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(B) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^{\circ} = -RT \ln K$ o

Vedclass · Accepted Answer

$K > 1$

Vedclass · Answer

(B) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^{\circ} = -RT \ln K$ or $\Delta G^{\circ} = -2.303 RT \log K$. Given that $\Delta G^{\circ} = -7.64 	imes 10^4 \ J \ mol^{-1}$. Since $\Delta G^{\circ}$ is negative,the reaction is spontaneous in the forward direction. For a spontaneous reaction,the equilibrium constant $K$ must be greater than $1$. Mathematically,$\log K = -\frac{\Delta G^{\circ}}{2.303 RT} = -\frac{-7.64 	imes 10^4}{2.303 	imes 8.314 	imes 298} > 0$. Since $\log K > 0$,it implies $K > 10^0$,so $K > 1$. Thus,option $(B)$ is correct.

For a chemical reaction,the standard Gibbs energy change,$\Delta G^{\circ}$ is $-7.64 \times 10^4 \ J \ mol^{-1}$. What is the value of equilibrium constant $(K)$?

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