300 पर एक अभिक्रिया के लिए साम्य स्थिरांक 10 है| `DeltaG^(ө)` का मान क्या होगा ?
`R=8.314 JK^(-1) "mol"^(-1)`

Options

A. `-5.74kJ`
B. `-574kJ`
C. `+11.48kJ`
D. `+5.74kJ`

Explanation

To find the standard Gibbs free energy change (ΔG°) for the reaction at 300 K with an equilibrium constant (K) of 10, we can use the following equation:

\[ \Delta G° = -2.303 \cdot R \cdot T \cdot \log K \]

### Step-by-Step Solution:

1. **Identify the given values:**
- Temperature (T) = 300 K
- Equilibrium constant (K) = 10
- Gas constant (R) = 8.314 J K⁻¹ mol⁻¹

2. **Convert the equilibrium constant to logarithmic form:**
- We need to calculate \(\log K\):
\[ K = 10 \implies \log K = \log 10 = 1 \]

3. **Substitute the values into the equation:**
\[ \Delta G° = -2.303 \cdot (8.314 \, \text{J K}^{-1} \text{mol}^{-1}) \cdot (300 \, \text{K}) \cdot \log(10) \]
\[ \Delta G° = -2.303 \cdot 8.314 \cdot 300 \cdot 1 \]

4. **Calculate the product:**
- First, calculate \(2.303 \cdot 8.314\):
\[ 2.303 \cdot 8.314 \approx 19.15 \]
- Now multiply by 300:
\[ 19.15 \cdot 300 \approx 5745 \, \text{J mol}^{-1} \]

5. **Convert Joules to kilojoules:**
\[ \Delta G° = -5745 \, \text{J mol}^{-1} = -5.745 \, \text{kJ mol}^{-1} \]

6. **Final answer:**
\[ \Delta G° \approx -5.74 \, \text{kJ mol}^{-1} \]

### Conclusion:
The value of \(\Delta G°\) is approximately \(-5.74 \, \text{kJ mol}^{-1}\).