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 find the value of ΔS° at 300 K for the isomerization reaction \( A \rightleftharpoons B \) given the temperature dependence of the equilibrium constant, we can follow these steps:

### Step 1: Understand the relationship between K and ΔS°
We start with the relationship between the equilibrium constant \( K \) and the standard entropy change \( ΔS° \):
\[ \ln K = \frac{ΔS°}{R} - \frac{ΔH°}{RT} \]

### Step 2: Convert the equation to logarithmic form
Given the problem states:
\[ \log K = 4 - \frac{2000}{T} \]
We can convert this logarithmic equation to natural logarithm (ln) form:
\[ \ln K = 2.303 \cdot \log K \]
Thus,
\[ \ln K = 2.303 \left( 4 - \frac{2000}{T} \right) \]

### Step 3: Expand the equation
Expanding the equation gives:
\[ \ln K = 2.303 \cdot 4 - \frac{2.303 \cdot 2000}{T} \]
\[ \ln K = 9.212 - \frac{4606}{T} \]

### Step 4: Compare coefficients
Now, we can compare this with the general equation:
\[ \ln K = \frac{ΔS°}{R} - \frac{ΔH°}{RT} \]
From this, we can identify:
\[ \frac{ΔS°}{R} = 9.212 \quad \text{and} \quad -\frac{ΔH°}{R} = -\frac{4606}{T} \]

### Step 5: Calculate ΔS°
To find \( ΔS° \), we need to use the value of \( R \). The value of \( R \) in calories per mole per Kelvin is approximately \( 2 \, \text{cal/mol·K} \).
Thus,
\[ ΔS° = 9.212 \cdot R = 9.212 \cdot 2 \, \text{cal/mol·K} = 18.424 \, \text{cal/mol} \]

### Final Answer
Therefore, the value of \( ΔS° \) at 300 K is:
\[ ΔS° = 18.424 \, \text{cal/mol} \]