Which is / are correct ?

Options

A. 2.303 log K = `(-Delta H^(@))/(RT) + (Delta S^(@))/(R)`
B. `Delta G^(@) = - 2.303 ` RT log K
C. `-2.303 "log K" = (-Delta H^(@))/(RT^(2)) + (Delta S^(@))/(R)`
D. 2.303 log K = `(1)/(RT) (Delta H^(@) + Delta S^(@))`

Answer

To solve the question regarding which equations are correct, we will analyze the relationships between Gibbs free energy (ΔG°), enthalpy (ΔH°), entropy (ΔS°), and the equilibrium constant (K).

### Step-by-Step Solution:

1. **Understand the Fundamental Equations**:
The relationship between Gibbs free energy and the equilibrium constant is given by:
\[ \Delta G° = -RT \ln K \]
where R is the universal gas constant and T is the temperature in Kelvin.

2. **Convert Natural Logarithm to Logarithm Base 10**:
We can convert the natural logarithm to base 10 using the conversion factor \( \ln K = 2.303 \log K \). Therefore, we can rewrite the equation as:
\[ \Delta G° = -2.303RT \log K \]

3. **Relate Gibbs Free Energy to Enthalpy and Entropy**:
The Gibbs free energy can also be expressed in terms of enthalpy and entropy:
\[ \Delta G° = \Delta H° - T \Delta S° \]

4. **Equate the Two Expressions for ΔG°**:
Setting the two expressions for ΔG° equal gives:
\[ -2.303RT \log K = \Delta H° - T \Delta S° \]

5. **Rearranging the Equation**:
Rearranging the equation leads to:
\[ 2.303RT \log K = \Delta H° - T \Delta S° \]
This can be rearranged to:
\[ 2.303 \log K = \frac{\Delta H°}{RT} - \frac{\Delta S°}{R} \]

6. **Identifying the Correct Equations**:
The equations presented in the question can be analyzed:
- **Equation 1**: \( 2.303 \log K = -\frac{\Delta H°}{RT} + \frac{\Delta S°}{R} \) is correct.
- **Equation 2**: \( \Delta G° = -2.303RT \log K \) is also correct.
- **Equation 3**: \( -2.303 \log K = -\frac{\Delta H°}{RT^2} + \frac{\Delta S°}{R} \) is incorrect.
- **Equation 4**: \( 2.303 \log K = \frac{1}{R} \left( \frac{\Delta H°}{T} + \Delta S° \right) \) is incorrect.

### Conclusion:
From the analysis, the correct equations are:
- \( 2.303 \log K = -\frac{\Delta H°}{RT} + \frac{\Delta S°}{R} \)
- \( \Delta G° = -2.303RT \log K \)

### Final Answer:
The correct answers are A and B.