Question

A certain temperature, the degree of dissociation of ${PCl}_5$ was found to be 0.25 under a total pressure of 15 atm. The value of $Kp$ for the dissociation of ${PCl}_5$ is:

• A. 1
• B. 0.25
• C. 0.5
• D. 0.75

Answer 1

The correct option is A 1

$\alpha =0.25,P=15atm$
$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$
$1$ $0$ $0$
$1-\alpha$ $\alpha$ $\alpha$
$n_e=1+\alpha$
$Kp=\frac{P_{C{l}_2}.P_{PC{l}_3}}{P_{PC{l}_5}}=\frac{\left(\frac{\alpha }{1+\alpha }P\right)\left(\frac{\alpha }{1+\alpha }P\right)}{\left(\frac{1-\alpha }{1+\alpha }P\right)}$
$Kp=\frac{{\alpha }^{2}P}{1-{\alpha }^2}$

$=\frac{\left(0.25{)}^2\right(15)}{1-(0.25{)}^2}=\frac{0.9375}{0.975}=1$

$=1$