OnlineCBT
Chemistry · EQUILIBRIUM
हिन्दीFor the gaseous reaction \( \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) \), the equilibrium constant is \( K_c \). If the reaction is written as \( \text{NH}_3(\text{g}) \rightleftharpoons \frac{1}{2}\text{N}_2(\text{g}) + \frac{3}{2}\text{H}_2(\text{g}) \), what is the new equilibrium constant \( K'_c \) in terms of \( K_c \)?
Correct answer · Option 3
\( \frac{1}{\sqrt{K_c}} \)
Explanation
Reversing the reaction inverts \( K_c \) to \( 1/K_c \), and multiplying coefficients by \( 1/2 \) raises the constant to power \( 1/2 \), giving \( K'_c = (1/K_c)^{1/2} = 1/\sqrt{K_c} \).
