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. a) `2.8 xx 10^(-4)`
B. b) `4.44 xx 10^(-5)`
C. c) `3.0xx10^(-4)`
D. d) `6 xx 10^(-4)`

Answer

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

\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]

### Step 1: Determine Initial Moles
Given the volume ratio of \( N_2 \) to \( H_2 \) is 1:5, we can assume the initial moles of \( N_2 \) is 1 mole and \( H_2 \) is 5 moles.

### Step 2: Define Change in Moles
Let \( X \) be the number of moles of \( N_2 \) that reacts. According to the stoichiometry of the reaction:
- \( N_2 \) decreases by \( X \)
- \( H_2 \) decreases by \( 3X \)
- \( NH_3 \) increases by \( 2X \)

### Step 3: Calculate Moles at Equilibrium
At equilibrium, the moles of each species will be:
- Moles of \( N_2 = 1 - X \)
- Moles of \( H_2 = 5 - 3X \)
- Moles of \( NH_3 = 2X \)

### Step 4: 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: Mole Fraction of \( NH_3 \)
The mole fraction of \( NH_3 \) is given as 0.426. Thus, we can write:
\[ \text{Mole fraction of } NH_3 = \frac{2X}{6 - 2X} = 0.426 \]

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

### Step 7: Calculate Moles at Equilibrium
Substituting \( X \) back into the moles:
- Moles of \( N_2 = 1 - 0.896 = 0.104 \)
- Moles of \( H_2 = 5 - 3(0.896) = 5 - 2.688 = 2.312 \)
- Moles of \( NH_3 = 2(0.896) = 1.792 \)

### Step 8: Calculate Mole Fractions
Now we can calculate the mole fractions:
- Mole fraction of \( N_2 = \frac{0.104}{6 - 2(0.896)} = \frac{0.104}{2.208} \approx 0.047 \)
- Mole fraction of \( H_2 = \frac{2.312}{2.208} \approx 0.104 \)
- Mole fraction of \( NH_3 = \frac{1.792}{2.208} \approx 0.406 \)

### Step 9: Calculate Partial Pressures
Using the total pressure of \( 10^3 \) atm:
- \( P_{N_2} = 0.047 \times 1000 \approx 47 \) atm
- \( P_{H_2} = 0.104 \times 1000 \approx 104 \) atm
- \( P_{NH_3} = 0.406 \times 1000 \approx 406 \) atm

### Step 10: Calculate \( K_p \)
Using the formula for \( K_p \):
\[ K_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \]
Substituting the values:
\[ K_p = \frac{(406)^2}{(47)(104)^3} \]
Calculating:
\[ K_p = \frac{164836}{47 \times 1124864} \approx 4.44 \times 10^{-5} \text{ atm}^{-2} \]

### Final Answer
Thus, the equilibrium constant \( K_p \) is:
\[ K_p \approx 4.44 \times 10^{-5} \text{ atm}^{-2} \]

---