Q. The relation between equilibrium constant $K_{p}$ and $K_{c}$ is
A) $K_{p} = K_{c}(RT)^{\mathrm{\Delta}n_{g}}$
B) $K_{c} = K_{p}(RT)^{\mathrm{\Delta}n_{g}}$
C) $K_{p} = \left( \frac{K_{c}}{RT} \right)^{\mathrm{\Delta}n_{g}}$
D) $K_{p} - K_{c} = (RT)^{\mathrm{\Delta}n_{g}}$
Explanation
The correct answer is Option A: $K_{p} = K_{c}(RT)^{\mathrm{\Delta}n_{g}}$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.