Consider the following reaction in a sealed vessel at equilibrium with concentrations of and .
If of is taken in a closed vessel, what will be degree of dissociation () of at equilibrium?
- A0.00889
- B0.0889
- C0.8889
- D0.717
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Correct answer: D
$$ \begin{aligned} \& 2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}\\ \& \begin{aligned} \mathrm{K}_c \& =\frac{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]}{[\mathrm{NO}]^2} \\ \& =\frac{3 \times 10^{-3} \times 4.2 \times 10^{-3}}{2.8 \times 10^{-3} \times 2.8 \times 10^{-3}} \\ \& =1.607 \end{aligned} \end{aligned} $$

$$ \mathrm{K}_{\mathrm{c}}=\frac{0.05 \alpha \times 0.05 \alpha}{(0.1-0.1 \alpha)^2} $$
$$ \begin{gathered} \mathrm{K}_{\mathrm{c}}=\frac{0.05 \alpha \times 0.05 \alpha}{0.01(1-\alpha)^2} \\ 1.607=\frac{(0.05)^2 \alpha^2}{0.01(1-\alpha)^2} \\ \frac{\alpha^2}{(1-\alpha)^2}=\frac{1.607 \times(0.1)^2}{(0.05)^2} \\ \frac{\alpha}{1-\alpha}=\frac{1.27 \times 0.1}{0.05} \\ \frac{\alpha}{1-\alpha}=2.54 \\ \alpha=2.54-2.54 \alpha \\ 3.54 \alpha=2.54 \\ \alpha=\frac{2.54}{3.54}=0.717 \end{gathered} $$
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