$STATEMENT-1$: 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.
- 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
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Similar Questions
The correct relationship between standard free energy change $(\Delta G^o)$ and equilibrium constant $(K)$ is:
Medium
View SolutionOf the following reactions:
$(i) \, A \rightleftharpoons B, \Delta G^{\circ} = 250 \, kJ \, mol^{-1}$
$(ii) \, D \rightleftharpoons E, \Delta G^{\circ} = -100 \, kJ \, mol^{-1}$
$(iii) \, F \rightleftharpoons G, \Delta G^{\circ} = -150 \, kJ \, mol^{-1}$
$(iv) \, M \rightleftharpoons N, \Delta G^{\circ} = 150 \, kJ \, mol^{-1}$
The reaction with the largest equilibrium constant is:
$(i) \, A \rightleftharpoons B, \Delta G^{\circ} = 250 \, kJ \, mol^{-1}$
$(ii) \, D \rightleftharpoons E, \Delta G^{\circ} = -100 \, kJ \, mol^{-1}$
$(iii) \, F \rightleftharpoons G, \Delta G^{\circ} = -150 \, kJ \, mol^{-1}$
$(iv) \, M \rightleftharpoons N, \Delta G^{\circ} = 150 \, kJ \, mol^{-1}$
The reaction with the largest equilibrium constant is:
DifficultKVPY 2014
View SolutionWhich of the following equations relates the temperature of a reaction with $\Delta H^{\circ}$ and $\Delta S^{\circ}$ at equilibrium?
For the reversible reaction,$A_{(s)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$,$\Delta G^{\circ} = -350 \ kJ$,which one of the following statements is true?
MediumKCET 2011
View SolutionAssuming ideal behaviour,the magnitude of $\log \, K$ for the following reaction at $25^{\circ} C$ is $x \times 10^{-1}$. The value of $x$ is $......$. (Integer answer)
$3 HC \equiv CH_{(g)} \rightleftharpoons C_6H_{6(\ell)}$
[Given: $\Delta_f G^{\circ}(HC \equiv CH) = -2.04 \times 10^5 \, J \, mol^{-1}$
$\Delta_f G^{\circ}(C_6H_6) = -1.24 \times 10^5 \, J \, mol^{-1}; R = 8.314 \, J \, K^{-1} \, mol^{-1}$]
$3 HC \equiv CH_{(g)} \rightleftharpoons C_6H_{6(\ell)}$
[Given: $\Delta_f G^{\circ}(HC \equiv CH) = -2.04 \times 10^5 \, J \, mol^{-1}$
$\Delta_f G^{\circ}(C_6H_6) = -1.24 \times 10^5 \, J \, mol^{-1}; R = 8.314 \, J \, K^{-1} \, mol^{-1}$]
DifficultJEE MAIN 2021
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