The incorrect expression among the following is

Options

A. `(DeltaG_("system"))/(DeltaS_("total"))=-T`
B. In isothermal process, `w_("reversible")=-nRT "In" (V_(f))/(V_(i))`
C. Ink `= (DeltaH^(@)-TDeltaS^(@))/(RT)`
D. `K=e^(-DeltaG^(@)//RT)`

Explanation

To determine the incorrect expression among the given options, we will analyze each option step by step.

### Step 1: Analyze the first expression
**Expression:** \(\Delta G_{\text{system}} / \Delta S_{\text{total}} = -T\)

**Analysis:**
This expression is derived from the relationship between Gibbs free energy change and entropy change. The correct relationship is:
\[ \Delta G = -T \Delta S_{\text{total}} \]
Thus, rearranging gives:
\[ \frac{\Delta G}{\Delta S_{\text{total}}} = -T \]
This expression is **correct**.

### Step 2: Analyze the second expression
**Expression:** In an isothermal process, the reversible work done is given by \(-nRT \ln \frac{V_{\text{final}}}{V_{\text{initial}}}\)

**Analysis:**
In an isothermal process, the work done can be expressed as:
\[ W = -\int P \, dV \]
Using the ideal gas law \(PV = nRT\), we can express pressure in terms of volume:
\[ P = \frac{nRT}{V} \]
Substituting this into the work equation gives:
\[ W = -\int_{V_{\text{initial}}}^{V_{\text{final}}} \frac{nRT}{V} \, dV = -nRT \ln \frac{V_{\text{final}}}{V_{\text{initial}}} \]
Thus, this expression is **correct**.

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

**Analysis:**
This expression is derived from the Gibbs free energy equation:
\[ \Delta G = \Delta H - T \Delta S \]
And the relationship:
\[ \Delta G = -RT \ln K \]
From these, we can equate:
\[ \Delta H - T \Delta S = -RT \ln K \]
Rearranging gives:
\[ RT \ln K = -(\Delta H - T \Delta S) \]
Thus:
\[ \ln K = -\frac{\Delta H - T \Delta S}{RT} \]
This means the expression should actually be:
\[ \ln K = \frac{-\Delta H + T \Delta S}{RT} \]
The provided expression is incorrect as it has a sign error. Therefore, this expression is **incorrect**.

### Step 4: Analyze the fourth expression
**Expression:** \(K = e^{-\Delta G / RT}\)

**Analysis:**
This expression is derived from the equation:
\[ \Delta G = -RT \ln K \]
Rearranging gives:
\[ \frac{\Delta G}{-RT} = \ln K \implies K = e^{-\Delta G / RT} \]
Thus, this expression is **correct**.

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