Which of the following relations is incorrect | vedclass

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(NONE) The relationship between standard Gibbs free energy change $(\Delta G^o)$ and the equilibrium constant $(Kp)$ is given by the equation: $\Delta

Vedclass · Accepted Answer

$Kp = (e)^{-\frac{\Delta G^o}{RT}}$

Vedclass · Answer

(NONE) The relationship between standard Gibbs free energy change $(\Delta G^o)$ and the equilibrium constant $(Kp)$ is given by the equation: $\Delta G^o = -RT \ln Kp$. Converting the natural logarithm $(\ln)$ to base $10$ logarithm $(\log)$,we get: $\Delta G^o = -2.303 \, RT \log Kp$. Rearranging this for $\log Kp$,we get: $\log Kp = \frac{-\Delta G^o}{2.303 \, RT}$. From $\Delta G^o = -RT \ln Kp$,we can write $\ln Kp = -\frac{\Delta G^o}{RT}$,which implies $Kp = e^{-\frac{\Delta G^o}{RT}}$. Both options $A$ and $B$ represent the same correct mathematical expression $Kp = e^{-\frac{\Delta G^o}{RT}}$. Since the question asks for the incorrect relation and all provided options $A, B, C,$ and $D$ are mathematically equivalent and correct representations of the thermodynamic relationship,there is no incorrect option provided.

Which of the following relations is incorrect?

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