Chemistry · EQUILIBRIUM
हिन्दीFor the reaction \( \text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI}(\text{g}) \), if the equilibrium constant is \( K_c \), then the equilibrium constant \( K_c' \) for the reaction \( \frac{1}{2}\text{H}_2(\text{g}) + \frac{1}{2}\text{I}_2(\text{g}) \rightleftharpoons \text{HI}(\text{g}) \) is:
Correct answer · Option 2
\( \sqrt{K_c} \)
Explanation
When a chemical equation is multiplied by a factor \( n \), the new equilibrium constant becomes \( K' = K^n \). Here \( n = 1/2 \), so \( K_c' = K_c^{1/2} = \sqrt{K_c} \).