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 adove reaction can be ( in `" mol"^(-2) L^(-2))`

Options

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

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Write the Balanced Chemical Equation
The reaction between nitrogen and hydrogen to form ammonia is given by:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

### Step 2: Calculate Initial Moles of Reactants
We need to calculate the number of moles of \( N_2 \) and \( H_2 \) using their given masses.

- Molar mass of \( N_2 = 28 \, g/mol \)
- Molar mass of \( H_2 = 2 \, g/mol \)

Calculating moles:
- Moles of \( N_2 = \frac{28 \, g}{28 \, g/mol} = 1 \, mol \)
- Moles of \( H_2 = \frac{6 \, g}{2 \, g/mol} = 3 \, mol \)

### Step 3: Set Up the Change in Moles at Equilibrium
Let \( x \) be the amount of \( N_2 \) that reacts. According to the stoichiometry of the balanced equation:
- \( N_2 \) decreases by \( x \)
- \( H_2 \) decreases by \( 3x \)
- \( NH_3 \) increases by \( 2x \)

At equilibrium:
- Moles of \( N_2 = 1 - x \)
- Moles of \( H_2 = 3 - 3x \)
- Moles of \( NH_3 = 2x \)

### Step 4: Use the Given Information to Find \( x \)
We know that the equilibrium mixture contains 27.54 g of \( NH_3 \). First, we calculate the moles of \( NH_3 \):
- Molar mass of \( NH_3 = 17 \, g/mol \)
- Moles of \( NH_3 = \frac{27.54 \, g}{17 \, g/mol} \approx 1.62 \, mol \)

Since \( 2x \) moles of \( NH_3 \) are formed:
\[ 2x = 1.62 \]
\[ x = 0.81 \]

### Step 5: Calculate Moles at Equilibrium
Now we can find the moles of each component at equilibrium:
- Moles of \( N_2 = 1 - 0.81 = 0.19 \)
- Moles of \( H_2 = 3 - 3(0.81) = 3 - 2.43 = 0.57 \)
- Moles of \( NH_3 = 2(0.81) = 1.62 \)

### Step 6: Calculate Concentrations at Equilibrium
Since the volume of the vessel is 1 L, the concentrations are equal to the number of moles:
- \([N_2] = 0.19 \, mol/L\)
- \([H_2] = 0.57 \, mol/L\)
- \([NH_3] = 1.62 \, mol/L\)

### Step 7: Calculate \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]
Substituting the concentrations:
\[ K_c = \frac{(1.62)^2}{(0.19)(0.57)^3} \]

Calculating:
- \( (1.62)^2 = 2.6244 \)
- \( (0.57)^3 = 0.185193 \)
- \( K_c = \frac{2.6244}{(0.19)(0.185193)} \)
- \( K_c = \frac{2.6244}{0.0358} \approx 73.2 \)

### Step 8: Approximate Value of \( K_c \)
The approximate value of \( K_c \) is around 75 \( mol^{-2} L^{-2} \).

### Final Answer
The approximate value of \( K_c \) is \( 75 \, mol^{-2} L^{-2} \).

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