The incorrect expression among the following is

Options

A. `(Delta G_("system"))/(Delta S_("total"))=-T`
B. In isothermal process `w_("reversible") = -"nRT In"(V_(f))/(V_(i))`
C. `InK=(Delta H^(0)-T Delta S^(0))/(RT)`
D. `K = e^(- Delta G^(0)//RT)`

Answer

To solve the problem of identifying the incorrect expression among the given options, we will analyze each expression step by step.

### Step 1: Analyze the First Expression
**Expression:** \(\frac{\Delta G_{system}}{\Delta S_{total}} = -T\)

- We know from thermodynamics that for a system, the relationship between Gibbs free energy change (\(\Delta G\)), enthalpy change (\(\Delta H\)), and entropy change (\(\Delta S\)) is given by:
\[ \Delta G = \Delta H - T \Delta S \]
- If we assume \(\Delta H = 0\) for a specific case, then:
\[ \Delta G = -T \Delta S \]
- Dividing both sides by \(\Delta S_{total}\) gives:
\[ \frac{\Delta G_{system}}{\Delta S_{total}} = -T \]
- **Conclusion:** This expression is **correct**.

### Step 2: Analyze the Second Expression
**Expression:** \(W_{reversible} = -nRT \ln\left(\frac{V_f}{V_i}\right)\)

- For an isothermal reversible process, the work done is given by:
\[ W = -\int P \, dV \]
- For an ideal gas, this can be expressed as:
\[ W = -nRT \ln\left(\frac{V_f}{V_i}\right) \]
- **Conclusion:** This expression is **correct**.

### Step 3: Analyze the Third Expression
**Expression:** \(\ln K = \frac{\Delta H^\circ - T \Delta S^\circ}{RT}\)

- We know that:
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]
- Also, from thermodynamics:
\[ \Delta G^\circ = -RT \ln K \]
- Therefore, we can equate:
\[ -RT \ln K = \Delta H^\circ - T \Delta S^\circ \]
- Rearranging gives:
\[ \ln K = -\frac{\Delta H^\circ - T \Delta S^\circ}{RT} \]
- Thus, the expression should actually be:
\[ \ln K = \frac{-\Delta H^\circ + T \Delta S^\circ}{RT} \]
- **Conclusion:** This expression is **incorrect**.

### Step 4: Analyze the Fourth Expression
**Expression:** \(K = e^{-\frac{\Delta G^\circ}{RT}}\)

- From the relation:
\[ \Delta G^\circ = -RT \ln K \]
- We can express \(K\) as:
\[ K = e^{-\frac{\Delta G^\circ}{RT}} \]
- **Conclusion:** This expression is **correct**.

### Final Conclusion
The incorrect expression among the options is:
\[ \ln K = \frac{\Delta H^\circ - T \Delta S^\circ}{RT} \]