When reaction is at standard state at equilibrium, then

Options

A. `DeltaH^(@)=0`
B. `DeltaS^(@)=0`
C. equilibrium constant K=0
D. equilibrium constant K=1

Explanation

To solve the question regarding the behavior of a reaction at standard state at equilibrium, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Standard State and Equilibrium**:
- At standard state, the concentrations of reactants and products are defined under standard conditions (1 M concentration, 1 atm pressure, etc.).
- At equilibrium, the rates of the forward and reverse reactions are equal, leading to constant concentrations of reactants and products.

2. **Setting Up the Reaction**:
- Consider a generic reaction: \( A \rightleftharpoons B + C \).
- Let the initial concentration of A be \( x \) and the change at equilibrium be \( y \). Thus, at equilibrium:
- Concentration of A = \( x - y \)
- Concentration of B = \( y \)
- Concentration of C = \( y \)

3. **Writing the Equilibrium Constant Expression**:
- The equilibrium constant \( K \) is defined as:
\[ K = \frac{[B][C]}{[A]} = \frac{y \cdot y}{x - y} = \frac{y^2}{x - y} \]

4. **Condition at Standard State**:
- At standard state, it is given that the concentrations of products and reactants are equal at equilibrium. This implies:
\[ [B] = [C] = [A] \]
- Therefore, at equilibrium, \( y = x - y \) or \( 2y = x \).

5. **Determining the Value of K**:
- If \( [B] = [C] = [A] \), then substituting \( y = \frac{x}{2} \) into the equilibrium expression gives:
\[ K = \frac{y^2}{x - y} = \frac{\left(\frac{x}{2}\right)^2}{x - \frac{x}{2}} = \frac{\frac{x^2}{4}}{\frac{x}{2}} = \frac{x}{2} = 1 \]
- Hence, the equilibrium constant \( K \) at standard state is equal to 1.

6. **Relating ΔG and K**:
- The change in Gibbs free energy (\( \Delta G \)) at equilibrium is given by:
\[ \Delta G = \Delta G^0 + RT \ln K \]
- Since \( K = 1 \), we have:
\[ \Delta G = \Delta G^0 + RT \ln(1) = \Delta G^0 \]
- At equilibrium, \( \Delta G = 0 \), thus \( \Delta G^0 = 0 \).

7. **Conclusion**:
- Therefore, when a reaction is at standard state at equilibrium, the equilibrium constant \( K \) is equal to 1, and the change in Gibbs free energy \( \Delta G^0 \) is also 0.

### Final Answer:
When a reaction is at standard state at equilibrium, the equilibrium constant \( K \) is equal to 1.