Question

The degree of dissociation of $PC{l}_5$ at a certain temperature and $1\text{atm}$ pressure is $0.40$. Calculate the value of equilibrium constant $K_p$.
$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$

• A. $0.19$
• B. $0.26$
• C. $0.15$
• D. $0.29$

Answer 1

The correct option is A $0.19$

Let, the degree of dissociation be $\alpha$.
$P$ be the total pressure at equilibrium.
$$\begin{array}{cccccc}& PC{l}_5& \rightleftharpoons & PC{l}_3& +& C{l}_2\\ \text{Initial moles}& 1& & 0& 0& \\ \text{At equilibrium}& 1-\alpha & & \alpha & \alpha & \end{array}$$
Total number of moles at equilibrium = $1-\alpha +\alpha +\alpha =1+\alpha$
Now, we know $K_p$ = $\frac{P_{PC{l}_3}\times P_{C{l}_2}}{PC{l}_5}$
$P_{PC{l}_5}=\left(\frac{1-\alpha }{1+\alpha }\right)$ $P$
$P_{PC{l}_3}=\left(\frac{\alpha }{1+\alpha }\right)$ $P$
$P_{C{l}_2}=\left(\frac{\alpha }{1+\alpha }\right)$ $P$
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, $P=1\text{atm}$, $\alpha =0.4$
On substituting the values, we get,
$\Rightarrow K_p=0.19\text{atm}$