Question

Sol. [Key Concept: Equilibrium constant and degree of dissociation]

$$2{NO}_{(g)} \rightleftharpoons N_{2(\text{ }g)} + O_{2(\text{ }g)}$$

$\begin{matrix} K_{c} & = \frac{\left\lbrack N_{2} \right\rbrack\left\lbrack O_{2} \right\rbrack}{\left\lbrack {NO}^{2} \right.\ } \\ & = \frac{3 \times 10^{- 3} \times 4.2 \times 10^{- 3}}{2.8 \times 10^{- 3} \times 2.8 \times 10^{- 3}} \\ & = 1.607 \\ & 2{NO}_{(g)} \rightleftharpoons N_{2(\text{ }g)} + O_{2(\text{ }g)} \\ t = 0\ 0.(3)\ 0\ 0 & 0.05\alpha \\ 0.1 & - 0.1\alpha\ 0.05\alpha\ 0.05\alpha \times 0.05\alpha \\ {\text{ }K}_{c} & = \frac{0.1 - 0.1\alpha)^{2}}{(0.1} \\ K_{c} = \frac{0.05\alpha \times 0.05\alpha}{0.01(1 - \alpha)^{2}} & \\ 1.607 = \frac{(0.05)^{2}\alpha^{2}}{0.01(1 - \alpha)^{2}} & \\ & \frac{\alpha^{2}}{(1 - \alpha)^{2}} = \frac{1.607 \times (0.1)^{2}}{(0.05)^{2}} \\ & \frac{\alpha}{1 - \alpha} = \frac{1.27 \times 0.1}{0.05} \\ & \frac{\alpha}{1 - \alpha} = 2.54 \\ & \alpha = 2.54 - 2.54\alpha \\ & 3.54\alpha = 2.54 \\ & \alpha = \frac{2.54}{3.54} = 0.717 \end{matrix}$.