For a reversible system at equilibrium, which of the following is correct

Options

A. `K = e^((-Delta G^(@))/(RT))`
B. `K = e^((-Delta G)/(RT))`
C. `K = 10^((-Delta G)/(RT))`
D. `10^((-Delta G)/(RT))`

Answer

To solve the question regarding the correct statement for a reversible system at equilibrium, we will analyze the relationship between Gibbs free energy change (ΔG°) and the equilibrium constant (K).

### Step-by-Step Solution:

1. **Understanding the Relationship**:
For a reversible reaction at equilibrium, the Gibbs free energy change (ΔG°) is related to the equilibrium constant (K) by the equation:
\[ \Delta G° = -RT \ln K \]
where R is the universal gas constant and T is the temperature in Kelvin.

2. **Rearranging the Equation**:
We can rearrange the equation to express K in terms of ΔG°:
\[ K = e^{-\Delta G° / RT} \]
This shows that the equilibrium constant K can be calculated from the standard Gibbs free energy change.

3. **Considering the Options**:
Now, we need to evaluate the provided options based on the derived equation:
- Option 1: \( K = e^{-\Delta G° / RT} \) (This matches our derived equation)
- Option 2: \( K = e^{-\Delta G / RT} \) (Incorrect, as it uses ΔG instead of ΔG°)
- Option 3: \( K = 10^{-\Delta G / (2.303RT)} \) (Incorrect, as it uses ΔG instead of ΔG°)
- Option 4: \( K = 10^{-\Delta G / RT} \) (Incorrect, as it uses ΔG instead of ΔG°)

4. **Conclusion**:
From the analysis, we find that the only correct option is:
\[ \text{Option 1: } K = e^{-\Delta G° / RT} \]

### Final Answer:
The correct statement for a reversible system at equilibrium is:
**Option 1: \( K = e^{-\Delta G° / RT} \)**