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Chemistry · EQUILIBRIUM
हिन्दीAt \( 500\text{ K} \), the equilibrium concentrations for the reaction \( \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) \) are \( [\text{N}_2] = 1.5 \times 10^{-2}\text{ M} \), \( [\text{H}_2] = 3.0 \times 10^{-2}\text{ M} \), and \( [\text{NH}_3] = 1.2 \times 10^{-2}\text{ M} \). The equilibrium constant \( K_c \) is:
Correct answer · Option 3
\( 3.55 \times 10^2\text{ M}^{-2} \)
Explanation
\( K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(1.2 \times 10^{-2})^2}{(1.5 \times 10^{-2})(3.0 \times 10^{-2})^3} = \frac{1.44 \times 10^{-4}}{(1.5 \times 10^{-2})(2.7 \times 10^{-5})} = \frac{1.44 \times 10^{-4}}{4.05 \times 10^{-7}} \approx 3.55 \times 10^2 \).
