The equilibrium constant for the reaction: (g) + (g) ⇌ 2(g) is 0.09 at 3500 K. The fraction of equimolar mixture of and converted into is:

Question

The equilibrium constant for the reaction: $N_2$(g) + $O_2$(g) ⇌ 2$NO$(g) is 0.09 at 3500 K. The fraction of equimolar mixture of $N_2$ and $O_2$ converted into $NO$ is:

Prashant Kotian · Accepted Answer

Correct answer: A. Reaction: ; \quad at 3500 K Starting from equimolar mixture: 1 mol each of N and O. \textbf{ICE table} (let mol of each react): \begin{center} \begin{tabular}{lccc} & N & O & NO \ Initial & 1 & 1 & 0 \ Change & & & \ Equil. & & & \ \end{tabular} \end{center} $$ K_c = \frac{[NO]^2}{[N_2][O_2]} = \frac{(2x)^2}{(1-x)^2} = \left(\frac{2x}{1-x}\right)^2 = 0.09 $$ $$ \frac{2x}{1-x} = \sqrt{0.09} = 0.3 $$ $$ 2x = 0.3(1-x) = 0.3 - 0.3x \implies 2.3x = 0.3 \implies x = \frac{0.3}{2.3} \approx 0.1304 $$ \textbf{Fraction converted} \textbf{Answer: A} —

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