The correct relation between equilibrium constant (K), standard free energy `(DeltaG^(@))` and temperature (T) is

Options

A. `DeltaG^(@)=RT lnK`
B. `K=e^(-DeltaG^(@)//2.303 RT)`
C. `DeltaG^(@)=-RTlog_(10)K`
D. `K=10^(-DeltaG^(@)//2.303RT)`

Answer

To find the correct relation between the equilibrium constant (K), standard free energy change (ΔG°), and temperature (T), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Gibbs Free Energy Equation**:
The Gibbs free energy change (ΔG) for a reaction can be expressed as:
\[ \Delta G = \Delta G^\circ + RT \ln Q \]
where:
- ΔG is the Gibbs free energy change at any point.
- ΔG° is the standard Gibbs free energy change.
- R is the universal gas constant (8.314 J/(mol·K)).
- T is the temperature in Kelvin.
- Q is the reaction quotient.

2. **Apply the Condition at Equilibrium**:
At equilibrium, the Gibbs free energy change (ΔG) is zero, and the reaction quotient (Q) becomes the equilibrium constant (K):
\[ 0 = \Delta G^\circ + RT \ln K \]

3. **Rearranging the Equation**:
Rearranging the equation gives:
\[ \Delta G^\circ = -RT \ln K \]

4. **Convert Natural Logarithm to Base 10**:
To express K in terms of base 10 logarithm, we can use the conversion between natural logarithm and logarithm base 10:
\[ \ln K = 2.303 \log K \]
Substituting this into the equation gives:
\[ \Delta G^\circ = -RT (2.303 \log K) \]

5. **Final Relation**:
Rearranging this gives us the final relation:
\[ \log K = -\frac{\Delta G^\circ}{2.303 RT} \]
This indicates that the logarithm of the equilibrium constant K is directly related to the standard free energy change ΔG° and inversely related to the temperature T.

### Conclusion:
The correct relation between the equilibrium constant (K), standard free energy change (ΔG°), and temperature (T) is:
\[ \log K = -\frac{\Delta G^\circ}{2.303 RT} \]