A reaction carried out by `1` mol of `N_(2)` and `3` mol of `H_(2)` shows at equilibrium the mole fraction of `NH_(3)` as `0.012` at `500^(@)C` and `10` atm pressure. Calculate `K_(p)` Also report the pressure at which "mole" `%` of `NH_(3)` in equilibrium mixture is increased to `10.4`.

Explanation

To solve the problem step by step, we will follow the outlined procedure to calculate \( K_p \) and the pressure at which the mole percent of \( NH_3 \) increases to \( 10.4\% \).

### Step 1: Write the Balanced Chemical Equation
The balanced chemical equation for the reaction is:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]

### Step 2: Determine Initial Moles and Change in Moles
Initially, we have:
- \( n(N_2) = 1 \) mol
- \( n(H_2) = 3 \) mol
- \( n(NH_3) = 0 \) mol

Let \( x \) be the amount of \( N_2 \) that reacts. Then, at equilibrium, the moles will be:
- \( n(N_2) = 1 - x \)
- \( n(H_2) = 3 - 3x \)
- \( n(NH_3) = 2x \)

### Step 3: Calculate Total Moles at Equilibrium
The total number of moles at equilibrium is:
\[ n_{total} = (1 - x) + (3 - 3x) + 2x = 4 - 2x \]

### Step 4: Use Mole Fraction to Find \( x \)
We know the mole fraction of \( NH_3 \) at equilibrium is given as \( 0.012 \):
\[ \text{Mole fraction of } NH_3 = \frac{n(NH_3)}{n_{total}} = \frac{2x}{4 - 2x} = 0.012 \]

Cross-multiplying gives:
\[ 2x = 0.012(4 - 2x) \]
\[ 2x = 0.048 - 0.024x \]
\[ 2x + 0.024x = 0.048 \]
\[ 2.024x = 0.048 \]
\[ x = \frac{0.048}{2.024} \approx 0.0237 \]

### Step 5: Calculate \( K_p \)
Using the expression for \( K_p \):
\[ K_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \]

Where:
- \( P_{NH_3} = \frac{2x}{4 - 2x} \times P \)
- \( P_{N_2} = \frac{1 - x}{4 - 2x} \times P \)
- \( P_{H_2} = \frac{3 - 3x}{4 - 2x} \times P \)

Substituting \( P = 10 \) atm:
1. Calculate \( n(NH_3) = 2x = 2 \times 0.0237 \approx 0.0474 \)
2. Calculate \( n(N_2) = 1 - x = 1 - 0.0237 \approx 0.9763 \)
3. Calculate \( n(H_2) = 3 - 3x = 3 - 3 \times 0.0237 \approx 2.9289 \)

Now substituting these values into the \( K_p \) expression:
\[ K_p = \frac{(0.0474)^2}{(0.9763)(2.9289)^3} \times 10^2 \]

Calculating gives:
\[ K_p \approx 1.431 \times 10^{-5} \text{ atm}^{-2} \]

### Step 6: Find Pressure for \( 10.4\% \) Mole Fraction of \( NH_3 \)
Set the mole fraction of \( NH_3 \) to \( 0.104 \):
\[ \frac{2x}{4 - 2x} = 0.104 \]

Cross-multiplying gives:
\[ 2x = 0.104(4 - 2x) \]
\[ 2x = 0.416 - 0.208x \]
\[ 2x + 0.208x = 0.416 \]
\[ 2.208x = 0.416 \]
\[ x = \frac{0.416}{2.208} \approx 0.1884 \]

### Step 7: Substitute \( x \) Back to Find New Pressure
Using the same \( K_p \) expression:
\[ 1.431 \times 10^{-5} = \frac{4(0.1884)^2(4 - 2(0.1884))^2}{(1 - 0.1884)(3 - 3(0.1884))^3} \times P^2 \]

Solving for \( P \) gives:
\[ P \approx 105.41 \text{ atm} \]

### Final Answers
- \( K_p \approx 1.431 \times 10^{-5} \text{ atm}^{-2} \)
- Pressure for \( 10.4\% \) mole fraction of \( NH_3 \) is approximately \( 105.41 \text{ atm} \).