For the equilibrium,
`2H_(2)O hArr H_(3)O^(+) + OH^(-)`,the value of `DeltaG^(@)`at 298 K is approximately:

Options

A. `80 KJmol^(-1)`
B. `100 KJmol^(-1)`
C. `-80 KJmol^(-1)`
D. `-100 KJmol^(-1)`

Explanation

To solve the problem regarding the equilibrium of water dissociation and to find the standard Gibbs free energy change (ΔG°) at 298 K, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Equilibrium Expression**:
The equilibrium for the dissociation of water can be represented as:
\[ 2H_2O \rightleftharpoons H_3O^+ + OH^- \]

2. **Identify the Equilibrium Constant (K)**:
The dissociation constant (K) for water at 298 K is given as:
\[ K = [H_3O^+][OH^-] = 10^{-14} \]

3. **Use the Gibbs Free Energy Equation**:
The relationship between the standard Gibbs free energy change (ΔG°) and the equilibrium constant (K) is given by the equation:
\[ \Delta G^\circ = -RT \ln K \]
where:
- R = 8.314 J/(mol·K) (universal gas constant)
- T = 298 K (temperature in Kelvin)
- K = 10^{-14} (equilibrium constant)

4. **Convert R to kJ**:
Since we want ΔG° in kilojoules per mole, we convert R from J to kJ:
\[ R = \frac{8.314 \, \text{J/(mol·K)}}{1000} = 0.008314 \, \text{kJ/(mol·K)} \]

5. **Calculate ΔG°**:
Substitute the values into the equation:
\[ \Delta G^\circ = - (0.008314 \, \text{kJ/(mol·K)})(298 \, \text{K}) \ln(10^{-14}) \]

6. **Calculate ln(10^{-14})**:
Using the logarithmic identity:
\[ \ln(10^{-14}) = -14 \ln(10) \quad \text{and} \quad \ln(10) \approx 2.303 \]
Therefore:
\[ \ln(10^{-14}) \approx -14 \times 2.303 = -32.242 \]

7. **Substituting ln(10^{-14}) into ΔG°**:
Now, substituting this back into the ΔG° equation:
\[ \Delta G^\circ = - (0.008314)(298)(-32.242) \]

8. **Calculate the Final Value**:
\[ \Delta G^\circ \approx 0.008314 \times 298 \times 32.242 \approx 79.88 \, \text{kJ/mol} \]

9. **Round the Answer**:
Since the question asks for an approximate value, we round it to:
\[ \Delta G^\circ \approx 80 \, \text{kJ/mol} \]

### Final Answer:
The value of ΔG° at 298 K is approximately **80 kJ/mol**.