The standard enthaply for the reaction `H_(2)(g) + 1//2 O_(2)(g) to H_(2)O(l)` is - 285.76 kJ at 298 K.
Calculate the value of `DeltaH` at 373 K .The molar heat capcities at constant pressure `(C_(P))` in the given temperature range of `H_(2)(g),O_(2)(g) and H_(2)O(l)` are respectively 38.83,16 and `75.312 JK^(-1) mol^(-1)`.

Explanation

To calculate the value of ΔH at 373 K for the reaction \( H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \), we can use Kirchhoff's equation, which relates the change in enthalpy at different temperatures to the heat capacities of the reactants and products.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Standard enthalpy at 298 K, \( \Delta H_{298} = -285.76 \, \text{kJ} \)
- Molar heat capacities:
- \( C_{P,H_2} = 38.83 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( C_{P,O_2} = 16 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( C_{P,H_2O} = 75.312 \, \text{J K}^{-1} \text{mol}^{-1} \)

2. **Calculate the Change in Heat Capacity (ΔCp) for the Reaction:**
\[ \Delta C_p = C_{P,H_2O} - \left( C_{P,H_2} + \frac{1}{2} C_{P,O_2} \right) \]
\[ \Delta C_p = 75.312 - \left( 38.83 + \frac{1}{2} \times 16 \right) \]
\[ \Delta C_p = 75.312 - (38.83 + 8) = 75.312 - 46.83 = 28.482 \, \text{J K}^{-1} \text{mol}^{-1} \]

3. **Calculate the Temperature Change (ΔT):**
\[ \Delta T = T_2 - T_1 = 373 \, \text{K} - 298 \, \text{K} = 75 \, \text{K} \]

4. **Apply Kirchhoff's Equation:**
\[ \Delta H_{T_2} = \Delta H_{T_1} + \Delta C_p \cdot \Delta T \]
Convert ΔH from kJ to J for consistency:
\[ \Delta H_{298} = -285.76 \, \text{kJ} = -285760 \, \text{J} \]
Now substitute the values:
\[ \Delta H_{373} = -285760 \, \text{J} + (28.482 \, \text{J K}^{-1} \text{mol}^{-1}) \cdot (75 \, \text{K}) \]
\[ \Delta H_{373} = -285760 \, \text{J} + 2136.15 \, \text{J} \]
\[ \Delta H_{373} = -283623.85 \, \text{J} \]
Convert back to kJ:
\[ \Delta H_{373} = -283.62 \, \text{kJ} \]

5. **Final Result:**
\[ \Delta H_{373} \approx -283.62 \, \text{kJ} \]