28 g of `N_(2)` and 6 g of `H_(2)` were kept at `400^(@)C` in 1 litre vessel, the equilibrium mixture contained `27.54` g of `NH_(3)`. The approximate value of `K_(c)` for the above reaction can be `("in mole"^(-2) "litre"^(2))`

Options

A. 75
B. 50
C. 25
D. 100

Explanation

To find the equilibrium constant \( K_c \) for the reaction between nitrogen and hydrogen to form ammonia, we will follow these steps:

### Step 1: Calculate the moles of reactants and products

1. **Calculate moles of \( N_2 \)**:
\[ \text{Molar mass of } N_2 = 14 \times 2 = 28 \text{ g/mol} \]
\[ \text{Moles of } N_2 = \frac{28 \text{ g}}{28 \text{ g/mol}} = 1 \text{ mole} \]

2. **Calculate moles of \( H_2 \)**:
\[ \text{Molar mass of } H_2 = 1 \times 2 = 2 \text{ g/mol} \]
\[ \text{Moles of } H_2 = \frac{6 \text{ g}}{2 \text{ g/mol}} = 3 \text{ moles} \]

3. **Calculate moles of \( NH_3 \)** at equilibrium:
\[ \text{Molar mass of } NH_3 = 14 + 1 \times 3 = 17 \text{ g/mol} \]
\[ \text{Moles of } NH_3 = \frac{27.54 \text{ g}}{17 \text{ g/mol}} \approx 1.62 \text{ moles} \]

### Step 2: Set up the reaction and the ICE table

The balanced chemical equation is:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]

| Species | Initial (mol) | Change (mol) | Equilibrium (mol) |
|-----------|----------------|---------------|--------------------|
| \( N_2 \) | 1 | -x | \( 1 - x \) |
| \( H_2 \) | 3 | -3x | \( 3 - 3x \) |
| \( NH_3 \)| 0 | +2x | \( 2x \) |

### Step 3: Determine the value of \( x \)

From the equilibrium moles of \( NH_3 \):
\[ 2x = 1.62 \implies x = 0.81 \]

Now substitute \( x \) back into the equilibrium expressions:
- Moles of \( N_2 \) at equilibrium:
\[ 1 - x = 1 - 0.81 = 0.19 \text{ moles} \]
- Moles of \( H_2 \) at equilibrium:
\[ 3 - 3x = 3 - 3(0.81) = 3 - 2.43 = 0.57 \text{ moles} \]

### Step 4: Calculate concentrations

Since the volume of the vessel is 1 L, the concentrations (in mol/L) are equal to the number of moles:
- Concentration of \( N_2 \):
\[ [N_2] = 0.19 \text{ M} \]
- Concentration of \( H_2 \):
\[ [H_2] = 0.57 \text{ M} \]
- Concentration of \( NH_3 \):
\[ [NH_3] = 1.62 \text{ M} \]

### Step 5: Write the expression for \( K_c \)

The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]

### Step 6: Substitute the concentrations into the \( K_c \) expression

Substituting in the values:
\[ K_c = \frac{(1.62)^2}{(0.19)(0.57)^3} \]

### Step 7: Calculate \( K_c \)

Calculating the values:
\[ K_c = \frac{2.6244}{0.19 \times 0.185193} \approx \frac{2.6244}{0.0351} \approx 74.7 \]

Thus, the approximate value of \( K_c \) is:
\[ K_c \approx 75 \, \text{(in mole}^{-2} \text{ litre}^{2}\text{)} \]