When a mixture of `N_2 and H_2` in the volume ratio of 1:5 is allowed to react at 100 k and `10^3` atm pressure , 0.426 mole fraction of `NH_3` is formed at equilibrium . The `K_p` for the reaction is

Options

A. `2.8 xx 10^(-4)`
B. `6.6 xx 10^(-4)`
C. `3.0xx10(-4)`
D. `6 xx 10^(-4)`

Answer

To solve for the equilibrium constant \( K_p \) for the reaction of nitrogen and hydrogen forming ammonia, we will follow these steps:

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

### Step 2: Determine the initial moles based on the volume ratio
Given the volume ratio of \( N_2 \) to \( H_2 \) is 1:5, we can assume:
- Let the moles of \( N_2 \) be \( 1 \) mole.
- Let the moles of \( H_2 \) be \( 5 \) moles.

### Step 3: Set up the change in moles at equilibrium
Let \( x \) be the amount of \( N_2 \) that reacts. The changes in moles can be expressed as:
- Moles of \( N_2 \) at equilibrium: \( 1 - x \)
- Moles of \( H_2 \) at equilibrium: \( 5 - 3x \)
- Moles of \( NH_3 \) at equilibrium: \( 2x \)

### Step 4: Calculate the total moles at equilibrium
The total moles at equilibrium can be calculated as:
\[ \text{Total moles} = (1 - x) + (5 - 3x) + 2x = 6 - 2x \]

### Step 5: Write the expression for mole fractions
The mole fraction of \( NH_3 \) is given as \( 0.426 \). Thus,
\[ \text{Mole fraction of } NH_3 = \frac{2x}{6 - 2x} = 0.426 \]

### Step 6: Solve for \( x \)
Cross-multiplying gives:
\[ 2x = 0.426(6 - 2x) \]
Expanding and rearranging:
\[ 2x = 2.556 - 0.852x \]
\[ 2.852x = 2.556 \]
\[ x = \frac{2.556}{2.852} \approx 0.896 \]

### Step 7: Calculate moles at equilibrium
Substituting \( x \) back into the expressions for moles at equilibrium:
- Moles of \( N_2 \): \( 1 - 0.896 = 0.104 \)
- Moles of \( H_2 \): \( 5 - 3(0.896) = 2.312 \)
- Moles of \( NH_3 \): \( 2(0.896) = 1.792 \)

### Step 8: Calculate the total moles at equilibrium
Total moles at equilibrium:
\[ \text{Total moles} = 0.104 + 2.312 + 1.792 = 4.208 \]

### Step 9: Calculate the mole fractions of each component
- Mole fraction of \( N_2 \):
\[ \frac{0.104}{4.208} \approx 0.0247 \]
- Mole fraction of \( H_2 \):
\[ \frac{2.312}{4.208} \approx 0.549 \]
- Mole fraction of \( NH_3 \):
\[ \frac{1.792}{4.208} \approx 0.425 \]

### Step 10: Calculate partial pressures
Given the total pressure \( P = 10^3 \) atm:
- Partial pressure of \( N_2 \):
\[ P_{N_2} = 0.0247 \times 1000 \approx 24.7 \text{ atm} \]
- Partial pressure of \( H_2 \):
\[ P_{H_2} = 0.549 \times 1000 \approx 549 \text{ atm} \]
- Partial pressure of \( NH_3 \):
\[ P_{NH_3} = 0.425 \times 1000 \approx 425 \text{ atm} \]

### Step 11: 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 values:
\[ K_p = \frac{(425)^2}{(24.7)(549)^3} \]

### Step 12: Calculate \( K_p \)
Calculating the values:
\[ K_p = \frac{180625}{24.7 \times 165813849} \]
\[ K_p \approx 4.419 \times 10^{-5} \]

### Final Answer
Thus, the value of \( K_p \) is approximately:
\[ K_p \approx 4.419 \times 10^{-5} \]