Acertain rection has equilibrium constant 10 and 100 at 300K and 400K respectively. The ratio of `Delta G^(@)` at 300K and 400K respectively

Options

A. `(1)/(10)`
B. `(3)/(8)`
C. `(1)/(2)`
D. `(3)/(4)`

Answer

To solve the problem, we need to find the ratio of the standard Gibbs free energy change (\( \Delta G^\circ \)) at two different temperatures (300 K and 400 K) for a reaction with given equilibrium constants.

### Step-by-Step Solution:

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

2. **Identify the given values**:
- At 300 K, the equilibrium constant \( K_1 = 10 \).
- At 400 K, the equilibrium constant \( K_2 = 100 \).
- The universal gas constant \( R \) is approximately \( 8.314 \, \text{J/(mol K)} \).

3. **Calculate \( \Delta G^\circ \) at 300 K**:
Using the formula:
\[ \Delta G^\circ_{300} = -R \cdot 300 \cdot \ln(10) \]

4. **Calculate \( \Delta G^\circ \) at 400 K**:
Similarly, for 400 K:
\[ \Delta G^\circ_{400} = -R \cdot 400 \cdot \ln(100) \]

5. **Set up the ratio**:
We need to find the ratio \( \frac{\Delta G^\circ_{300}}{\Delta G^\circ_{400}} \):
\[ \frac{\Delta G^\circ_{300}}{\Delta G^\circ_{400}} = \frac{-R \cdot 300 \cdot \ln(10)}{-R \cdot 400 \cdot \ln(100)} \]

6. **Simplify the ratio**:
The \( -R \) cancels out:
\[ \frac{\Delta G^\circ_{300}}{\Delta G^\circ_{400}} = \frac{300 \cdot \ln(10)}{400 \cdot \ln(100)} \]

7. **Further simplify**:
Since \( \ln(100) = \ln(10^2) = 2 \ln(10) \):
\[ \frac{\Delta G^\circ_{300}}{\Delta G^\circ_{400}} = \frac{300 \cdot \ln(10)}{400 \cdot 2 \ln(10)} = \frac{300}{800} = \frac{3}{8} \]

8. **Final result**:
Therefore, the ratio of \( \Delta G^\circ \) at 300 K to that at 400 K is:
\[ \frac{\Delta G^\circ_{300}}{\Delta G^\circ_{400}} = 0.375 \]

### Conclusion:
The correct answer is \( 0.375 \).