Question

Consider the following reaction in a sealed vessel at equilibrium with concentrations of \(N_2 = 3.0 \times 10^{-3} M,\) \(O_2 = 4.2 \times 10^{-3} M\) and \(NO = 2.8 \times 10^{-3} M.\)

\(2NO _{(g)} \rightleftharpoons N_{2(g)} + O _{2(g)}\)

If 0.1 mol \(L^{-1}\) of \(NO_{(g)}\) is taken in a closed vessel, what will be degree of dissociation \((\alpha)\) of \(NO_{(g)}\) at equilibrium? 

(1) 0.00889 

(2) 0.0889 

(3) 0.8889 

(4) 0.717

Answer 1

Correct option is (4) 0.717

\( 2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}\)

\(\mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]}{[\mathrm{NO}]^2} \)

\(=\frac{3 \times 10^{-3} \times 4.2 \times 10^{-3}}{2.8 \times 10^{-3} \times 2.8 \times 10^{-3}}\)

\(=1.607\)

\(\mathrm{K}_c=\frac{0.05 \alpha \times 0.05 \alpha}{(0.1-0.1 \alpha)^2}\)

\(\mathrm{~K}_c=\frac{0.05 \alpha \times 0.05 \alpha}{0.01(1-\alpha)^2} \)

\(1.607=\frac{(0.05)^2 \alpha^2}{0.01(1-\alpha)^2}\)

\(\frac{\alpha^2}{(1-\alpha)^2}=\frac{1.607 \times(0.1)^2}{(0.05)^2} \)

\(\frac{\alpha}{1-\alpha}=\frac{1.27 \times 0.1}{0.05}\)

\(\frac{\alpha}{1-\alpha}=2.54\)

\( \alpha=2.54-2.54 \alpha\)

\(3.54 \alpha=2.54\)

\(\alpha=\frac{2.54}{3.54}=0.717\)