`K_(p)` for the reaction `N_(2)+3H_(2) hArr 2NH_(3)` is `1.6xx10^(-4) atm^(-2)` at `400^(@)C`. What will be `K_(p)` at `500^(@)C`? The heat of reaction on this temperature is `-25.14` kcal?

Explanation

To find the equilibrium constant \( K_p \) at \( 500^\circ C \) given the \( K_p \) at \( 400^\circ C \) and the heat of reaction, we will use the van 't Hoff equation in the following steps:

### Step 1: Convert temperatures to Kelvin
First, we need to convert the temperatures from Celsius to Kelvin.

\[ T_1 = 400^\circ C + 273.15 = 673.15 \, K \]
\[ T_2 = 500^\circ C + 273.15 = 773.15 \, K \]

### Step 2: Use the van 't Hoff equation
The van 't Hoff equation relates the change in the equilibrium constant with temperature:

\[ \log \left( \frac{K_{p2}}{K_{p1}} \right) = \frac{\Delta H}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

Where:
- \( K_{p1} = 1.6 \times 10^{-4} \) atm\(^-2\) (at \( T_1 \))
- \( \Delta H = -25.14 \) kcal (convert to J: \( -25.14 \times 4184 \) J)
- \( R = 8.314 \) J/mol·K

### Step 3: Calculate \( \Delta H \) in J
Convert the heat of reaction from kcal to J:

\[ \Delta H = -25.14 \, \text{kcal} \times 4184 \, \text{J/kcal} = -105,000.56 \, \text{J} \]

### Step 4: Substitute values into the van 't Hoff equation
Now, substitute the values into the equation:

\[ \log \left( \frac{K_{p2}}{1.6 \times 10^{-4}} \right) = \frac{-105000.56}{2.303 \times 8.314} \left( \frac{1}{673.15} - \frac{1}{773.15} \right) \]

### Step 5: Calculate the right-hand side
First, calculate the term \( \left( \frac{1}{673.15} - \frac{1}{773.15} \right) \):

\[ \frac{1}{673.15} \approx 0.001484 \, K^{-1} \]
\[ \frac{1}{773.15} \approx 0.001292 \, K^{-1} \]
\[ \frac{1}{673.15} - \frac{1}{773.15} \approx 0.001484 - 0.001292 = 0.000192 \, K^{-1} \]

Now calculate the entire right-hand side:

\[ \frac{-105000.56}{2.303 \times 8.314} \approx -5,669.05 \]

Multiply by \( 0.000192 \):

\[ -5,669.05 \times 0.000192 \approx -1.090 \]

### Step 6: Solve for \( K_{p2} \)
Now we can solve for \( K_{p2} \):

\[ \log \left( \frac{K_{p2}}{1.6 \times 10^{-4}} \right) \approx -1.090 \]

Exponentiating both sides gives:

\[ \frac{K_{p2}}{1.6 \times 10^{-4}} \approx 10^{-1.090} \]

Calculating \( 10^{-1.090} \):

\[ 10^{-1.090} \approx 0.081 \]

Thus:

\[ K_{p2} \approx 0.081 \times 1.6 \times 10^{-4} \approx 1.296 \times 10^{-5} \, \text{atm}^{-2} \]

### Final Answer
\[ K_{p2} \approx 1.3 \times 10^{-5} \, \text{atm}^{-2} \]

---