Question

The degree of dissociation of PCl5 at 1 atm is 0.2. How can one calculate the pressure at which PCl5 is dissociated to 50%?

• A. $0.134$
• B. $0.125$
• C. $1.2$
• D. $0.14$

Answer 1

Answer: (B)

$0.125$

$PC{l}_5={PCl}_3+{Cl}_2$

Then total number of moles at eqilibrium $=1+\alpha$

$i=\frac{1+\alpha }{1}$

let pressure be P.

$\frac{1-\alpha }{1+\alpha }p\frac{\alpha }{1+2}\frac{\alpha }{1+\alpha }p$

$k_p=\frac{{\alpha }^2}{(1+\alpha {)}^2}\frac{\left(1\alpha \right)}{(1-\alpha )}p$

$k_p=\frac{{\alpha }^2}{1-{\alpha }^2}p\alpha =0\cdot 2,p=1$

$k_p=0.04167atm$

Now when dissociation is $50\%$

$\alpha =0.5$ and let the pressure the $P$

$\begin{array}{cc}{K}_P=& \frac{{\alpha }^{2}P}{1{\alpha }^2}\\ 0.04167& =\frac{(0.5{)}^2}{1-(0.5{)}^2}\times P\\ P& =0.04167\times 3\\ P& =0.125atm\end{array}$

Correct answer $-\underset{–}{B}$