Q. In the reaction, $PCl_{5}(g) \rightleftharpoons PCl_{3}(g) + Cl_{2}(g),$ the equilibrium concentration of $PCl_{5}$ and $PCl_{3}$ are 0.4 and 0.2 mol/L respectively. If the value of $K_{c}$ is 0.5 what is the concentration of$Cl_{2}$ in mol/L?
A) 0.5
B) 1.0
C) 1.5
D) 2.0
Explanation
(b)
$K_{c} = \frac{\lbrack PCl_{3}\rbrack\lbrack Cl_{2}\rbrack}{\lbrack PCl_{5}\rbrack} = \frac{0.2 \times x}{0.4} = 0.5$
$\therefore\ x = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.