(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

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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.

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6-1.Equilibrium (Chemical Equilibrium)

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Standard free energy

The equilibrium constant of a reaction is related to

EasyAIIMS 1991

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Calculate the standard Gibbs free energy change $\Delta G^o$ at $298 \ K$ for the conversion of oxygen to ozone,given by the reaction: $\frac{3}{2} O_{2(g)} \rightleftharpoons O_{3(g)}$. The equilibrium constant $K_p$ for this conversion is $3 \times 10^{-29}$.

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The free energy change for a reversible reaction at equilibrium is

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For the reaction $A \rightleftharpoons B$,find the value of $log_{10}K$. Given: $\Delta_rH^o_{298\,K} = -54.07\, kJ\, mol^{-1}$,$\Delta_rS^o_{298\,K} = 10\, J\, K^{-1}\, mol^{-1}$,$R = 8.314\, J\, K^{-1}\, mol^{-1}$,$2.303 \times 8.314 \times 298 = 5705$.

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For the chemical reaction $X \rightleftharpoons Y$,the standard reaction Gibbs energy depends on temperature $T$ (in $K$) as
${\Delta_r}{G^o}$ (in $kJ \ mol^{-1}$) $= 120 - \frac{3}{8} \ T$
The major component of the reaction mixture at $T$ is

MediumJEE MAIN 2019

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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.

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The equilibrium constant of a reaction is related to

Calculate the standard Gibbs free energy change $\Delta G^o$ at $298 \ K$ for the conversion of oxygen to ozone,given by the reaction: $\frac{3}{2} O_{2(g)} \rightleftharpoons O_{3(g)}$. The equilibrium constant $K_p$ for this conversion is $3 \times 10^{-29}$.

The free energy change for a reversible reaction at equilibrium is

For the reaction $A \rightleftharpoons B$,find the value of $log_{10}K$. Given: $\Delta_rH^o_{298\,K} = -54.07\, kJ\, mol^{-1}$,$\Delta_rS^o_{298\,K} = 10\, J\, K^{-1}\, mol^{-1}$,$R = 8.314\, J\, K^{-1}\, mol^{-1}$,$2.303 \times 8.314 \times 298 = 5705$.

For the chemical reaction $X \rightleftharpoons Y$,the standard reaction Gibbs energy depends on temperature $T$ (in $K$) as${\Delta_r}{G^o}$ (in $kJ \ mol^{-1}$) $= 120 - \frac{3}{8} \ T$The major component of the reaction mixture at $T$ is

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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.

For the chemical reaction $X \rightleftharpoons Y$,the standard reaction Gibbs energy depends on temperature $T$ (in $K$) as
${\Delta_r}{G^o}$ (in $kJ \ mol^{-1}$) $= 120 - \frac{3}{8} \ T$
The major component of the reaction mixture at $T$ is