Question

Pure $PC{l}_5$ is introduced into an evacuated chamber and it comes to equilibrium at ${250}^{o}C$ and $2atm$. The equilibrium contains $40.7\%$ chlorine by volume. Calculate the partial pressures of all the gases present at equilibrium.

Answer 1

$V_{sol}=\frac{nRT}{P}=1\times .0821\times 5232=.05L$ or $50ml$

Now in $100ml$ of solution chlorine is $40.7ml$

$\therefore$ in $50ml$ chlorine is $=\frac{40.7\times 50}{100}=20.35ml$

$PC{l}_5⟶PC{l}_3+C{l}_2$

As the stochiometry coffiecient of $PC{l}_3$ and $C{l}_2$ is same so the volumes will be same.

So $V_{PC{l}_3}=V_{C{l}_2}=20.35ml$ $Moles\left(n\right)=\frac{V}{22.4}$

$n_{PC{l}_3}or{n}_{C{l}_2}=\frac{20.35}{22400}=9\times {10}^{-4}$

$V_{PC{l}_5}=100-40.7=59.3ml$

so $n_{PC{l}_5}=\frac{59.3}{22400}=2.6\times {10}^{-3}$

Now

$P_{PC{l}_5}=\frac{2.6\times {10}^{-3}}{4.4\times {10}^{-3}}$

$P_{PC{l}_3}or{P}_{C{l}_2}=\frac{.9\times {10}^{-3}}{4.4\times {10}^{-3}}$