Q. Two moles of $PCl_{5}$ is heated in a closed vessel of 2 L capacity. When the equilibrium is attained 40% of it has been found to be dissociated. What is the $K_{c}$ in $\frac{mol}{dm^{3}}$?
A) 0.532
B) 0.266
C) 0.133
D) 0.174
Explanation
(b)
Number of moles of $PCl_{5}$ dissociated at equilibrium
$= 2 \times \frac{40}{100} = 0.8$
$PCl_{5}\ \ \ \rightleftharpoons \ \ PCl_{3}\ \ \ \ \ \ \ + \ \ \ \ \ \ Cl_{2}$
(Initial)2 0 0
(At equilibrium) (2-0.8) mol 0.8 mol 0.8 mol
$\left\lbrack PCl_{5} \right\rbrack = \frac{1.2}{2} = 0.6\ ML^{- 1}$
$\left\lbrack PCl_{3} \right\rbrack = \left\lbrack Cl_{2} \right\rbrack = \frac{0.8}{2} = 0.4\ ML^{- 1}$
$\therefore$ $K_{c} = \frac{\left\lbrack PCl_{3} \right\rbrack\left\lbrack Cl_{2} \right\rbrack}{\left\lbrack PCl_{5} \right\rbrack} = \frac{0.4 \times 0.4}{0.6}$
$\ \ \ \ = 0.267\ mol/{dm}^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.