Equilibrium constants (K) for the reaction `3/2 H_2 (g) + 1/2N_2 (g) Leftrightarrow NH_3 (g) `are 0.0266 and 0.0129 at `350^@C` and `400^@C` respectively. Calculate the heat of formation of gaseous ammonia.

Answer

To calculate the heat of formation of gaseous ammonia (NH₃) using the equilibrium constants at two different temperatures, we can use the Van 't Hoff equation. Here’s a step-by-step solution:

### Step 1: Write down the given data
We have the following equilibrium constants:
- \( K_1 = 0.0266 \) at \( T_1 = 350^\circ C = 350 + 273 = 623 \, K \)
- \( K_2 = 0.0129 \) at \( T_2 = 400^\circ C = 400 + 273 = 673 \, K \)

### Step 2: Use the Van 't Hoff equation
The Van 't Hoff equation relates the change in the equilibrium constant with temperature to the enthalpy change (\( \Delta H \)):
\[ \ln \left( \frac{K_2}{K_1} \right) = -\frac{\Delta H}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
Where:
- \( R \) is the universal gas constant, which we will use in terms of calories: \( R = 1.987 \, \text{cal/(mol K)} \)

### Step 3: Calculate \( \frac{K_2}{K_1} \)
Calculate the ratio of the equilibrium constants:
\[ \frac{K_2}{K_1} = \frac{0.0129}{0.0266} \approx 0.485 \]

### Step 4: Calculate the natural logarithm
Now, take the natural logarithm of the ratio:
\[ \ln \left( 0.485 \right) \approx -0.718 \]

### Step 5: Calculate \( \frac{1}{T_2} - \frac{1}{T_1} \)
Calculate the difference in the reciprocals of the temperatures:
\[ \frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{673} - \frac{1}{623} \approx -0.000074 \]

### Step 6: Substitute values into the Van 't Hoff equation
Substituting the values into the Van 't Hoff equation:
\[ -0.718 = -\frac{\Delta H}{1.987} \left( -0.000074 \right) \]

### Step 7: Solve for \( \Delta H \)
Rearranging gives:
\[ \Delta H = \frac{0.718 \times 1.987}{0.000074} \approx 19488.5 \, \text{cal/mol} \]

### Step 8: Convert to kJ/mol
Since \( 1 \, \text{kJ} = 1000 \, \text{cal} \):
\[ \Delta H \approx 19.49 \, \text{kJ/mol} \]

### Final Answer
The heat of formation of gaseous ammonia (\( \Delta H \)) is approximately:
\[ \Delta H \approx 19.49 \, \text{kJ/mol} \]