`K_(p)` for `3//2H_(2)+1//2N_(2) hArr NH_(3)` are `0.0266` and `0.0129 atm^(-1)`, respectively, at `350^(@)C` and `400^(@)C`. Calculate the heat of formation of `NH_(3)`.

Explanation

To calculate the heat of formation of NH₃ using the given equilibrium constants (Kp) at two different temperatures, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Kp₁ at 350°C = 0.0266 atm⁻¹
- Kp₂ at 400°C = 0.0129 atm⁻¹
- T₁ = 350°C = 623 K (350 + 273)
- T₂ = 400°C = 673 K (400 + 273)

2. **Use the Van 't Hoff Equation**:
The Van 't Hoff equation relates the change in equilibrium constant with temperature to the enthalpy change (ΔH) of the reaction:
\[ \frac{d(\ln K)}{dT} = \frac{\Delta H}{R T^2} \]
For two temperatures, it can be rearranged to:
\[ \log \frac{K_{p2}}{K_{p1}} = \frac{\Delta H}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

3. **Substitute the Values**:
We can rearrange the equation to solve for ΔH:
\[ \Delta H = 2.303 R \cdot \log \frac{K_{p2}}{K_{p1}} \cdot \left( \frac{T_1 T_2}{T_2 - T_1} \right) \]
Here, R (the gas constant) can be taken as 2 cal/mol·K.

4. **Calculate the Logarithm**:
\[ \log \frac{K_{p2}}{K_{p1}} = \log \frac{0.0129}{0.0266} \]
Calculate this value:
\[ \log \frac{0.0129}{0.0266} \approx -0.194 \]

5. **Calculate ΔH**:
Substitute the values into the ΔH equation:
\[ \Delta H = 2.303 \times 2 \times (-0.194) \times \left( \frac{623 \times 673}{673 - 623} \right) \]

First, calculate the term inside the parentheses:
\[ \frac{623 \times 673}{673 - 623} = \frac{419399}{50} = 8387.98 \]

Now substitute this back:
\[ \Delta H = 2.303 \times 2 \times (-0.194) \times 8387.98 \]
\[ \Delta H \approx -12140 \text{ cal} \]

6. **Final Answer**:
Since the heat of formation is conventionally expressed as a negative value for exothermic reactions:
\[ \Delta H = -12140 \text{ cal} \]

### Conclusion:
The heat of formation of NH₃ is approximately -12140 cal.