Predict the standard reaction enthalpy of `2NO_(2)(g)rarrN_(2)O_(4)(g)` at `100^(@)C`. `Delta H^(@)` at `25^(@)C` is `-57.2 kj.mol^(-1)C_(p)(NO_(2))=37.2j.mol^(-1)K^(-1)C_(p)(N_(2)O_(4))=77.28 J. mol^(-1)k^(-1)`

Explanation

To predict the standard reaction enthalpy of the reaction \( 2NO_2(g) \rightarrow N_2O_4(g) \) at \( 100^\circ C \), we can use Kirchhoff's law of thermodynamics. This law relates the change in enthalpy at two different temperatures to the heat capacities of the reactants and products.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Standard reaction enthalpy at \( 25^\circ C \) (or \( 298 K \)):
\[ \Delta H^\circ_{298} = -57.2 \, \text{kJ/mol} \]
- Heat capacity of \( NO_2 \):
\[ C_p(NO_2) = 37.2 \, \text{J/mol·K} \]
- Heat capacity of \( N_2O_4 \):
\[ C_p(N_2O_4) = 77.28 \, \text{J/mol·K} \]
- Temperature change:
\[ T_1 = 298 \, K \quad (25^\circ C) \\ T_2 = 373 \, K \quad (100^\circ C) \]

2. **Calculate the Change in Heat Capacity (\( \Delta C_p \)):**
\[ \Delta C_p = C_p(N_2O_4) - 2 \times C_p(NO_2) \]
\[ \Delta C_p = 77.28 \, \text{J/mol·K} - 2 \times 37.2 \, \text{J/mol·K} \]
\[ \Delta C_p = 77.28 \, \text{J/mol·K} - 74.4 \, \text{J/mol·K} = 2.88 \, \text{J/mol·K} \]

3. **Calculate the Change in Enthalpy (\( \Delta H \)) at \( 100^\circ C \):**
Using Kirchhoff's equation:
\[ \Delta H^\circ_{T_2} = \Delta H^\circ_{T_1} + \Delta C_p \times (T_2 - T_1) \]
Substitute the values:
\[ \Delta H^\circ_{373} = -57.2 \, \text{kJ/mol} + (2.88 \, \text{J/mol·K} \times (373 \, K - 298 \, K)) \]
\[ = -57.2 \, \text{kJ/mol} + (2.88 \, \text{J/mol·K} \times 75 \, K) \]
\[ = -57.2 \, \text{kJ/mol} + 216 \, \text{J/mol} \]
Convert \( 216 \, \text{J/mol} \) to \( \text{kJ/mol} \):
\[ 216 \, \text{J/mol} = 0.216 \, \text{kJ/mol} \]
Now, substitute back:
\[ \Delta H^\circ_{373} = -57.2 \, \text{kJ/mol} + 0.216 \, \text{kJ/mol} = -56.984 \, \text{kJ/mol} \]

4. **Final Answer:**
\[ \Delta H^\circ_{100^\circ C} \approx -56.98 \, \text{kJ/mol} \]