P\left(g\right)=\frac{1}{16}\times initialpressure=\frac{1}{16}\times 80=5mmofHg$
So, $p-2y=5\Rightarrow 80-2y=5\Rightarrow y=37.5mmofHg$
So, total pressure of gases without vapour pressure $=p+3y=80+3\times 37.5=192.5$
And total pressure= 192.5+20.25=212.75 mm of Hg