Q. At a given temperature the equilibrium constant for the reaction of $PCl_{5} \rightleftharpoons PCl_{3} + Cl_{2}\ is\ 2.4 \times 10^{- 3}$. At the same temperature, the equilibrium constant for the reaction $PCl_{3}(g) + Cl_{2}(g) \rightleftharpoons PCl_{5}(g)$ is
A) $2.4 \times 10^{- 3}$
B) $- 2.4 \times 10^{- 3}$
C) $4.2 \times 10^{2}$
D) $4.8 \times 10^{- 2}$
Explanation
(c)
Given, $PCl_{5} \rightleftharpoons PCl_{3} + Cl_{2},\ \ K = \ 2.4 \times 10^{- 3}$
$PCl_{3} + Cl_{2} \rightleftharpoons PCl_{5},\ \ K^{'} = ?$
$\therefore$ $K^{'} = \frac{1}{K} = \frac{1}{2.4 \times 10^{- 3}} = 4.2 \times 10^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.