The incorrect expression among the following is :

The question asks which one of the four given thermodynamic formulae is wrong. We check each statement one by one.

Statement A
For every process at constant temperature and pressure, the relation between the system’s Gibbs free-energy change and the total entropy change is $$\Delta G_{\text{system}} = -T\,\Delta S_{\text{total}}$$ Dividing both sides by $$\Delta S_{\text{total}}$$ gives $$\frac{\Delta G_{\text{system}}}{\Delta S_{\text{total}}} = -T$$ Hence Statement A is correct.

Statement B
For an ideal gas undergoing a reversible isothermal change, work is obtained from $$w_{\text{reversible}} = -\int_{V_i}^{V_f} P\,dV$$ With $$P = \dfrac{nRT}{V}$$ (ideal-gas equation at constant $$T$$), this becomes $$w_{\text{reversible}} = -nRT \int_{V_i}^{V_f} \frac{dV}{V} = -nRT \ln\!\left(\frac{V_f}{V_i}\right)$$ Thus Statement B is correct.

Statement D
The fundamental relation between standard Gibbs free energy and the equilibrium constant is $$\Delta G^{0} = -RT \ln K \; \Longrightarrow \; K = e^{-\Delta G^{0}/RT}$$ Hence Statement D is correct.

Statement C
Start with $$\Delta G^{0} = \Delta H^{0} - T\Delta S^{0}$$. Substituting in $$\Delta G^{0} = -RT \ln K$$ gives $$-RT \ln K = \Delta H^{0} - T\Delta S^{0}$$ Rearranging, $$\ln K = -\,\frac{\Delta H^{0} - T\Delta S^{0}}{RT} = -\frac{\Delta H^{0}}{RT} + \frac{\Delta S^{0}}{R}$$ The expression quoted in option C, $$\ln K = \dfrac{\Delta H^{0} - T\Delta S^{0}}{RT}$$, is missing the overall negative sign, so it is incorrect.

Therefore the incorrect expression is: Option C which is: $$\ln K = \dfrac{\Delta H^{0} - T\Delta S^{0}}{RT}$$