For the auto-ionization of water at `25^(@)C, H_(2)O(l)iff H^(+)(aq)+OH^(-)` (aq) equilibrium constant is `10^(-14)`.
What is `DeltaG^(@)` for the process?

Options

A. `~=8xx10^(4)J`
B. `~=3.5xx10^(4)J`
C. `~=10^(4)J`
D. None of these

Explanation

To find the standard Gibbs free energy change (ΔG°) for the auto-ionization of water at 25°C, we can use the relationship between ΔG°, the universal gas constant (R), the temperature (T), and the equilibrium constant (K). The equation is given by:

\[ \Delta G^\circ = -RT \ln K \]

### Step-by-Step Solution:

1. **Identify the values needed**:
- The equilibrium constant (K) for the auto-ionization of water at 25°C is given as \( K = 10^{-14} \).
- The universal gas constant \( R = 8.314 \, \text{J/(mol K)} \).
- The temperature \( T = 25°C = 298 \, \text{K} \) (since we convert Celsius to Kelvin by adding 273).

2. **Convert the equilibrium constant to natural logarithm**:
- We need to calculate \( \ln K \). Since \( K = 10^{-14} \), we can use the relationship between natural logarithm and base 10 logarithm:
\[ \ln K = \log_{10} K \times 2.303 \]
- Therefore,
\[ \ln(10^{-14}) = -14 \times 2.303 = -32.242 \]

3. **Substitute the values into the equation**:
- Now we can substitute \( R \), \( T \), and \( \ln K \) into the equation for ΔG°:
\[ \Delta G^\circ = - (8.314 \, \text{J/(mol K)}) \times (298 \, \text{K}) \times (-32.242) \]

4. **Calculate ΔG°**:
- Performing the multiplication:
\[ \Delta G^\circ = 8.314 \times 298 \times 32.242 \]
- Calculating this gives:
\[ \Delta G^\circ \approx 840 \, \text{J/mol} \]

5. **Final Result**:
- Therefore, the standard Gibbs free energy change for the auto-ionization of water at 25°C is:
\[ \Delta G^\circ \approx 840 \, \text{J/mol} \]