In the Haber 's process of ammonia manufacture:
`N_(2) (g)+3H_(2) (g) to 2NH_(3) (g), triangleH_(298K)^(@)=-92.2kJ`
If `N_(2) (g), H_(2) (g) and NH_(3) (g)" have "C_(p) (JK^(-1) mol^(-1))` values 29.1, 28.8 and 35.1 respectively then if `C_p` is independent of temperature, the reaction at `100^@C` as compared to that of `25^@C` will be

Options

A. More endothermic
B. Less endothermic
C. Less exothermic
D. More exothermic

Answer

To solve the problem, we need to determine how the enthalpy change (ΔH) of the reaction changes when the temperature is increased from 25°C (298 K) to 100°C (373 K). We will use the provided heat capacities (C_p) of the reactants and products to find the change in heat capacity (ΔC_p) and then apply it to calculate the new enthalpy change (ΔH_2) at 100°C.

### Step-by-Step Solution:

1. **Identify the Reaction and Given Data**:
The reaction is:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]
Given:
- ΔH at 298 K = -92.2 kJ
- C_p values:
- N2 = 29.1 J/(K·mol)
- H2 = 28.8 J/(K·mol)
- NH3 = 35.1 J/(K·mol)

2. **Calculate ΔC_p**:
The change in heat capacity (ΔC_p) for the reaction is calculated as:
\[ ΔC_p = C_p(\text{products}) - C_p(\text{reactants}) \]
For the products (2 moles of NH3):
\[ C_p(\text{products}) = 2 \times 35.1 = 70.2 \text{ J/(K·mol)} \]
For the reactants (1 mole of N2 and 3 moles of H2):
\[ C_p(\text{reactants}) = 29.1 + 3 \times 28.8 = 29.1 + 86.4 = 115.5 \text{ J/(K·mol)} \]
Therefore:
\[ ΔC_p = 70.2 - 115.5 = -45.3 \text{ J/(K·mol)} \]

3. **Apply the Enthalpy Change Formula**:
The change in enthalpy at the new temperature (100°C or 373 K) can be calculated using the formula:
\[ ΔH_2 - ΔH_1 = ΔC_p \times (T_2 - T_1) \]
Where:
- \( ΔH_1 = -92.2 \text{ kJ} \) (at 298 K)
- \( T_1 = 298 \text{ K} \)
- \( T_2 = 373 \text{ K} \)

Plugging in the values:
\[ ΔH_2 - (-92.2) = -45.3 \times (373 - 298) \]
\[ ΔH_2 + 92.2 = -45.3 \times 75 \]
\[ ΔH_2 + 92.2 = -3397.5 \text{ J} = -3.3975 \text{ kJ} \]
(Note: Convert J to kJ by dividing by 1000)

4. **Calculate ΔH_2**:
\[ ΔH_2 = -3.3975 - 92.2 = -95.5975 \text{ kJ} \]
Rounding off gives:
\[ ΔH_2 \approx -95.6 \text{ kJ} \]

5. **Determine the Nature of the Reaction**:
Since ΔH_2 is more negative than ΔH_1, the reaction at 100°C is more exothermic compared to at 25°C.

### Conclusion:
The reaction at 100°C is more exothermic compared to that at 25°C.