The equilibrium constant of a reaction is `0.008` at `298 K`. The standard free energy change of the reaction at the same temperture is

Options

A. `+11.96 kJ`
B. `-11.96 kJ`
C. `-5.43 kJ`
D. `-8.46 kJ`

Explanation

To find the standard free energy change (ΔG°) of the reaction at 298 K given the equilibrium constant (K) of 0.008, we can use the following formula:

\[ \Delta G^\circ = -2.303 \times R \times T \times \log K \]

Where:
- \( R \) is the universal gas constant, which is approximately \( 8.314 \, \text{J/mol·K} \)
- \( T \) is the temperature in Kelvin (298 K in this case)
- \( K \) is the equilibrium constant (0.008)

### Step-by-Step Solution:

1. **Identify the values:**
- \( K = 0.008 \)
- \( R = 8.314 \, \text{J/mol·K} \)
- \( T = 298 \, \text{K} \)

2. **Calculate the logarithm of K:**
- First, express \( K \) in a more manageable form:
\[ K = 0.008 = 8 \times 10^{-3} \]
- Now, calculate \( \log K \):
\[ \log K = \log(8 \times 10^{-3}) = \log 8 + \log(10^{-3}) = \log 8 - 3 \]
- We know that \( \log 8 \approx 0.903 \) and \( \log(10^{-3}) = -3 \):
\[ \log K = 0.903 - 3 = -2.097 \]

3. **Substitute the values into the ΔG° formula:**
\[ \Delta G^\circ = -2.303 \times 8.314 \, \text{J/mol·K} \times 298 \, \text{K} \times (-2.097) \]

4. **Calculate ΔG°:**
- First, calculate the product:
\[ -2.303 \times 8.314 \times 298 \times (-2.097) \]
- Calculate \( 2.303 \times 8.314 \times 298 \):
\[ 2.303 \times 8.314 \approx 19.184 \quad \text{(approx)} \]
\[ 19.184 \times 298 \approx 5725.792 \quad \text{(approx)} \]
- Now multiply by \( -2.097 \):
\[ \Delta G^\circ \approx 5725.792 \times -2.097 \approx -11964.6 \, \text{J/mol} \]

5. **Convert to kilojoules:**
\[ \Delta G^\circ \approx -11.9646 \, \text{kJ/mol} \approx -11.96 \, \text{kJ/mol} \]

### Final Answer:
The standard free energy change of the reaction at 298 K is approximately:
\[ \Delta G^\circ \approx -11.96 \, \text{kJ/mol} \]