Which is the correct relationship between `DeltaG^(@)` and equilibrium constant `K_(p)`?

Options

A. a. `K_(p)=-RT log DeltaG^(@)`
B. b. `K_(p)=[E//RT]^(DeltaG^(@))`
C. c. `K_(p)=-DeltaG^(@)//RT`
D. d. `K_(p)=e^(-DeltaG^(@)//RT)`

Answer

To find the correct relationship between the standard Gibbs free energy change (ΔG°) and the equilibrium constant (Kp), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definitions**:
- ΔG° (standard Gibbs free energy change) indicates the change in free energy under standard conditions (1 bar, 25°C).
- Kp is the equilibrium constant for a reaction expressed in terms of partial pressures.

2. **Use the Gibbs Free Energy Equation**:
- The general Gibbs free energy equation is:
\[ \Delta G = \Delta G° + RT \ln Q \]
- Here, ΔG is the Gibbs free energy change at any point in the reaction, R is the gas constant, T is the temperature in Kelvin, and Q is the reaction quotient.

3. **Consider the Reaction at Equilibrium**:
- At equilibrium, ΔG = 0 and Q becomes Kp (the equilibrium constant).
- Therefore, we can rewrite the equation as:
\[ 0 = \Delta G° + RT \ln Kp \]

4. **Rearrange the Equation**:
- Rearranging gives:
\[ \Delta G° = -RT \ln Kp \]

5. **Express Kp in Terms of ΔG°**:
- To express Kp in terms of ΔG°, we can exponentiate both sides:
\[ Kp = e^{-\Delta G°/(RT)} \]

6. **Conclusion**:
- The correct relationship between ΔG° and Kp is:
\[ Kp = e^{-\Delta G°/(RT)} \]

### Final Answer:
The correct relationship is:
\[ Kp = e^{-\Delta G°/(RT)} \]