For the reaction at 298 K
`A(g) + B(g) iff C(g)+ D(g)`
`DeltaH^(@) = -29.8 Kcal, DeltaS^(@) = - 0.1 Kcal //K`
Calculate `DeltaG^(@)` and K.

Explanation

To solve the problem, we need to calculate the standard Gibbs free energy change (ΔG°) and the equilibrium constant (K) for the given reaction at 298 K.

### Step-by-Step Solution:

1. **Identify the given values**:
- ΔH° = -29.8 Kcal
- ΔS° = -0.1 Kcal/K
- Temperature (T) = 298 K

2. **Convert ΔH° and ΔS° to consistent units**:
Since ΔH° and ΔS° are given in Kcal, we can use them directly. However, it is often useful to convert them to Joules for standard calculations:
- ΔH° = -29.8 Kcal × 4184 J/Kcal = -124,000 J
- ΔS° = -0.1 Kcal/K × 4184 J/Kcal = -418.4 J/K

3. **Use the Gibbs free energy equation**:
The relationship between ΔG°, ΔH°, and ΔS° is given by the equation:
\[ ΔG° = ΔH° - TΔS° \]

4. **Substitute the values into the equation**:
\[ ΔG° = -124,000 J - (298 K)(-418.4 J/K) \]

5. **Calculate TΔS°**:
\[ TΔS° = 298 K × -418.4 J/K = -124,000 J \]

6. **Calculate ΔG°**:
\[ ΔG° = -124,000 J + 124,000 J = 0 J \]

7. **Calculate the equilibrium constant (K)**:
The relationship between ΔG° and K is given by:
\[ ΔG° = -RT \ln K \]
Where R = 8.314 J/(mol·K).

Since ΔG° = 0, we can rearrange the equation:
\[ 0 = -RT \ln K \]
This implies:
\[ \ln K = 0 \]

8. **Solve for K**:
Since \(\ln K = 0\), we can exponentiate both sides:
\[ K = e^0 = 1 \]

### Final Results:
- ΔG° = 0 J
- K = 1