Consider the following reaction in a sealed vessel at equilibrium with concentrations of N_2=3.0 × 10^-3 M, O_2=4.2 × 10^-3 M and NO=2.8 × 10^-3 M. 2…

$$\begin{aligned} & 2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \\ & \mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]}{[\mathrm{NO}]^2} \\ &=\frac{3 \times 10^{-3} \times 4.2 \times 10^{-3}}{2.8 \times 10^{-3} \times 2.8 \times 10^{-3}} \\ &=1.607 \\ & \mathrm{t}=0 \quad 2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \\ & 0.1-0.1 \alpha \quad 0.05 \alpha \quad 0.05 \alpha \\ & \mathrm{K}_{\mathrm{c}}=\frac{0.05 \alpha \times 0.05 \alpha}{(0.1-0.1 \alpha)^2} \\ & \mathrm{~K}_{\mathrm{c}}=\frac{0.05 \alpha \times 0.05 \alpha}{0.01(1-\alpha)^2} \\ & 1.607=\frac{(0.05)^2 \alpha^2}{0.01(1-\alpha)^2} \\ & \frac{\alpha^2}{(1-\alpha)^2}=\frac{1.607 \times(0.1)^2}{(0.05)^2}\end{aligned}$$​2NO(g)​⇌N2( g)​+O2( g)​Kc​=[NO]2[N2​][O2​]​=2.8×10−3×2.8×10−33×10−3×4.2×10−3​=1.607t=02NO(g)​⇌N2( g)​+O2( g)​0.1−0.1α0.05α0.05αKc​=(0.1−0.1α)20.05α×0.05α​ Kc​=0.01(1−α)20.05α×0.05α​1.607=0.01(1−α)2(0.05)2α2​(1−α)2α2​=(0.05)21.607×(0.1)2​​ $$\begin{gathered}\frac{\alpha}{1-\alpha}=\frac{1.27 \times 0.1}{0.05} \\ \frac{\alpha}{1-\alpha}=2.54 \\ \alpha=2.54-2.54 \alpha \\ 3.54 \alpha=2.54 \\ \alpha=\frac{2.54}{3.54}=0.717\end{gathered}$$1−αα​=0.051.27×0.1​1−αα​=2.54α=2.54−2.54α3.54α=2.54α=3.542.54​=0.717​