For an isomerisation reaction, `A hArr B`, the temperature dependence of equilibrium constant is given by
`Log_0K=4.0 -(2000)/T`
Find the value of `DeltaS^0` at 300 k in cal.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between Gibbs free energy and equilibrium constant
The Gibbs free energy change (ΔG°) for a reaction at equilibrium is related to the equilibrium constant (K) by the equation:
\[ \Delta G° = -RT \ln K \]
We can also express this in terms of logarithm base 10:
\[ \Delta G° = -2.303RT \log K \]

### Step 2: Use the given temperature dependence of the equilibrium constant
We are given:
\[ \log K = 4.0 - \frac{2000}{T} \]
This equation shows how the equilibrium constant changes with temperature.

### Step 3: Relate ΔG° to ΔH° and ΔS°
From thermodynamics, we know:
\[ \Delta G° = \Delta H° - T \Delta S° \]
At equilibrium, we can equate the two expressions for ΔG°:
\[ -2.303RT \log K = \Delta H° - T \Delta S° \]

### Step 4: Substitute the expression for log K
Substituting the expression for log K into the equation gives:
\[ -2.303RT \left(4.0 - \frac{2000}{T}\right) = \Delta H° - T \Delta S° \]

### Step 5: Simplify the equation
Distributing the left side:
\[ -2.303RT \cdot 4.0 + 2.303R \cdot 2000 = \Delta H° - T \Delta S° \]
This simplifies to:
\[ -2.303 \cdot 4.0RT + 2.303 \cdot 2000R = \Delta H° - T \Delta S° \]

### Step 6: Rearranging to find ΔS°
Now, we can rearrange the equation to isolate ΔS°:
\[ T \Delta S° = \Delta H° + 2.303 \cdot 4.0RT - 2.303 \cdot 2000R \]
\[ \Delta S° = \frac{\Delta H° + 2.303 \cdot 4.0RT - 2.303 \cdot 2000R}{T} \]

### Step 7: Find ΔH° using the given information
From the comparison of coefficients:
1. The constant term (4) corresponds to \(\frac{\Delta H°}{2.303R}\)
2. The coefficient of \(\frac{1}{T}\) corresponds to \(\frac{2000}{2.303R}\)

Thus, we can express ΔH°:
\[ \Delta H° = 4 \cdot 2.303R \]
Substituting R = 2 cal/mol·K:
\[ \Delta H° = 4 \cdot 2.303 \cdot 2 = 18.424 \text{ cal/mol} \]

### Step 8: Substitute ΔH° back to find ΔS°
Now substituting ΔH° back into the equation for ΔS°:
\[ \Delta S° = \frac{18.424 + 2.303 \cdot 4.0 \cdot 2 - 2.303 \cdot 2000}{300} \]
Calculating the terms:
1. \(2.303 \cdot 4.0 \cdot 2 = 18.424\)
2. \(2.303 \cdot 2000 = 4606\)

Thus:
\[ \Delta S° = \frac{18.424 + 18.424 - 4606}{300} \]
\[ \Delta S° = \frac{36.848 - 4606}{300} = \frac{-4569.152}{300} \approx -15.23 \text{ cal/mol·K} \]

### Final Answer
The value of ΔS° at 300 K is approximately -15.23 cal/mol·K.