A mixture of `N_2 and H_2` in the molar ratio of 1:3 reacts to give `NH_3` developing an equilibrium pressure of 50 atm and `650 ^@C`. `NH_3` present at equilibrium at equilibrium is 25% by weight, calculate `K_p` for `N_2(g) +3H_3(g) hArr 2NH_3(g)`

Answer

To solve the problem, we need to calculate the equilibrium constant \( K_p \) for the reaction:

\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

Given:
- Molar ratio of \( N_2 \) to \( H_2 \) is 1:3.
- Total equilibrium pressure = 50 atm.
- Temperature = 650 °C.
- \( NH_3 \) is 25% by weight at equilibrium.

### Step-by-Step Solution:

**Step 1: Define Initial Moles**
Assume we start with 1 mole of \( N_2 \) and 3 moles of \( H_2 \). Therefore, the initial moles are:
- \( N_2 = 1 \)
- \( H_2 = 3 \)
- \( NH_3 = 0 \)

**Step 2: Change in Moles at Equilibrium**
Let \( x \) be the moles of \( N_2 \) that react. According to the stoichiometry of the reaction:
- \( N_2 \) at equilibrium = \( 1 - x \)
- \( H_2 \) at equilibrium = \( 3 - 3x \)
- \( NH_3 \) at equilibrium = \( 2x \)

**Step 3: Calculate Weight Percent of \( NH_3 \)**
We know that \( NH_3 \) is 25% by weight at equilibrium. The total weight of the system can be expressed as:
\[ \text{Weight of } N_2 + \text{Weight of } H_2 + \text{Weight of } NH_3 = 28(1 - x) + 2(3 - 3x) + 17(2x) \]
Where:
- Molar mass of \( N_2 \) = 28 g/mol
- Molar mass of \( H_2 \) = 2 g/mol
- Molar mass of \( NH_3 \) = 17 g/mol

The total weight of the system can be simplified to:
\[ 28(1 - x) + 6 - 6x + 34x = 28 - 28x + 6 - 6x + 34x = 34 + 0x \]
Thus, the total weight is \( 34 \) g.

Since \( NH_3 \) is 25% by weight:
\[ \frac{17(2x)}{34} = 0.25 \implies 17(2x) = 8.5 \implies 2x = 0.5 \implies x = 0.25 \]

**Step 4: Calculate Moles at Equilibrium**
Substituting \( x \) back into the equilibrium expressions:
- \( N_2 = 1 - 0.25 = 0.75 \)
- \( H_2 = 3 - 3(0.25) = 2.25 \)
- \( NH_3 = 2(0.25) = 0.5 \)

**Step 5: Calculate Total Moles at Equilibrium**
Total moles at equilibrium:
\[ 0.75 + 2.25 + 0.5 = 3.5 \text{ moles} \]

**Step 6: Calculate Mole Fractions**
Now, we calculate the mole fractions:
- Mole fraction of \( N_2 = \frac{0.75}{3.5} = 0.214 \)
- Mole fraction of \( H_2 = \frac{2.25}{3.5} = 0.643 \)
- Mole fraction of \( NH_3 = \frac{0.5}{3.5} = 0.143 \)

**Step 7: Calculate Partial Pressures**
Using the total pressure of 50 atm:
- Partial pressure of \( N_2 = 0.214 \times 50 = 10.7 \text{ atm} \)
- Partial pressure of \( H_2 = 0.643 \times 50 = 32.15 \text{ atm} \)
- Partial pressure of \( NH_3 = 0.143 \times 50 = 7.15 \text{ atm} \)

**Step 8: Calculate \( K_p \)**
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \]
Substituting the values:
\[ K_p = \frac{(7.15)^2}{(10.7)(32.15)^3} \]
Calculating this gives:
\[ K_p \approx 21.37 \times 10^{-4} \]

### Final Answer
\[ K_p \approx 2.137 \times 10^{-3} \]