At a certain temperature, the equilibrium constant of the reaction: (g) + (g) ⇌ 2(g) is 0.0004. Assuming air to be a mixture of four volumes of nitrogen with one volume of oxygen, the percentage of ni

Question

At a certain temperature, the equilibrium constant of the reaction: $N_2$(g) + $O_2$(g) ⇌ 2$NO$(g) is 0.0004. Assuming air to be a mixture of four volumes of nitrogen with one volume of oxygen, the percentage of nitric oxide, by volume, in the gas produced by allowing air to reach equilibrium at this temperature, is:

Prashant Kotian · Accepted Answer

Correct answer: B. Reaction: ; \quad Air composition: 4 volumes N : 1 volume O → basis 4 mol N, 1 mol O. \textbf{ICE table} (let mol of N and mol of O react): \begin{center} \begin{tabular}{lccc} & N & O & NO \ Initial & 4 & 1 & 0 \ Change & & & \ Equil. & & & \ \end{tabular} \end{center} Total moles unchanged (since ): . $$ K_c = \frac{(2x)^2}{(4-x)(1-x)} \approx \frac{4x^2}{4 \times 1} = x^2 = 0.0004 \quad (\text{since } x \ll 1) $$ $$ x = \sqrt{0.0004} = 0.02\ \text{mol} $$ Moles of NO at equilibrium: \textbf{Volume (mole) percent of NO:} $$ \%\text{NO} = \frac{0.04}{5} \times 100 = 0.8\% $$ \textbf{Answer: B} — 0.8\%

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