Consider the following reaction in a sealed vessel at equilibrium with concentrations of $\mathrm{N}_2=3.0…

Consider the following reaction in a sealed vessel at equilibrium with concentrations of $\mathrm{N}_2=3.0 \times 10^{-3} \mathrm{M}, \mathrm{O}_2=4.2 \times 10^{-3} \mathrm{M}$ and $\mathrm{NO}=2.8 \times 10^{-3} \mathrm{M}$. $2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$ If 0.1 mol $\mathrm{L}^{-1}$ of $\mathrm{NO}_{(\mathrm{g})}$ is taken in a closed vessel, what will be degree of dissociation ( $\alpha$ ) of $\mathrm{NO}_{(\mathrm{g})}$ at equilibrium?
  1. 0.0889
  2. 0.8889
  3. 0.717
  4. 0.00889

Solution

$\begin{aligned} & 2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \\ & \mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]}{[\mathrm{NO}]^2} \\ &=\frac{3 \times 10^{-3} \times 4.2 \times 10^{-3}}{2.8 \times 10^{-3} \times 2.8 \times 10^{-3}} \\ &=1.607 \\ & \mathrm{t}=0 \quad 2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \\ & 0.1-0.1 \alpha \quad 0.05 \alpha \quad 0.05 \alpha \\ & \mathrm{K}_{\mathrm{c}}=\frac{0.05 \alpha \times 0.05 \alpha}{(0.1-0.1 \alpha)^2} \\ & \mathrm{~K}_{\mathrm{c}}=\frac{0.05 \alpha \times 0.05 \alpha}{0.01(1-\alpha)^2} \\ & 1.607=\frac{(0.05)^2 \alpha^2}{0.01(1-\alpha)^2} \\ & \frac{\alpha^2}{(1-\alpha)^2}=\frac{1.607 \times(0.1)^2}{(0.05)^2}\end{aligned}$ $\begin{gathered}\frac{\alpha}{1-\alpha}=\frac{1.27 \times 0.1}{0.05} \\ \frac{\alpha}{1-\alpha}=2.54 \\ \alpha=2.54-2.54 \alpha \\ 3.54 \alpha=2.54 \\ \alpha=\frac{2.54}{3.54}=0.717\end{gathered}$

Asked in: NEET 2024

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