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.

Options

A. `-92.843 kJ`
B. `-47.77723 kJ`
C. `-132.5 kJ`
D. `-176.114 kJ`

Explanation

To find the heat of reaction for the equation \(N_2 + 3H_2 \rightarrow 2NH_3\) at \(50^\circ C\), we can use Kirchhoff's equation, which relates the change in enthalpy (\(\Delta H\)) of a reaction to the heat capacities of the reactants and products.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Heat of reaction at \(27^\circ C\) (\(\Delta H_1\)): \(-91.94 \, \text{kJ}\)
- Molar heat capacities at \(27^\circ C\):
- \(C_{p, N_2} = 28.45 \, \text{J/mol·K}\)
- \(C_{p, H_2} = 28.32 \, \text{J/mol·K}\)
- \(C_{p, NH_3} = 37.07 \, \text{J/mol·K}\)
- Temperature change: from \(T_1 = 27^\circ C\) to \(T_2 = 50^\circ C\)

2. **Convert Temperatures to Kelvin:**
- \(T_1 = 27 + 273 = 300 \, \text{K}\)
- \(T_2 = 50 + 273 = 323 \, \text{K}\)

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

4. **Convert \(\Delta C_p\) to kJ:**
\[ \Delta C_p = -39.27 \, \text{J/mol·K} = -0.03927 \, \text{kJ/mol·K} \]

5. **Apply Kirchhoff's Equation:**
\[ \Delta H_2 = \Delta H_1 + \Delta C_p \times (T_2 - T_1) \]
- Substitute the values:
\[ \Delta H_2 = -91.94 \, \text{kJ} + (-0.03927 \, \text{kJ/mol·K}) \times (323 \, \text{K} - 300 \, \text{K}) \]
- Calculate the temperature difference:
\[ T_2 - T_1 = 323 - 300 = 23 \, \text{K} \]
- Now calculate:
\[ \Delta H_2 = -91.94 \, \text{kJ} - 0.03927 \times 23 \]
\[ \Delta H_2 = -91.94 \, \text{kJ} - 0.90321 \, \text{kJ} = -92.84321 \, \text{kJ} \]

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

### Conclusion:
The heat of reaction for \(N_2 + 3H_2 \rightarrow 2NH_3\) at \(50^\circ C\) is approximately \(-92.84 \, \text{kJ}\).