Question

Two moles of $P{Cl}_5$ are heated in a closed vessel of $2$ litre capacity. When the equilibrium is attained $40$% of it has been found to be dissociated. What is the value of $K_c$ in $mol/{dm}^3$?

• A. $0.532$
• B. $0.266$
• C. $0.133$
• D. $0.174$
• E. $0.25$

Answer 1

The correct option is B $0.266$

Two moles of $P{Cl}_5$ are heated in a closed vessel of $2$ litre capacity. When the equilibrium is attained $40$% of it has been found to be dissociated. The value of $K_c$ is $0.266$ $mol/{dm}^3$.
$P{Cl}_5\left(g\right)\rightleftharpoons P{Cl}_3\left(g\right)+{Cl}_2\left(g\right)$
Initial $\left[P{Cl}_5\right]=\frac{\text{2 mol}}{\text{2 L}}=\text{1 M}$
When the equilibrium is attained $40$% of $P{Cl}_5$ has been found to be dissociated.
Change in $\left[P{Cl}_5\right]=\frac{\text{40}}{\text{100}}\times \text{1 M}=\text{0.40 M}$
Equilibrium $\left[P{Cl}_5\right]=\text{1 M}-\text{0.40 M}=\text{0.60 M}$
Equilibrium $\left[P{Cl}_3\right]=\text{0.40 M}$
Equilibrium $\left[{Cl}_2\right]=\text{0.40 M}$
$K_c=\frac{\left[P{Cl}_3\right]\left[{Cl}_2\right]}{\left[P{Cl}_5\right]}$
$K_c=\frac{\text{0.40 M}\times \text{0.40 M}}{\text{0.60 M}}$
$K_c=0.266$ $mol/{dm}^3$.