Question

$9.2$g $N_2{O}_4$ is heated in $1$L vessel till equilibrium state is established.
$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
In equilibrium state, $50\%$ $N_2{O}_4$ was dissociated, equilibrium constant will be :

[Molecular wt. of $N_2{O}_4=92$]

• A. $0.1$
• B. $0.4$
• C. $0.3$
• D. $0.2$

Answer 1

The correct option is D $0.2$

$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$

Molar concentration of $\left[N_2{O}_4\right]=\frac{9.2}{92}=0.1$mol$/$L

In equilibrium state,

when it $50\%$ dissociates

$\left[N_2{O}_4\right]=0.05$M

$\left[N{O}_2\right]=0.1$M

Now, from $K_C=\frac{[N{O}_2{]}^2}{\left[N_2{O}_4\right]}$

$K_C=\frac{0.1\times 0.1}{0.05}=0.2$.