The standard Gibbs free energy change $\Delta G^o$ is related to the equilibrium constant $K_p$ as:
- A$K_p = - RT \ln \Delta G^o$
- B$K_p = \left( \frac{e}{RT} \right)^{\Delta G^o}$
- C$K_p = - \frac{\Delta G^o}{RT}$
- D$K_p = e^{- \frac{\Delta G^o}{RT}}$
Explore More
Similar Questions
The free energy change for a reversible reaction at equilibrium is
Easy
View SolutionWhich of the following relations is incorrect?
Medium
View SolutionCalculate $\Delta_{r} G^{\ominus}$ for the conversion of oxygen to ozone,$\frac{3}{2} O_{2(g)} \rightarrow O_{3(g)}$ at $298 \, K$,if $K_{p}$ for this conversion is $2.47 \times 10^{-29}$.
Easy
View SolutionCalculate $\Delta G^\circ$ for the reaction, $CH_4(g) + H_2(g) \rightarrow C_2H_6(g)$ at $298 \text{ K}$, given $K_p = 2 \times 10^{17}$ and $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$.
DifficultMHT CET 2026
View SolutionFor a reaction,$\Delta G^{\circ} = -115 \, kJ$. What is the value of $\log \, K_p$ at $298 \, K$?
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo