Q. 9.2 g of $N_{2}O_{4}(g)$ is taken in a closed 1 L vessel and heated till the following equilibrium is reached $N_{2}O_{4}(g) \rightleftharpoons 2{NO}_{2}(g)$ At equilibrium, 50% $N_{2}O_{4}(g)$ is dissociated. What is the equilibrium constant $(in\ molL^{- 1})$? (Molecular weight of $N_{2}O_{4} = 92$)
A) 0.1
B) 0.2
C) 0.3
D) 0.4
Explanation
(b)
$K_{c} = \frac{\left\lbrack NO_{2} \right\rbrack^{2}}{\left\lbrack N_{2}O_{4} \right\rbrack} = \frac{4 \times (0.05)^{2}}{0.05} = 4 \times 0.05 = \mathbf{0.2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.