Question

$18.4$g of $N_2{O}_4$ is taken in a $1$ L closed vessel and heated till the equilibrium is reached.
$N_2{O}_{4\left(g\right)}\rightleftharpoons 2N{O}_{2\left(g\right)}$
At equilibrium it is found that $50\%$ of $N_2{O}_4$ is dissociated. What will be the value of equilibrium constant?

• A. $0.2$
• B. $2$
• C. $0.4$
• D. $0.8$

Answer 1

Answer: (C)

$0.4$

The given reaction is :-

$N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)$

Initial moles. $\frac{18.4}{92}=0.2$ $0$ $(\because no.ofmoles=\frac{wt.given}{molarmass})$

At eqm $(0.2-x)$ $2x$ (let)

Given that at eqm $50$% of mass is dissociated.

so, $x=\frac{50}{100}\times 0.2=0.1$

So, no. of moles of $N_2{O}_4$ at equilibrium $=0.1$

no. of moles of ${NO}_2$ at equilibrium $=0.2$

$\therefore K_C=\frac{{\left[{NO}_2\right]}^2}{\left[N_2{O}_4\right]}=\frac{0.2\times 0.2}{0.1}=0.4$