A reaction attains equilibrium state under standard conditions, then what is incorrect for this?

Options

A. Equilibrium constant K = 0
B. Equilibrium constant K = 1
C. `DeltaG^(@)=0 and DeltaH^(@)=TDeltaS^(@)`
D. `DeltaS=0 and DeltaH=TDeltaS`

Explanation

To determine the incorrect statement regarding a reaction that attains equilibrium under standard conditions, we will analyze the properties of Gibbs free energy and equilibrium constants.

### Step-by-Step Solution:

1. **Understanding Equilibrium Conditions**:
- At equilibrium, the Gibbs free energy change (ΔG) is equal to zero. This is a fundamental principle of thermodynamics.

2. **Gibbs Free Energy Equation**:
- The relationship between Gibbs free energy change and standard conditions can be expressed as:
\[ \Delta G = \Delta H - T \Delta S \]
- Where ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy.

3. **Setting ΔG to Zero**:
- Since ΔG = 0 at equilibrium, we can set the equation to:
\[ 0 = \Delta H - T \Delta S \]
- Rearranging gives us:
\[ \Delta H = T \Delta S \]

4. **Gibbs Free Energy and Equilibrium Constant**:
- Under standard conditions, the relationship between Gibbs free energy and the equilibrium constant (K) is given by:
\[ \Delta G^\circ = -RT \ln K \]
- Where R is the gas constant and T is the temperature.

5. **Equilibrium Constant at Equilibrium**:
- Setting ΔG to zero in the equation gives:
\[ 0 = -RT \ln K \]
- This implies:
\[ \ln K = 0 \quad \Rightarrow \quad K = e^0 = 1 \]

6. **Analyzing the Options**:
- Now we analyze the provided options:
- **Option 1**: K = 0 (This is incorrect because K cannot be zero at equilibrium).
- **Option 2**: K = 1 (This is correct as derived).
- **Option 3**: ΔG = 0 (This is correct as it is a condition for equilibrium).
- **Option 4**: ΔS = 0 (This is incorrect because if ΔS = 0, ΔG = ΔH, which does not represent equilibrium unless ΔH is also zero).

7. **Conclusion**:
- The incorrect statement regarding the equilibrium state under standard conditions is **Option 1: K = 0**.