The INCORRECT match in the following is

Options

A. `DeltaG^@ = 0, K = 1`
B. `DeltaG^@ lt 0, K lt 1`
C. `DeltaG^@ gt 0, K lt 1`
D. `DeltaG^@ lt 0, K gt 1`

Explanation

To solve the question regarding the incorrect match of thermodynamic conditions, we will analyze the relationships between Gibbs free energy (ΔG), standard Gibbs free energy (ΔG°), and the equilibrium constant (K).

### Step-by-Step Solution:

1. **Understanding Gibbs Free Energy**:
- The relationship between ΔG, ΔG°, and K is given by the equation:
\[ \Delta G = \Delta G^\circ + RT \ln K \]
- At equilibrium, ΔG = 0, which leads us to:
\[ 0 = \Delta G^\circ + RT \ln K \]
- Rearranging gives:
\[ \Delta G^\circ = -RT \ln K \]

2. **Analyzing Conditions**:
- **Condition 1**: If ΔG° = 0, then:
- From the equation, we have:
\[ 0 = -RT \ln K \implies \ln K = 0 \implies K = 1 \]
- This condition is correct.

- **Condition 2**: If ΔG° < 0, then:
- This implies:
\[ -RT \ln K < 0 \implies \ln K > 0 \implies K > 1 \]
- This condition is also correct.

- **Condition 3**: If ΔG° > 0, then:
- This implies:
\[ -RT \ln K > 0 \implies \ln K < 0 \implies K < 1 \]
- This condition is correct.

3. **Identifying the Incorrect Match**:
- We have established the correctness of the first three conditions.
- The question states that the second condition is an incorrect match. This means that the statement associated with ΔG° < 0 and K < 1 is incorrect because we have shown that ΔG° < 0 corresponds to K > 1.

4. **Conclusion**:
- Therefore, the incorrect match is the second condition, which states that when ΔG° is less than 0, K is less than 1. This is incorrect because ΔG° < 0 implies K > 1.

### Final Answer:
The incorrect match is the second condition.

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