Question

$9.2\, g\, N_2O_4$ is heated in a $1L$ vessel till equilibrium state is established $N_2O_{4(g)} {<=>}2NO_{2(g)}$ In equilibrium state $50\% N_2O_4$ was dissociated, equilibrium constant will be (mol. wt. of $N_2O_4 = 92$)

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

Answer 1

Answer: (B)

$0.2$

Answer 2

Moles of $N_{2}O_{4} =\frac{9.2}{92.0} =0.1$ mole $\begin{matrix}&N_{2}O_{4}& {<=>}&2NO_{2}(g)&&\\\ \text{Initiailly }&0.1 {\text{moles}}&&0&\\\ \text{At eqm.}&0.05&&0.05 \times 2 \end{matrix}$ $K=\frac{\left[NO_{2}\right]^{2}}{\left[N_{2}O_{4}\right]}$ $=\frac{0.1\times0.1}{0.05}$ $=0.2$