Find the value of the equilibrium constant $(K)$ of a reaction at $300 \ K$, when standard Gibbs free energy change is $-25 \ kJ \ mol^{-1}$? (Consider $R = 8.33 \ J \ mol^{-1} \ K^{-1}$)
- A$e^{8}$
- B$e^{9}$
- C$e^{10}$
- D$e^{11}$
Explore More
Similar Questions
The equilibrium constant of a reaction is $0.008$ at $298 \ K$. The standard free energy change of the reaction at the same temperature is
MediumKCET 2012
View SolutionWhich of the following expressions is correct?
In an equilibrium reaction for which $\Delta G^o = 0$,the equilibrium constant $K = $
Easy
View SolutionAt $300 \ K$,$\Delta_{r} G^{\circ}$ for the reaction $A_{2(g)} \rightleftharpoons B_{2(g)}$ is $-11.5 \ kJ \ mol^{-1}$. The equilibrium constant at $300 \ K$ is approximately $(R=8.314 \ J \ mol^{-1} \ K^{-1})$.
For a reaction at $298 \ K$,the equilibrium constant is ${K_p} = 0.17 \times {10^{12}}$. Find the standard Gibbs free energy change $\Delta {G^\Theta }$. (Given: $R = 8.314 \ J \ mol^{-1} \ K^{-1}$)
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