Find the incorrect equation among the following.

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$\frac{Δ G_{system}}{Δ S_{total}} = - T$

$In isothermal process, W_{reversible} = - n R T \ln \frac{V_{f}}{V_{i}}$

$\ln K = \frac{Δ H^{\circ} - T Δ S^{\circ}}{R T}$

$K = e^{- Δ G^{o} / R T}$

answer is C.

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Detailed Solution

Applying gibbs energy,
$Δ G = Δ H - T Δ S$
Where, $Δ G$ is the change in gibbs energy
$Δ H$ is the change in enthalpy
$Δ S$ is the change in entropy
T is temperature
a) For a system, total entropy change $= Δ S_{total}$
$\begin{array}{l} Δ H_{total} = 0 \\ ∴ Δ G_{system} = - T Δ S_{total} \\ ∴ \frac{Δ G_{system}}{Δ S_{total}} = - T \end{array}$

Thus (a) is correct
b) For isothermal reversible process, $Δ E = 0$
By first law of thermodynamics, $Δ E = q + W$
$\begin{array}{l} ∴ W_{reversible} = - q = - \int_{V_{i}}^{V_{f}} p d V \\ \Rightarrow W_{reversible} = - n R T \ln \frac{V_{f}}{V_{i}} \end{array}$
Thus, (b) is correct.
c) $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$
Also, $Δ G^{\circ} = - R T \ln K$
$\begin{array}{l} \Rightarrow \ln K = \frac{- Δ G^{\circ}}{R T} \\ \Rightarrow \ln K = \frac{(Δ H^{\circ} - T Δ S^{\circ})}{R T} [from Equation (1)] \end{array}$
Thus (c) is incorrect.
d) The standard free energy $(Δ G^{\circ})$ is released to equilibrium constant K as
$\begin{array}{l} Δ G^{\circ} = - R T \ln K \\ ∴ \ln K = \frac{Δ G^{\circ}}{R T} \Rightarrow K = - e^{\frac{- Δ G^{\circ}}{R T}} \end{array}$
Thus, (d) is also correct.