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Chemistry · EQUILIBRIUM
हिन्दीFor the formation of ammonia from its elements, \( \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) \), the equilibrium constant is \( K_c = 3.55 \times 10^2 \) at \( 500\text{ K} \). What is the equilibrium constant \( K_c^* \) at the same temperature for the reaction: \( \text{NH}_3(\text{g}) \rightleftharpoons \frac{1}{2}\text{N}_2(\text{g}) + \frac{3}{2}\text{H}_2(\text{g}) \)?
Correct answer · Option 4
\( 5.31 \times 10^{-2} \)
Explanation
Reversing and multiplying the stoichiometry by 1/2 gives \( K_c^* = \frac{1}{\sqrt{K_c}} = \frac{1}{\sqrt{355}} \approx \frac{1}{18.84} \approx 0.0531 = 5.31 \times 10^{-2} \).
