Consider the following reaction in a sealed vessel at equilibrium: $2\text{NO}(g) \rightleftharpoons \text{N}_2(g) + \text{O}_2(g)$ [Given: $[\text{N}_2]=3.0 \times 10^{-3}\text{ M}$, $[\text{O}_2]=4.2 \times 10^{-3}\text{ M}$ and $[\text{NO}]=2.8 \times 10^{-3}\text{ M}$] If $0.1\text{ mol L}^{-1}$ of $\text{NO}(g)$ is taken in a closed vessel, what will be the degree of dissociation ($\alpha$) of $\text{NO}(g)$ at equilibrium?

First, calculate the equilibrium constant ($K_c$) using the given equilibrium concentrations: $2\text{NO}(g) \rightleftharpoons \text{N}_2(g) + \text{O}_2(g)$ $K_c = \frac{[\text{N}_2][\text{O}_2]}{[\text{NO}]^2}$ $K_c = \frac{(3.0 \times 10^{-3})(4.2 \times 10^{-3})}{(2.8 \times 10^{-3})^2}$ $K_c = \frac{12.6 \times 10^{-6}}{7.84 \times 10^{-6}} = \frac{12.6}{7.84} = 1.607$

Now, $0.1\text{ mol L}^{-1}$ of $\text{NO}$ is taken in a closed vessel. Let $\alpha$ be its degree of dissociation. Initial concentration: $[\text{NO}] = 0.1\text{ M}$ $[\text{N}_2] = 0$ $[\text{O}_2] = 0$

At equilibrium: $[\text{NO}] = 0.1 - 0.1\alpha = 0.1(1 - \alpha)$ $[\text{N}_2] = \frac{0.1\alpha}{2} = 0.05\alpha$ $[\text{O}_2] = \frac{0.1\alpha}{2} = 0.05\alpha$

Using the calculated $K_c$ value: $K_c = \frac{(0.05\alpha)(0.05\alpha)}{(0.1(1 - \alpha))^2}$ $1.607 = \frac{0.0025\alpha^2}{0.01(1 - \alpha)^2} = 0.25 \left(\frac{\alpha}{1 - \alpha}\right)^2$ $\left(\frac{\alpha}{1 - \alpha}\right)^2 = \frac{1.607}{0.25} = 6.428$ Taking the square root on both sides: $\frac{\alpha}{1 - \alpha} = \sqrt{6.428} \approx 2.535$ $\alpha = 2.535(1 - \alpha)$ $\alpha = 2.535 - 2.535\alpha$ $\alpha + 2.535\alpha = 2.535$ $3.535\alpha = 2.535$ $\alpha = \frac{2.535}{3.535} \approx 0.717$

Thus, the degree of dissociation ($\alpha$) is $0.717$.