Q. 9.2g $N_{2}O_{4}$ is heated in a 1L vessel till equilibrium state is established $N_{2}O_{4}(g) \rightleftharpoons 2NO_{2}(g)$ In equilibrium state 50% $N_{2}O_{4}$ was dissociated, Equilibrium constant will be (mol. wt. of $N_{2}O_{4} = 92)$
A) 0.1
B) 0.4
C) 0.3
D) 0.2
Explanation
(d)
$N_{2}O_{4}(g) \rightleftharpoons 2NO_{2}(g)$
Molar concentration of
$\left\lbrack N_{2}O_{4} \right\rbrack = \frac{9.2}{92} = 0.1$ mol/L
In equilibrium state,
(When it 50% dissociates)
$\left\lbrack N_{2}O_{4} \right\rbrack = 0.05\ M$
$\left\lbrack NO_{2} \right\rbrack = 0.1\ M$
$\because$ $K_{c} = \frac{\left\lbrack NO_{2} \right\rbrack^{2}}{\lbrack N_{2}O_{4}\rbrack}$
$\therefore$ $K_{c} = \frac{0.1 \times 0.1}{0.05}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0.2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.