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 $PC{l}_5$ is $0.4$ and mole fraction of $C{l}_2$ is $0.3$. Then find out mole fraction of $PC{l}_3$.

• A. $0.3$
• B. $0.7$
• C. $0.4$
• D. $0.6$

Answer 1

Answer: (A)

$0.3$

$\underset{(a-x)}{{P{Cl}_5}_{\left(g\right)}}\rightleftharpoons \underset{\left(x\right)}{{P{Cl}_3}_{\left(g\right)}}+\underset{\left(x\right)}{{{Cl}_2}_{\left(g\right)}}$

Total moles of equillibrium $=a-x+x+x=a+x$

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

Given that mole fraction of $P{Cl}_3$ is 0.3.

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