Consider the following reaction in a sealed vessel at equilibrium with concentrations \mathrm{N2} = 3.0 imes 10^{-3} M, \mathrm{O2} = 4.2 imes 10^{-3} M and \mathrm{NO} = 2.8 imes 10^{-3} M:

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Question

Consider the following reaction in a sealed vessel at equilibrium with concentrations \mathrm{N2} = 3.0 	imes 10^{-3} M, \mathrm{O2} = 4.2 	imes 10^{-3} M and \mathrm{NO} = 2.8 	imes 10^{-3} M:

\mathrm{2NO(g) ightleftharpoons N2(g) + O2(g)}

If 0.1\ \mathrm{mol\,L^{-1}} of NO(g) is taken in a closed vessel, what will be the degree of dissociation (\alpha) of NO(g) at equilibrium?

Vishal Gupta · Accepted Answer

C. Kc = \frac{[\mathrm{N2}][\mathrm{O2}]}{[\mathrm{NO}]^2} = \frac{3.0	imes10^{-3} 	imes 4.2	imes10^{-3}}{(2.8	imes10^{-3})^2} \approx 1.607. Starting from 0.1 M NO: Kc = \frac{(0.05\alpha)^2}{(0.1(1-\alpha))^2} = \frac{\alpha^2}{4(1-\alpha)^2}, so \frac{\alpha}{2(1-\alpha)} = 1.268 \Rightarrow \alpha \approx 0.717.

Equilibrium

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