(D) The relationship between standard Gibbs energy change $\Delta G^{\circ}$ and equilibrium constant $K$ is given by $\Delta G^{\circ} = -RT \ln K$.A

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Assertion is incorrect but Reason is correct.

(D) The relationship between standard Gibbs energy change $\Delta G^{\circ}$ and equilibrium constant $K$ is given by $\Delta G^{\circ} = -RT \ln K$.At equilibrium,$\Delta G = 0$,but $\Delta G^{\circ}$ is not necessarily zero unless $K = 1$.Therefore,the Assertion is incorrect.However,a chemical reaction is spontaneous in the direction of decreasing Gibbs energy $(\Delta G < 0)$ at constant temperature and pressure,which makes the Reason correct.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Standard free energy

Medium

Assertion: For every chemical reaction at equilibrium,the standard Gibbs energy change is zero.
Reason: At constant temperature and pressure,a chemical reaction is spontaneous in the direction of decreasing Gibbs energy.

A
Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
B
Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
C
Assertion is correct but Reason is incorrect.
D
Assertion is incorrect but Reason is correct.

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Chemistry Questions

Chapter

6-1.Equilibrium (Chemical Equilibrium)

Subtopic

Standard free energy

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

Easy

View Solution

For the following reaction at $50^\circ$ $C$ and at $2 \text{ atm}$ pressure, $2N_2O_5(g) \rightleftharpoons 2N_2O_4(g) + O_2(g)$. $N_2O_5$ is $50\%$ dissociated. The magnitude of standard free energy change at this temperature is $x$. $x = . . . . . . \text{ J mol}^{-1}$.

DifficultJEE MAIN 2026

View Solution

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

Easy

View Solution

The equilibrium constant for a reaction is $20$. What is the value of $\Delta G^{\circ}$ at $300 \ K$? (Given: $R = 8 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$,$\ln(20) \approx 2.996$)

EasyMHT CET 2020

View Solution

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 Solution

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

Assertion: For every chemical reaction at equilibrium,the standard Gibbs energy change is zero.Reason: At constant temperature and pressure,a chemical reaction is spontaneous in the direction of decreasing Gibbs energy.

Similar Questions

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

For the following reaction at $50^\circ$ $C$ and at $2 \text{ atm}$ pressure, $2N_2O_5(g) \rightleftharpoons 2N_2O_4(g) + O_2(g)$. $N_2O_5$ is $50\%$ dissociated. The magnitude of standard free energy change at this temperature is $x$. $x = . . . . . . \text{ J mol}^{-1}$.

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

The equilibrium constant for a reaction is $20$. What is the value of $\Delta G^{\circ}$ at $300 \ K$? (Given: $R = 8 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$,$\ln(20) \approx 2.996$)

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion: For every chemical reaction at equilibrium,the standard Gibbs energy change is zero.
Reason: At constant temperature and pressure,a chemical reaction is spontaneous in the direction of decreasing Gibbs energy.

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