Question

Pure $PC{l}_5$ is introduced into an evacuated chamber and comes to equilibrium at ${247}^{\circ }\text{C}$ and $2.0atm$. The equilibrium gaseous mixture contains $40\%$ chlorine by volume. Calculate $K_p$ at ${247}^{\circ }\text{C}$ for the reaction
$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$

• A. $0.625atm$
• B. $4atm$
• C. $1.6atm$
• D. None of these

Answer 1

The correct option is C $1.6atm$

The gaseous mixture contains $40\%C{l}_2$ and $40\%PC{l}_3$, since they are produced in $1:1$ mole ratio. The $PC{l}_5$ is $20$.
For ideal gases mole % = volume %
$P_{C{l}_2}=P_{PC{l}_3}$
$\Rightarrow 2\times 0.40=0.80\text{atm}$
$PC{l}_5=2\times 0.2=0.40\text{atm}$
$\therefore K_p=\frac{P_{PC{l}_3}.P_{C{l}_2}}{P_{PC{l}_5}}$
$=\frac{0.80\times 0.80}{0.40}$
$=1.6\text{atm}$