The incorrect expression among the following is:

We inspect each of the four given thermodynamic relations and compare them with the standard formulae that you would have learnt in class.

Option A states that for a reversible isothermal process of an ideal gas, the work done is

$$w_{\text{reversible}}=-nRT\ln\frac{V_f}{V_i}.$$

We recall the general expression for reversible work in an isothermal change of an ideal gas:

$$w_{\text{rev}}=\int_{V_i}^{V_f} -P\,dV=\int_{V_i}^{V_f} -\frac{nRT}{V}\,dV=-nRT\left[\ln V\right]_{V_i}^{V_f}=-nRT\ln\frac{V_f}{V_i}.$$

The given expression matches the derivation exactly, so Option A is correct.

Option B claims

$$\ln K=\frac{\Delta H^\circ-T\Delta S^\circ}{RT}.$$

To check the sign, we begin with two fundamental relationships:

1. The Gibbs-Helmholtz definition of standard free energy change: $$\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ.$$

2. The link between free energy and the equilibrium constant: $$\Delta G^\circ=-RT\ln K.$$

Now we substitute the first equation into the second:

$$-RT\ln K=\Delta H^\circ-T\Delta S^\circ.$$

Multiplying both sides by $$-1$$ gives

$$RT\ln K=-\Delta H^\circ+T\Delta S^\circ.$$

Dividing every term by $$RT$$ yields

$$\ln K=\frac{-\Delta H^\circ+T\Delta S^\circ}{RT}=-\frac{\Delta H^\circ-T\Delta S^\circ}{RT}.$$

Thus the correct formula contains an overall negative sign. The expression in Option B lacks this negative sign, so Option B is incorrect.

Option C gives the ratio

$$\frac{\Delta G_{\text{system}}}{\Delta S_{\text{total}}}=-T\quad(\text{at constant }P).$$

We know the general relation for any process at constant temperature and pressure:

$$\Delta G_{\text{system}}=-T\Delta S_{\text{universe}}=-T\Delta S_{\text{total}}.$$

Dividing both sides by $$\Delta S_{\text{total}}$$ immediately provides

$$\frac{\Delta G_{\text{system}}}{\Delta S_{\text{total}}}=-T,$$

exactly matching Option C. Hence Option C is correct.

Option D presents

$$K=e^{-\Delta G^\circ/RT}.$$

Starting once more from $$\Delta G^\circ=-RT\ln K,$$ we isolate $$K$$ by exponentiation:

$$\ln K=-\frac{\Delta G^\circ}{RT}\quad\Longrightarrow\quad K=e^{-\Delta G^\circ/RT}.$$

The option is therefore correct.

Out of the four expressions, only Option B contains an incorrect sign. All the others conform to the standard thermodynamic identities.

Hence, the correct answer is Option B.