At equilibrium, Kp = 1 then the value of `DeltaG^(@)` will be equal to.....

Answer

To solve the problem, we need to determine the value of ΔG° (standard Gibbs free energy change) when Kp (equilibrium constant in terms of partial pressures) is equal to 1.

### Step-by-Step Solution:

1. **Understand the relationship between ΔG° and Kp**:
The relationship between standard Gibbs free energy change (ΔG°) and the equilibrium constant (Kp) is given by the formula:
\[ ΔG° = -RT \ln K \]
where:
- ΔG° is the standard Gibbs free energy change,
- R is the universal gas constant (approximately 8.314 J/(mol·K)),
- T is the temperature in Kelvin,
- K is the equilibrium constant (Kp in this case).

2. **Substitute the given value of Kp**:
We are given that Kp = 1. Substitute this value into the equation:
\[ ΔG° = -RT \ln(1) \]

3. **Calculate ln(1)**:
The natural logarithm of 1 is 0:
\[ \ln(1) = 0 \]

4. **Simplify the equation**:
Substitute ln(1) back into the equation:
\[ ΔG° = -RT \cdot 0 \]
This simplifies to:
\[ ΔG° = 0 \]

5. **Conclusion**:
Therefore, when Kp = 1, the value of ΔG° is:
\[ ΔG° = 0 \]

### Final Answer:
The value of ΔG° when Kp = 1 is **0**.