The value of `K_(p)` for the water gas reaction, `CO +H_(2)O hArr CO_(2) +H_(2)`is `1.06 xx 10^(5)` at `25^(@)C` . Calculate the standard free energy change for the reaction at `25^(@)C`.

Explanation

To calculate the standard free energy change (ΔG°) for the water gas reaction at 25°C, we can use the following equation:

\[ \Delta G° = -RT \ln K_p \]

Where:
- \( R \) is the universal gas constant (8.314 J/mol·K)
- \( T \) is the temperature in Kelvin
- \( K_p \) is the equilibrium constant (given as \( 1.06 \times 10^5 \))

### Step-by-Step Solution:

**Step 1: Convert the temperature to Kelvin**
- The given temperature is 25°C. To convert this to Kelvin:
\[ T(K) = 25 + 273.15 = 298.15 \, K \]

**Step 2: Substitute the values into the equation**
- We know \( R = 8.314 \, \text{J/mol·K} \) and \( K_p = 1.06 \times 10^5 \).
- Now, we can substitute these values into the equation:
\[ \Delta G° = - (8.314 \, \text{J/mol·K}) \times (298.15 \, K) \times \ln(1.06 \times 10^5) \]

**Step 3: Calculate \( \ln(K_p) \)**
- First, we calculate \( \ln(1.06 \times 10^5) \):
\[ \ln(1.06 \times 10^5) = \ln(1.06) + \ln(10^5) = \ln(1.06) + 5 \ln(10) \]
- Where \( \ln(10) \approx 2.303 \):
\[ \ln(1.06) \approx 0.058 (using a calculator) \]
\[ \ln(1.06 \times 10^5) \approx 0.058 + 5 \times 2.303 \approx 0.058 + 11.515 = 11.573 \]

**Step 4: Plug in the value of \( \ln(K_p) \)**
- Now substitute \( \ln(1.06 \times 10^5) \) back into the equation:
\[ \Delta G° = - (8.314) \times (298.15) \times (11.573) \]

**Step 5: Calculate \( \Delta G° \)**
- Performing the multiplication:
\[ \Delta G° = - (8.314 \times 298.15 \times 11.573) \approx - 24173.5 \, \text{J/mol} \]
- Converting this to kJ/mol:
\[ \Delta G° \approx -24.17 \, \text{kJ/mol} \]

### Final Answer:
\[ \Delta G° \approx -24.17 \, \text{kJ/mol} \]