Question

In the reaction $PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$ the partial pressure of $PC{l}_3,C{l}_2$ and $PC{l}_5$ are $0.3,0.2$ and $0.6atm$ respectively at equilibrium. If partial pressure of $PC{l}_3$ and $C{l}_2$ was increased twice, what will be the partial pressure of $PC{l}_5$ is in $atm$ at new equilibrium condition:

• A. $0.3$
• B. $1.2$
• C. $2.4$
• D. $0.15$

Answer 1

Answer: (C)

$2.4$

$PC{l}_5\stackrel{K_p}{\leftrightharpoons }PC{l}_3+C{l}_2$

$K_C=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}$,

Equilibrium constant of reaction remain same.

$K_P=\frac{\left[P_{PC{l}_3}\right]\left[P_{C{L}_2}\right]}{\left[P_{PC{l}_5}\right]}$

$K_P=\frac{\left(0.3\right)\left(0.2\right)}{0.6}=0.1$,

Now as concentrations are doubled for $PC{l}_3$ and $C{l}_2$

$K_p=0.1=\frac{\left(0.6\right)\left(0.4\right)}{\left[PC{l}_5\right]}$

$⟹P_{PC{l}_5}=2.4$

Hence, the correct option is C.