Standard free energy change for an equilibrium is zero, the value of `K_(p)` is

Options

A. zero
B. 1
C. 2
D. 100

Answer

To solve the question regarding the relationship between standard free energy change (ΔG°) and the equilibrium constant (Kp), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The relationship between standard free energy change (ΔG°) and the equilibrium constant (Kp) is given by the equation:
\[ \Delta G° = -2.303RT \log K_p \]
where:
- ΔG° is the standard free energy change,
- R is the ideal gas constant (8.314 J/mol·K),
- T is the temperature in Kelvin,
- Kp is the equilibrium constant at partial pressures.

2. **Given Condition**:
We are given that ΔG° = 0 at equilibrium.

3. **Substituting the Given Value**:
Substituting ΔG° = 0 into the equation:
\[ 0 = -2.303RT \log K_p \]

4. **Analyzing the Equation**:
Since the left side of the equation is zero, we can conclude:
\[ -2.303RT \log K_p = 0 \]
This implies that:
\[ \log K_p = 0 \]

5. **Finding Kp**:
If \(\log K_p = 0\), then we can find Kp by using the property of logarithms:
\[ K_p = 10^0 = 1 \]

6. **Conclusion**:
Therefore, the value of Kp when the standard free energy change is zero is:
\[ K_p = 1 \]

### Final Answer:
The value of \( K_p \) is **1**.

---