`K_(p)` for the reaction `N_(2) + 3H_(2) hArr 2NH_(3) `, is `1.6 xx 10^(-4) atm^(-2)` at `400^(@)` C. What will be `K_(p)` at `500^(@)`C ? Heat of reaction in this temperature range is-25.14 Kcal.

Options

A. `1.43 xx 10^(-5) atm^(-2)`
B. `1.43 xx 10^(5) atm^(-2)`
C. `1.43 xx 10^(-5) atm^(2)`
D. `1.43 xx 10^(5) atm^(2)`

Answer

To find the value of \( K_p \) at \( 500^\circ C \) for the reaction

\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]

given that \( K_p \) at \( 400^\circ C \) is \( 1.6 \times 10^{-4} \, \text{atm}^{-2} \) and the heat of reaction \( \Delta H \) is \( -25.14 \, \text{kcal} \), we can use the van 't Hoff equation:

\[ \log \frac{K_2}{K_1} = \frac{\Delta H}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

### Step 1: Convert temperatures to Kelvin
- \( T_1 = 400^\circ C = 400 + 273 = 673 \, K \)
- \( T_2 = 500^\circ C = 500 + 273 = 773 \, K \)

### Step 2: Convert \( \Delta H \) to calories
Since \( \Delta H \) is given in kcal, we convert it to calories:
\[ \Delta H = -25.14 \, \text{kcal} \times 1000 = -25140 \, \text{cal} \]

### Step 3: Use the value of \( R \)
The value of \( R \) in calories is \( 2 \, \text{cal/(mol K)} \).

### Step 4: Substitute the values into the van 't Hoff equation
We can now substitute \( K_1 = 1.6 \times 10^{-4} \), \( T_1 = 673 \, K \), \( T_2 = 773 \, K \), \( \Delta H = -25140 \, \text{cal} \), and \( R = 2 \, \text{cal/(mol K)} \):

\[ \log \frac{K_2}{1.6 \times 10^{-4}} = \frac{-25140}{2.303 \times 2} \left( \frac{1}{673} - \frac{1}{773} \right) \]

### Step 5: Calculate the right-hand side
First, calculate \( \frac{1}{673} - \frac{1}{773} \):
\[ \frac{1}{673} \approx 0.001484 \quad \text{and} \quad \frac{1}{773} \approx 0.001292 \]
\[ \frac{1}{673} - \frac{1}{773} \approx 0.001484 - 0.001292 = 0.000192 \]

Now, calculate the entire right-hand side:
\[ \frac{-25140}{2.303 \times 2} \times 0.000192 \]
Calculating \( 2.303 \times 2 = 4.606 \):
\[ \frac{-25140}{4.606} \approx -5467.23 \]
Now multiply by \( 0.000192 \):
\[ -5467.23 \times 0.000192 \approx -1.050 \]

### Step 6: Solve for \( K_2 \)
Now we have:
\[ \log K_2 - \log(1.6 \times 10^{-4}) = -1.050 \]
\[ \log K_2 = -1.050 + \log(1.6 \times 10^{-4}) \]
Calculating \( \log(1.6 \times 10^{-4}) \):
\[ \log(1.6) \approx 0.2041 \quad \text{and} \quad \log(10^{-4}) = -4 \]
\[ \log(1.6 \times 10^{-4}) \approx 0.2041 - 4 = -3.7959 \]
Now substituting back:
\[ \log K_2 = -1.050 - 3.7959 \approx -4.8459 \]
Converting back from log:
\[ K_2 = 10^{-4.8459} \approx 1.429 \times 10^{-5} \, \text{atm}^{-2} \]

### Final Answer
Thus, the value of \( K_p \) at \( 500^\circ C \) is approximately:
\[ K_p \approx 1.43 \times 10^{-5} \, \text{atm}^{-2} \]

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