If `K lt1.0`, what will be the value of `DeltaG^(@)` out of the following `?`

Options

A. `1.0`
B. Zero
C. Negative
D. Positive

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. **Understand 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. **Analyze the Condition \(K < 1\)**:
Since \(K < 1\), it implies that the natural logarithm of \(K\) will be negative because the logarithm of a number less than 1 is always negative:
\[ \ln K < 0 \]

3. **Substitute into the Equation**:
Substituting the negative value of \(\ln K\) into the equation for \(\Delta G^\circ\):
\[ \Delta G^\circ = -RT \ln K \]
Since \(\ln K < 0\), we have:
\[ \Delta G^\circ = -RT (\text{negative value}) = \text{positive value} \]

4. **Conclusion**:
Therefore, if \(K < 1\), \(\Delta G^\circ\) will be greater than 0, indicating that the reaction is non-spontaneous under standard conditions.

### Final Answer:
\(\Delta G^\circ > 0\)

---