Question: 89. 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 \mathrm{~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) 0.00889
(2) 0.0889
(3) 0.8889
(4) 0.717
Answer: Option (4)
Explanation:
For the equilibrium reaction
2\mathrm{NO} \rightleftharpoons \mathrm{N}_{2} + \mathrm{O}_{2}The equilibrium constant \mathrm{K}_{\mathrm{c}} is given by
\mathrm{K}_{\mathrm{c}}=\dfrac{[\mathrm{N}_{2}][\mathrm{O}_{2}]}{[\mathrm{NO}]^{2}}Substituting the given equilibrium concentrations,
\mathrm{K}_{\mathrm{c}}=\dfrac{(3.0 \times 10^{-3})(4.2 \times 10^{-3})}{(2.8 \times 10^{-3})^{2}} \mathrm{K}_{\mathrm{c}}=\dfrac{12.6 \times 10^{-6}}{7.84 \times 10^{-6}} \approx 1.61Let the initial concentration of \mathrm{NO} be 0.1 \ \mathrm{M}
and the degree of dissociation be \alpha.
At equilibrium, concentrations will be:
[\mathrm{NO}] = 0.1(1-\alpha) [\mathrm{N}_{2}] = \dfrac{0.1\alpha}{2} [\mathrm{O}_{2}] = \dfrac{0.1\alpha}{2}Substituting into the expression for \mathrm{K}_{\mathrm{c}},
\mathrm{K}_{\mathrm{c}}=\dfrac{\left(\dfrac{0.1\alpha}{2}\right)\left(\dfrac{0.1\alpha}{2}\right)}{[0.1(1-\alpha)]^{2}} \mathrm{K}_{\mathrm{c}}=\dfrac{0.0025\alpha^{2}}{(1-\alpha)^{2}}Substituting \mathrm{K}_{\mathrm{c}} = 1.61,
1.61=\dfrac{0.0025\alpha^{2}}{(1-\alpha)^{2}}Solving this equation gives
\alpha \approx 0.717Therefore, the degree of dissociation of \mathrm{NO}
at equilibrium is 0.717.