Which of the following expressions is correct?

  • A
    $\Delta G = -RT \ln K$
  • B
    $\Delta G = \frac{1}{RT^2 \ln K}$
  • C
    $\Delta G^0 = -RT \ln K$
  • D
    $\Delta G^0 = -\frac{1}{RT^2 \ln K}$

Explore More

Similar Questions

At $227^{\circ} C$,dinitrogen tetraoxide is $60 \%$ dissociated. What is the standard free energy change at this temperature and at $1 \ atm$ pressure?

Difficult
View Solution

Find the equilibrium constant for the reaction below at $25$ $^{\circ}C$:
$H_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons H_{2}O_{(g)}$ [ $\Delta_{f}G^{o} = -54.64 \ kcal$ ].

In the glycolysis process,during the phosphorylation of glucose,the equilibrium constant at $298 \ K$ is $3.6 \times 10^{-3}$. Find the value of $\Delta G^{\Theta}$. What does this indicate? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$

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 $R = 8.33 \ J \ mol^{-1} \ K^{-1}$)

$A$ reaction attains equilibrium when the free energy change accompanying it is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo