The correct relationship between the standard free energy change $(\Delta G^o)$ and the equilibrium constant $(K_p)$ is ......
- A$K_p = -RT \ln \Delta G^o$
- B$K_p = (\frac{e}{RT})^{\Delta G^o}$
- C$K_p = \frac{-\Delta G^o}{RT}$
- D$K_p = e^{\left( \frac{-\Delta G^o}{RT} \right)}$
Explore More
Similar Questions
For the reaction $CH_{4(g)} + H_{2(g)} \longrightarrow C_2H_{6(g)}$,$K_p = 3.356 \times 10^{17}$,calculate $\Delta G^{\circ}$ for the reaction at $298 \ K$.
When a reaction is carried out at standard states,then at equilibrium:
Write the formula relating the equilibrium constant $K$ and the standard Gibbs free energy change $\Delta G^{\circ}$.
Easy
View SolutionThe free energy change for a reversible reaction at equilibrium is
Easy
View SolutionFind the equilibrium constant for the reaction below at $298 \ K$. $2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$ given that $\Delta H^o = 18.4512 \ kcal$ and $\Delta S^o = 29.16 \ cal/K$.
Easy
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