In a mixture of `N_(2)` and `H_(2)` in the ratio `1:3` at `30` atm and `300^(@)C`, the `%` of `NH_(3)` at equilibrium is `17.8`. Calculate `K_(p)` for `N_(2)+3H_(2) hArr 2NH_(3)`.

Explanation

To calculate \( K_p \) for the reaction \( N_2 + 3H_2 \rightleftharpoons 2NH_3 \), we can follow these steps:

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

### Step 2: Define initial conditions
We are given that the initial ratio of \( N_2 \) to \( H_2 \) is \( 1:3 \). Let’s assume we have:
- 1 mole of \( N_2 \)
- 3 moles of \( H_2 \)

### Step 3: Set up the change in moles at equilibrium
Let \( x \) be the amount of \( N_2 \) that reacts at equilibrium. The changes in moles will be:
- \( N_2 \): \( 1 - x \)
- \( H_2 \): \( 3 - 3x \)
- \( NH_3 \): \( 2x \)

### Step 4: Calculate total moles at equilibrium
The total number of moles at equilibrium can be expressed as:
\[ \text{Total moles} = (1 - x) + (3 - 3x) + 2x = 4 - 2x \]

### Step 5: Use the percentage of \( NH_3 \) to find \( x \)
We are given that the percentage of \( NH_3 \) at equilibrium is \( 17.8\% \). This means:
\[ \frac{\text{moles of } NH_3}{\text{total moles}} = \frac{2x}{4 - 2x} = 0.178 \]

### Step 6: Set up the equation and solve for \( x \)
From the equation:
\[ 2x = 0.178(4 - 2x) \]
Expanding and rearranging gives:
\[ 2x = 0.712 - 0.356x \]
\[ 2.356x = 0.712 \]
\[ x = \frac{0.712}{2.356} \approx 0.302 \]

### Step 7: Calculate the equilibrium moles
Now we can calculate the moles at equilibrium:
- Moles of \( N_2 \) at equilibrium: \( 1 - x = 1 - 0.302 = 0.698 \)
- Moles of \( H_2 \) at equilibrium: \( 3 - 3x = 3 - 3(0.302) = 2.094 \)
- Moles of \( NH_3 \) at equilibrium: \( 2x = 2(0.302) = 0.604 \)

### Step 8: Calculate the partial pressures
The total pressure is given as \( 30 \, \text{atm} \). The partial pressures can be calculated as:
- \( P_{NH_3} = \frac{0.604}{4 - 2(0.302)} \times 30 \)
- \( P_{N_2} = \frac{0.698}{4 - 2(0.302)} \times 30 \)
- \( P_{H_2} = \frac{2.094}{4 - 2(0.302)} \times 30 \)

Calculating the total moles at equilibrium:
\[ 4 - 2(0.302) = 3.396 \]

Now substituting:
\[ P_{NH_3} = \frac{0.604}{3.396} \times 30 \approx 5.34 \, \text{atm} \]
\[ P_{N_2} = \frac{0.698}{3.396} \times 30 \approx 6.17 \, \text{atm} \]
\[ P_{H_2} = \frac{2.094}{3.396} \times 30 \approx 18.49 \, \text{atm} \]

### 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:
\[ K_p = \frac{(5.34)^2}{(6.17)(18.49)^3} \]

Calculating:
\[ K_p \approx \frac{28.51}{(6.17)(18.49^3)} \approx 7.29 \times 10^{-4} \, \text{atm}^{-2} \]

### Final Answer
Thus, the value of \( K_p \) is approximately:
\[ K_p \approx 7.29 \times 10^{-4} \, \text{atm}^{-2} \]

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