At `450^(@)C` the equilibrium constant `K_(p)` for the reaction `N_(2)+3H_(2) hArr 2NH_(3)` was found to be `1.6xx10^(-5)` at a pressure of `200` atm. If `N_(2)` and `H_(2)` are taken in `1:3` ratio. What is `%` of `NH_(3)` formed at this temperature?

Explanation

To solve the problem step by step, we will follow the outlined procedure based on the equilibrium reaction provided.

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

### Step 2: Set Up Initial Conditions
Given that we start with a 1:3 ratio of \( N_2 \) to \( H_2 \):
- Initial moles of \( N_2 = 1 \) mole
- Initial moles of \( H_2 = 3 \) moles
- Initial moles of \( NH_3 = 0 \)

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

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

### Step 5: Calculate Partial Pressures
Using the total pressure \( P = 200 \) atm, we can find the partial pressures:
- Partial pressure of \( N_2 \):
\[ P_{N_2} = \frac{(1 - x)}{(4 - 2x)} \times 200 \]
- Partial pressure of \( H_2 \):
\[ P_{H_2} = \frac{(3 - 3x)}{(4 - 2x)} \times 200 \]
- Partial pressure of \( NH_3 \):
\[ P_{NH_3} = \frac{(2x)}{(4 - 2x)} \times 200 \]

### Step 6: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \]
Substituting the expressions for partial pressures:
\[ K_p = \frac{\left(\frac{(2x)}{(4 - 2x)} \times 200\right)^2}{\left(\frac{(1 - x)}{(4 - 2x)} \times 200\right) \left(\frac{(3 - 3x)}{(4 - 2x)} \times 200\right)^3} \]

### Step 7: Substitute Known Values
Given \( K_p = 1.6 \times 10^{-5} \):
\[ 1.6 \times 10^{-5} = \frac{(2x)^2 \cdot 200^2}{(1 - x)(3 - 3x)^3 \cdot (4 - 2x)^4} \]

### Step 8: Solve for \( x \)
This equation can be simplified and solved for \( x \). After simplification, we find:
\[ x \approx 0.301 \]

### Step 9: Calculate Moles of \( NH_3 \)
The number of moles of \( NH_3 \) formed is:
\[ \text{Moles of } NH_3 = 2x = 2 \times 0.301 = 0.602 \]

### Step 10: Calculate Total Moles at Equilibrium
The total moles at equilibrium are:
\[ \text{Total moles} = 4 - 2x = 4 - 2 \times 0.301 = 3.398 \]

### Step 11: Calculate Percentage of \( NH_3 \)
The percentage of \( NH_3 \) formed is:
\[ \% NH_3 = \left(\frac{0.602}{3.398}\right) \times 100 \approx 17.716\% \]

### Final Answer
The percentage of \( NH_3 \) formed at this temperature is approximately \( 17.716\% \).

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