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Chemistry · EQUILIBRIUM
हिन्दीIn a \( 2.0\text{ L} \) closed vessel at \( 500\text{ K} \), the equilibrium mixture contains \( 0.030\text{ mol} \) of \( \text{N}_2 \) and \( 0.024\text{ mol} \) of \( \text{NH}_3 \). If \( K_c \) for the reaction \( \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) \) is \( 3.55 \times 10^2\text{ L}^2\text{ mol}^{-2} \), what is the equilibrium concentration of \( \text{H}_2 \) (in \( \text{mol L}^{-1} \))?
Correct answer · Option 1
\( 3.0 \times 10^{-2}\text{ M} \)
Explanation
\( [\text{N}_2] = \frac{0.030}{2.0} = 1.5 \times 10^{-2}\text{ M} \), \( [\text{NH}_3] = \frac{0.024}{2.0} = 1.2 \times 10^{-2}\text{ M} \). \( K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \implies [\text{H}_2]^3 = \frac{(1.2 \times 10^{-2})^2}{(1.5 \times 10^{-2})(3.55 \times 10^2)} = \frac{1.44 \times 10^{-4}}{5.325} \approx 2.704 \times 10^{-5} = 27.04 \times 10^{-6} \). Thus, \( [\text{H}_2] = (27.04 \times 10^{-6})^{1/3} \approx 3.0 \times 10^{-2}\text{ M} \).
