Question

$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
In above reaction at equilibrium condition mole fraction of the $PC{l}_5$ is 0.4 and mole fraction of the $C{l}_2$ is 0.3, then find out mole fraction of the $PC{l}_3$.

• A. 0.3
• B. 0.7
• C. 0.4
• D. 0.6

Answer 1

Answer: (A)

0.3

Solution:- (A) $0.3$

$\underset{\text{At equilibrium}}{\underset{\text{Initally}}{}}\underset{a-x}{\underset{a}{{P{Cl}_5}_{\left(g\right)}}}⟶\underset{x}{\underset{0}{{P{Cl}_3}_{\left(g\right)}}}+\underset{x}{\underset{0}{{{Cl}_2}_{\left(g\right)}}}$

Total moles at equilibrium $=(a-x)+x+x=a+x$

Now,

Mole fraction of ${Cl}_2=\frac{x}{a+x}=$ Mole fraction of $P{Cl}_3$

Mole fraction of ${Cl}_2=0.3\left(\text{Given}\right)$

Therefore,

Mole fraction of $P{Cl}_3=0.3$

Hence the mole fraction of $P{Cl}_3$ is $0.3$.