The heat of reaction for `N_(2)+3H_(2) rarr 2NH_(3)` at `27^(@)C` is `-91.94kJ` . What will be its value at `50^(@)C` ? The molar heat capacities at constant pressure and `27^(@)C` for `N_(2),H_(2)` and `NH_(3)` are `28.45, 28.32` and `37.07` joule respectively.

Answer

To find the heat of reaction at 50°C for the reaction \( N_2 + 3H_2 \rightarrow 2NH_3 \) given the heat of reaction at 27°C is \(-91.94 \, \text{kJ}\), we will use Kirchhoff's equation. This equation relates the change in enthalpy at two different temperatures to the heat capacities of the reactants and products.

### Step-by-step Solution:

1. **Write Kirchhoff's Equation**:
\[ \Delta H_2 = \Delta H_1 + \Delta C_p \cdot (T_2 - T_1) \]
where:
- \(\Delta H_2\) = heat of reaction at \(T_2\) (50°C)
- \(\Delta H_1\) = heat of reaction at \(T_1\) (27°C)
- \(\Delta C_p\) = change in heat capacity
- \(T_2\) = 50°C
- \(T_1\) = 27°C

2. **Convert Temperatures to Kelvin**:
- \(T_1 = 27 + 273.15 = 300.15 \, K\)
- \(T_2 = 50 + 273.15 = 323.15 \, K\)

3. **Calculate \(\Delta C_p\)**:
\[ \Delta C_p = C_p (\text{products}) - C_p (\text{reactants}) \]
- For the products (2 moles of \(NH_3\)):
\[ C_p (\text{products}) = 2 \times 37.07 \, \text{J/mol·K} = 74.14 \, \text{J/mol·K} \]
- For the reactants (\(N_2\) and \(3H_2\)):
\[ C_p (\text{reactants}) = 28.45 \, \text{J/mol·K} + 3 \times 28.32 \, \text{J/mol·K} = 28.45 + 84.96 = 113.41 \, \text{J/mol·K} \]
- Therefore:
\[ \Delta C_p = 74.14 - 113.41 = -39.27 \, \text{J/mol·K} \]

4. **Calculate the Temperature Difference**:
\[ T_2 - T_1 = 50 - 27 = 23 \, \text{°C} = 23 \, \text{K} \]

5. **Calculate the Change in Enthalpy**:
- Convert \(\Delta C_p\) to kJ:
\[ \Delta C_p = -39.27 \, \text{J/mol·K} \times \frac{1 \, \text{kJ}}{1000 \, \text{J}} = -0.03927 \, \text{kJ/mol·K} \]
- Now substitute into Kirchhoff's equation:
\[ \Delta H_2 = -91.94 \, \text{kJ} + (-0.03927 \, \text{kJ/mol·K}) \times 23 \, \text{K} \]
\[ \Delta H_2 = -91.94 \, \text{kJ} - 0.90321 \, \text{kJ} \]
\[ \Delta H_2 = -92.84321 \, \text{kJ} \]

6. **Final Answer**:
\[ \Delta H_2 \approx -92.84 \, \text{kJ} \]