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 step by step, we will follow the reaction given and the information provided in the question.

### Step 1: Write the balanced chemical equation
The reaction is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

### Step 2: Set up initial moles
Given the molar ratio of \(N_2\) to \(H_2\) is 1:3, we can assume:
- Moles of \(N_2 = 1\)
- Moles of \(H_2 = 3\)
- Moles of \(NH_3 = 0\) (initially)

### Step 3: Define the change in moles at equilibrium
Let \(x\) be the extent of the reaction (the amount of \(N_2\) that reacts). At equilibrium, the moles will be:
- Moles of \(N_2 = 1 - x\)
- Moles of \(H_2 = 3 - 3x\)
- Moles of \(NH_3 = 2x\)

### Step 4: Use the weight percentage of \(NH_3\) to find \(x\)
We know that at equilibrium, \(NH_3\) is 25% by weight.

The molecular weight of \(NH_3\) is 17 g/mol. The total mass at equilibrium can be calculated as follows:
- Mass of \(NH_3 = 2x \times 17\)
- Mass of \(N_2 = (1 - x) \times 28\)
- Mass of \(H_2 = (3 - 3x) \times 2\)

Total mass:
\[ \text{Total mass} = (1 - x) \times 28 + (3 - 3x) \times 2 + 2x \times 17 \]

Setting up the equation for weight percentage:
\[ \frac{2x \times 17}{\text{Total mass}} = 0.25 \]

### Step 5: Solve for \(x\)
Substituting the total mass into the equation and simplifying:
\[ \frac{34x}{(28(1 - x) + 6(3 - 3x) + 34x)} = 0.25 \]

Solving this equation gives us \(x = 0.75\).

### Step 6: Calculate the moles at equilibrium
Now substituting \(x\) back into the moles:
- Moles of \(N_2 = 1 - 0.75 = 0.25\)
- Moles of \(H_2 = 3 - 3(0.75) = 0.75\)
- Moles of \(NH_3 = 2(0.75) = 1.5\)

### Step 7: Calculate total moles
Total moles at equilibrium:
\[ \text{Total moles} = 0.25 + 0.75 + 1.5 = 2.5 \]

### Step 8: Calculate partial pressures
Using the total pressure of 50 atm:
- Partial pressure of \(N_2\):
\[ P_{N_2} = \left(\frac{0.25}{2.5}\right) \times 50 = 5 \text{ atm} \]
- Partial pressure of \(H_2\):
\[ P_{H_2} = \left(\frac{0.75}{2.5}\right) \times 50 = 15 \text{ atm} \]
- Partial pressure of \(NH_3\):
\[ P_{NH_3} = \left(\frac{1.5}{2.5}\right) \times 50 = 30 \text{ atm} \]

### Step 9: Calculate \(K_p\)
Using the expression for \(K_p\):
\[ K_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \]
Substituting the values:
\[ K_p = \frac{(30)^2}{(5)(15)^3} = \frac{900}{5 \times 3375} = \frac{900}{16875} = 0.0533 \]

### Final Result
Thus, the value of \(K_p\) is approximately \(0.0533\).

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