Question

The degree of dissociation of $PC{l}_5$ at a certain temperature and under atmospheric pressure is 0.2. Calculate the pressure at which it will be half dissociated at the same temperature:

• A. $2atm$
• B. $0.225atm$
• C. $0.12atm$
• D. $2.50atm$

Answer 1

The correct option is C $0.12atm$

Let$\alpha$ be the degree of dissociation at certain temperature and pressure P:
$\begin{array}{cccccc}& PC{l}_5& \rightleftharpoons & PC{l}_3& +& C{l}_2\\ Initialconc& 1& & 0& & 0\\ Atequilibrium& 1-\alpha & & \alpha & & \alpha \end{array}$
Total moles = $1+\alpha$
$\therefore K_p=\frac{{\alpha }^2}{1-{\alpha }^2}\times P$

Putting P = 1 atm and $\alpha =0.2$

$K_p=\frac{(0.2{)}^2}{1-(0.2{)}^2}\times 1=0.04$
When $\alpha =\frac{1}{2}=0.5$, then let pressure is P'
$K_p=\frac{{\alpha }^2}{1-{\alpha }^2}.P^{{}^{\prime }}$
$\Rightarrow 0.04=\frac{(0.5{)}^2{P}^{{}^{\prime }}}{1-(0.5{)}^2}$
$P^{{}^{\prime }}=\frac{\left(0.04\right)[1-(0.5{)}^2]}{(0.5{)}^2}$
$=0.12atm$