The standard free energy change of a reaction is `DeltaG^(@)=-115kJ//mol^(-1)` at `298 K.` Calculate the value of `log_(10)K_(p)` `(R=8.314JK^(-1)mol^(-1))`

Options

A. `20.16`
B. `2.303`
C. `2.016`
D. `13.83`

Answer

To solve the problem, we need to calculate the value of \( \log_{10} K_p \) using the given standard free energy change \( \Delta G^\circ \) and the relationship between \( \Delta G^\circ \) and \( K_p \).

### Step-by-Step Solution:

1. **Identify the given values:**
- Standard free energy change, \( \Delta G^\circ = -115 \, \text{kJ/mol} \)
- Temperature, \( T = 298 \, \text{K} \)
- Gas constant, \( R = 8.314 \, \text{J/(K·mol)} \)

2. **Convert \( \Delta G^\circ \) to Joules:**
Since \( R \) is in Joules, we need to convert \( \Delta G^\circ \) from kilojoules to joules:
\[ \Delta G^\circ = -115 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -115000 \, \text{J/mol} \]

3. **Use the relationship between \( \Delta G^\circ \) and \( K_p \):**
The formula relating \( \Delta G^\circ \) and \( K_p \) is:
\[ \Delta G^\circ = -2.303 \, R \, T \, \log_{10} K_p \]

4. **Rearrange the formula to solve for \( \log_{10} K_p \):**
\[ \log_{10} K_p = -\frac{\Delta G^\circ}{2.303 \, R \, T} \]

5. **Substitute the values into the equation:**
\[ \log_{10} K_p = -\frac{-115000}{2.303 \times 8.314 \times 298} \]

6. **Calculate the denominator:**
\[ 2.303 \times 8.314 \times 298 \approx 5730.78 \]

7. **Calculate \( \log_{10} K_p \):**
\[ \log_{10} K_p = \frac{115000}{5730.78} \approx 20.16 \]

8. **Final Result:**
\[ \log_{10} K_p \approx 20.16 \]