Question

The degree of dissociation of $PC{l}_5$ at a certain temperature and pressure= $1atm$ is 0.20. Calculate the value of equilibrium constant $K_p$.

• A. $0.084$ atm
• B. $0.84$ atm
• C. $0.42$ atm
• D. $0.042$ atm

Answer 1

The correct option is D $0.042$ atm

Let the degree of dissociation be $\alpha$.
$P$ be the total pressure at equilibrium.
$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$
Initial moles 1 0 0
At equi. $1-\alpha$ $\alpha$ $\alpha$
Total number of moles at equilibrium = $1-\alpha +\alpha +\alpha$ = $1+\alpha$
Now, we know $K_p$ = $\frac{P_{PC{l}_3}\times P_{C{l}_2}}{PC{l}_5}$
$P_{PC{l}_5}$ = $\frac{1-\alpha }{1+\alpha }$P
$P_{PC{l}_3}$ = $\frac{\alpha }{1+\alpha }$P
$P_{C{l}_2}$ = $\frac{\alpha }{1+\alpha }$P
So, $K_p$ = $\frac{\frac{\alpha }{1+\alpha }P\times \frac{\alpha }{1+\alpha }P}{\frac{1-\alpha }{1+\alpha }P}$
$K_p$ = $\frac{{\alpha }^2}{1-{\alpha }^2}P$
Given, P = 1 atm, $\alpha$ = 0.2
$K_p$ = 0.042 atm