Ths standard free energy change `Delta G^(@)` is related to K (equilibrium constant ) as`

Options

A. `Delta G^(@) = -2.303 RT log K`
B. `Delta G^(@) = 2.303 RT log K`
C. `Delta G^(@) = RT log K`
D. `Delta G^(@) = - RT log K`

Answer

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

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The standard free energy change (ΔG°) is related to the equilibrium constant (K) of a reaction at a given temperature. The relationship is given by the equation:
\[ \Delta G^\circ = -RT \ln K \]
where R is the universal gas constant and T is the temperature in Kelvin.

2. **Deriving the Equation**:
- Start with the Gibbs free energy change for a reaction:
\[ \Delta G = \Delta G^\circ + RT \ln Q \]
where Q is the reaction quotient.
- At equilibrium, the Gibbs free energy change (ΔG) is zero, so we set ΔG = 0:
\[ 0 = \Delta G^\circ + RT \ln K \]
Here, K is the equilibrium constant when the reaction reaches equilibrium.
- Rearranging the equation gives:
\[ \Delta G^\circ = -RT \ln K \]

3. **Using Logarithmic Properties**:
- The equation can also be expressed in terms of base 10 logarithm:
\[ \Delta G^\circ = -2.303 RT \log K \]
This is derived from the relationship between natural logarithm and logarithm base 10, where:
\[ \ln K = 2.303 \log K \]

4. **Final Formulation**:
- Thus, we conclude that the standard free energy change is directly related to the equilibrium constant by the equations:
\[ \Delta G^\circ = -RT \ln K \]
or
\[ \Delta G^\circ = -2.303 RT \log K \]

### Summary of the Solution:
The standard free energy change (ΔG°) is related to the equilibrium constant (K) through the equations:
\[ \Delta G^\circ = -RT \ln K \]
or
\[ \Delta G^\circ = -2.303 RT \log K \]