For the reaction at 298K: `A_((g)) +B_((g)) hArr C_((g)) + D_((g)) Delta H^(@) + `29.8kcal and `Delta S^(@) = 100cal K^(-1)`. Find the value of equilibrium constant.

Answer

To find the equilibrium constant \( K \) for the reaction at 298 K, we can use the Gibbs free energy change (\( \Delta G^\circ \)) and the relationship between \( \Delta G^\circ \) and the equilibrium constant \( K \).

### Step-by-Step Solution:

1. **Identify the given data**:
- \( \Delta H^\circ = 29.8 \, \text{kcal} \)
- \( \Delta S^\circ = 100 \, \text{cal K}^{-1} \)
- Temperature \( T = 298 \, \text{K} \)

2. **Convert units**:
- Convert \( \Delta H^\circ \) from kcal to cal:
\[ \Delta H^\circ = 29.8 \, \text{kcal} \times 1000 \, \text{cal/kcal} = 29800 \, \text{cal} \]

3. **Calculate \( \Delta G^\circ \)** using the formula:
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]
Substitute the values:
\[ \Delta G^\circ = 29800 \, \text{cal} - 298 \, \text{K} \times 100 \, \text{cal K}^{-1} \]
\[ \Delta G^\circ = 29800 \, \text{cal} - 29800 \, \text{cal} = 0 \, \text{cal} \]

4. **Relate \( \Delta G^\circ \) to the equilibrium constant \( K \)**:
The relationship is given by:
\[ \Delta G^\circ = -RT \ln K \]
Where \( R \) (the gas constant) is \( 1.987 \, \text{cal K}^{-1} \text{mol}^{-1} \).

5. **Substitute the values**:
Since \( \Delta G^\circ = 0 \):
\[ 0 = -RT \ln K \]
This implies:
\[ RT \ln K = 0 \]

6. **Solve for \( K \)**:
Since \( RT \) is not zero, we can conclude:
\[ \ln K = 0 \]
Therefore:
\[ K = e^0 = 1 \]

### Final Answer:
The equilibrium constant \( K \) is \( 1 \).