If chemical equilibrium is attained at standard states then what is the value of `DeltaG^(@)`?

Answer

To solve the question regarding the value of \(\Delta G^\circ\) when chemical equilibrium is attained at standard states, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Concept of Standard State**:
- Standard state refers to the conditions of 1 bar pressure and a specified temperature (usually 25°C) for gases, and 1 M concentration for solutes in solution.

2. **Defining Chemical Equilibrium**:
- At chemical equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant.

3. **Setting Up the Reaction**:
- Let's consider a generic reaction:
\[ A \rightleftharpoons B \]
- At equilibrium, we can express the equilibrium constant \( K_c \) as:
\[ K_c = \frac{[B]}{[A]} \]

4. **Concentrations at Standard State**:
- At standard states, the concentrations of reactants and products are often taken as unity (1 M). Therefore:
\[ [A] = 1 \text{ M} \quad \text{and} \quad [B] = 1 \text{ M} \]
- This leads to:
\[ K_c = \frac{1}{1} = 1 \]

5. **Using the Relationship Between \(\Delta G^\circ\) and \( K_c \)**:
- The relationship between the standard Gibbs free energy change (\(\Delta G^\circ\)) and the equilibrium constant (\( K_c \)) is given by the equation:
\[ \Delta G^\circ = -RT \ln K_c \]
- Here, \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin.

6. **Substituting the Value of \( K_c \)**:
- Since we found that \( K_c = 1 \):
\[ \Delta G^\circ = -RT \ln(1) \]

7. **Calculating \(\ln(1)\)**:
- The natural logarithm of 1 is 0:
\[ \ln(1) = 0 \]

8. **Final Calculation**:
- Therefore, substituting this back into the equation gives:
\[ \Delta G^\circ = -RT \cdot 0 = 0 \]

### Conclusion:
- The value of \(\Delta G^\circ\) when chemical equilibrium is attained at standard states is:
\[ \Delta G^\circ = 0 \]