(D) The standard Gibbs energy change of a reaction is given by $\Delta G^\circ = -RT \ln K$. At equilibrium,the reaction quotient $Q = K$,but the stan

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$(A)$ is not correct but $(R)$ is correct

(D) The standard Gibbs energy change of a reaction is given by $\Delta G^\circ = -RT \ln K$. At equilibrium,the reaction quotient $Q = K$,but the standard Gibbs energy change $\Delta G^\circ$ is not necessarily zero unless the equilibrium constant $K = 1$. Therefore,Assertion $(A)$ is incorrect. The reason statement is correct because for a spontaneous process at constant temperature and pressure,the Gibbs energy of the system must decrease $(\Delta G < 0)$. Thus,$(A)$ is incorrect but $(R)$ is correct.

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

Easy

Assertion $(A)$: For every chemical reaction at equilibrium,standard Gibbs energy change of the reaction is zero.
Reason $(R)$: At constant temperature and pressure,chemical reactions are spontaneous in the direction of decreasing Gibbs energy.

AP EAMCET 2017 All AP EAMCET

A
Both $(A)$ and $(R)$ are correct and $(R)$ is the correct explanation of $(A)$
B
Both $(A)$ and $(R)$ are correct but $(R)$ is not the correct explanation of $(A)$
C
$(A)$ is correct and $(R)$ is not correct
D
$(A)$ is not correct but $(R)$ is correct

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6-1.Equilibrium (Chemical Equilibrium)

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Standard free energy

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The equilibrium constant of the following given reaction is $K_p = 6.022 \times 10^{-5}$ at $298 \ K$ temperature. $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$. Calculate the value of $\Delta_r G^o$.

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For the reaction,$A_{(g)} + B_{(g)} \to C_{(g)} + D_{(g)}$,$\Delta H^o$ and $\Delta S^o$ are,respectively,$-29.8 \, kJ \, mol^{-1}$ and $-0.100 \, kJ \, K^{-1} \, mol^{-1}$ at $298 \, K$. The equilibrium constant for the reaction at $298 \, K$ is

DifficultJEE MAIN 2016

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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.

Medium

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The correct relationship between the standard Gibbs free energy change $(\Delta G^o)$ and the equilibrium constant $(K_c)$ for a reaction is .......

Easy

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The value of $\log _{10} K$ for a reaction $A \rightleftharpoons B$ is
(Given : $\Delta _{r} H_{298 K}^{\circ} = -54.07 \ kJ \ mol^{-1}$,$\Delta _{r} S_{298 K}^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$; $2.303 \times 8.314 \times 298 = 5705$)

DifficultIIT 2007

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Assertion $(A)$: For every chemical reaction at equilibrium,standard Gibbs energy change of the reaction is zero. Reason $(R)$: At constant temperature and pressure,chemical reactions are spontaneous in the direction of decreasing Gibbs energy.

Similar Questions

The equilibrium constant of the following given reaction is $K_p = 6.022 \times 10^{-5}$ at $298 \ K$ temperature. $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$. Calculate the value of $\Delta_r G^o$.

For the reaction,$A_{(g)} + B_{(g)} \to C_{(g)} + D_{(g)}$,$\Delta H^o$ and $\Delta S^o$ are,respectively,$-29.8 \, kJ \, mol^{-1}$ and $-0.100 \, kJ \, K^{-1} \, mol^{-1}$ at $298 \, K$. The equilibrium constant for the reaction at $298 \, K$ is

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.

The correct relationship between the standard Gibbs free energy change $(\Delta G^o)$ and the equilibrium constant $(K_c)$ for a reaction is .......

The value of $\log _{10} K$ for a reaction $A \rightleftharpoons B$ is(Given : $\Delta _{r} H_{298 K}^{\circ} = -54.07 \ kJ \ mol^{-1}$,$\Delta _{r} S_{298 K}^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$; $2.303 \times 8.314 \times 298 = 5705$)

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Assertion $(A)$: For every chemical reaction at equilibrium,standard Gibbs energy change of the reaction is zero.
Reason $(R)$: At constant temperature and pressure,chemical reactions are spontaneous in the direction of decreasing Gibbs energy.

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.

The value of $\log _{10} K$ for a reaction $A \rightleftharpoons B$ is
(Given : $\Delta _{r} H_{298 K}^{\circ} = -54.07 \ kJ \ mol^{-1}$,$\Delta _{r} S_{298 K}^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$; $2.303 \times 8.314 \times 298 = 5705$)