(D) At equilibrium,the change in Gibbs energy $\Delta G = 0$,but the standard Gibbs energy change $\Delta G^{\circ}$ is related to the equilibrium con

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

$STATEMENT-1$ is False,$STATEMENT-2$ is True

(D) At equilibrium,the change in Gibbs energy $\Delta G = 0$,but the standard Gibbs energy change $\Delta G^{\circ}$ is related to the equilibrium constant $K$ by the equation $\Delta G^{\circ} = -RT \ln K$. Thus,$\Delta G^{\circ}$ is zero only if $K = 1$. Therefore,$STATEMENT-1$ is False.At constant temperature and pressure,a chemical reaction is spontaneous if the Gibbs energy of the system decreases,i.e.,$\Delta G < 0$. Therefore,$STATEMENT-2$ is True.

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

Easy

$STATEMENT-1$: For every chemical reaction at equilibrium,standard Gibbs energy of reaction is zero. $STATEMENT-2$: At constant temperature and pressure,chemical reactions are spontaneous in the direction of decreasing Gibbs energy.

IIT 2008 All IIT

A
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is correct explanation for $STATEMENT-1$
B
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is $NOT$ a correct explanation for $STATEMENT-1$
C
$STATEMENT-1$ is True,$STATEMENT-2$ is False
D
$STATEMENT-1$ is False,$STATEMENT-2$ is True

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

Previous Year

IIT 2008 Paper All IIT years →

If the equilibrium constant of a process is $3.8 \times 10^{-3}$ at $25^{\circ} C$, what is the standard free energy change of the process? $(R = 8.314 \ J \ mol^{-1} \ K^{-1}, \log 0.0038 = -2.42)$

EasyTS EAMCET 2018

View Solution

Calculate $\Delta G^{\circ} \ (kJ/mol)$ at $127 \ ^{\circ}C$ for a reaction with $K_{equilibrium} = 10^5$.

Medium

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

Hydrolysis of sucrose gives,
$Sucrose + H_{2}O \rightleftharpoons Glucose + Fructose$
Equilibrium constant $K_{c}$ for the reaction is $2 \times 10^{13}$ at $300 \ K$. Calculate $\Delta G^{\ominus}$ at $300 \ K$.

Easy

View Solution

For a certain reaction at $300 \ K$,$K=10$,then $\Delta G^{\circ}$ for the same reaction is . . . . . . $\times 10^{-1} \ kJ \ mol^{-1}$. (Given $R=8.314 \ J \ K^{-1} \ mol^{-1}$)

DifficultJEE MAIN 2024

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

$STATEMENT-1$: For every chemical reaction at equilibrium,standard Gibbs energy of reaction is zero. $STATEMENT-2$: At constant temperature and pressure,chemical reactions are spontaneous in the direction of decreasing Gibbs energy.

Similar Questions

If the equilibrium constant of a process is $3.8 \times 10^{-3}$ at $25^{\circ} C$, what is the standard free energy change of the process? $(R = 8.314 \ J \ mol^{-1} \ K^{-1}, \log 0.0038 = -2.42)$

Calculate $\Delta G^{\circ} \ (kJ/mol)$ at $127 \ ^{\circ}C$ for a reaction with $K_{equilibrium} = 10^5$.

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

Hydrolysis of sucrose gives,$Sucrose + H_{2}O \rightleftharpoons Glucose + Fructose$Equilibrium constant $K_{c}$ for the reaction is $2 \times 10^{13}$ at $300 \ K$. Calculate $\Delta G^{\ominus}$ at $300 \ K$.

For a certain reaction at $300 \ K$,$K=10$,then $\Delta G^{\circ}$ for the same reaction is . . . . . . $\times 10^{-1} \ kJ \ mol^{-1}$. (Given $R=8.314 \ J \ K^{-1} \ mol^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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

Hydrolysis of sucrose gives,
$Sucrose + H_{2}O \rightleftharpoons Glucose + Fructose$
Equilibrium constant $K_{c}$ for the reaction is $2 \times 10^{13}$ at $300 \ K$. Calculate $\Delta G^{\ominus}$ at $300 \ K$.