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 calculate 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_w). Here’s a step-by-step solution:

### Step 1: Write down the given data
- Equilibrium constant (K_w) = \(10^{-14}\)
- Temperature (T) = \(25°C\) = \(25 + 273 = 298 K\)
- Universal gas constant (R) = \(8.314 \, J/(mol \cdot K)\)

### Step 2: Use the formula for ΔG°
The relationship between ΔG° and K_w is given by the equation:
\[ \Delta G° = -RT \ln K_w \]

### Step 3: Substitute the values into the equation
Substituting the known values into the equation:
\[ \Delta G° = - (8.314 \, J/(mol \cdot K)) \times (298 \, K) \times \ln(10^{-14}) \]

### Step 4: Calculate the natural logarithm
Calculate \(\ln(10^{-14})\):
\[ \ln(10^{-14}) = -14 \ln(10) \]
Using \(\ln(10) \approx 2.303\):
\[ \ln(10^{-14}) \approx -14 \times 2.303 \approx -32.242 \]

### Step 5: Substitute back into the equation
Now substitute \(\ln(10^{-14})\) back into the ΔG° equation:
\[ \Delta G° = - (8.314) \times (298) \times (-32.242) \]

### Step 6: Calculate ΔG°
Calculating the product:
\[ \Delta G° = 8.314 \times 298 \times 32.242 \]
Calculating \(8.314 \times 298 \approx 2477.572\):
\[ \Delta G° \approx 2477.572 \times 32.242 \approx 79801.55 \, J/mol \]
Converting to kJ/mol:
\[ \Delta G° \approx 79.8 \, kJ/mol \]

### Final Answer
\[ \Delta G° \approx 8 \times 10^4 \, J/mol \]