If `K lt 1` then the value of `Delta G^(@)` will be

Options

A. positive
B. negative
C. zero
D. 1

Explanation

To determine the value of \( \Delta G^\circ \) when the equilibrium constant \( K < 1 \), we can follow these steps:

### Step-by-Step Solution:

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

2. **Analyzing the Case When \( K < 1 \)**:
When \( K < 1 \), the natural logarithm of \( K \) will be negative:
\[ \ln K < 0 \]

3. **Substituting into the Equation**:
Since \( \ln K < 0 \), we can substitute this into the equation for \( \Delta G^\circ \):
\[ \Delta G^\circ = -RT \ln K \]
Since \( \ln K \) is negative, multiplying by \(-RT\) (where \( R \) and \( T \) are positive) will result in:
\[ \Delta G^\circ > 0 \]

4. **Conclusion**:
Therefore, if \( K < 1 \), then \( \Delta G^\circ \) is positive.

### Final Answer:
If \( K < 1 \), then \( \Delta G^\circ > 0 \) (i.e., \( \Delta G^\circ \) is positive).

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