One mole of nitrogen is mixed with 3 mole of hydrogen in a closed 3 litre vessel. `20%` of nitrogen is converted into `NH_(3)`. Then `K_(C)` for the `(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)hArr NH_(3)` is

Options

A. 0.36
B. 0.46
C. 0.5
D. 0.2

Answer

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

### Step 1: Determine the initial moles of reactants
We start with:
- 1 mole of \( N_2 \)
- 3 moles of \( H_2 \)

### Step 2: Calculate the amount of \( N_2 \) that reacts
According to the problem, 20% of nitrogen is converted into \( NH_3 \). Therefore:
\[ \text{Moles of } N_2 \text{ reacted} = 20\% \text{ of } 1 \text{ mole} = 0.2 \text{ moles} \]

### Step 3: Calculate the moles of \( H_2 \) that react
From the stoichiometry of the reaction:
\[ \frac{1}{2} N_2 + \frac{3}{2} H_2 \rightleftharpoons NH_3 \]
For every 1 mole of \( N_2 \) that reacts, 3 moles of \( H_2 \) are required. Therefore, for 0.2 moles of \( N_2 \):
\[ \text{Moles of } H_2 \text{ reacted} = 3 \times 0.2 = 0.6 \text{ moles} \]

### Step 4: Calculate the moles of \( NH_3 \) formed
According to the stoichiometry, 2 moles of \( NH_3 \) are produced for every mole of \( N_2 \) that reacts. Therefore:
\[ \text{Moles of } NH_3 \text{ formed} = 2 \times 0.2 = 0.4 \text{ moles} \]

### Step 5: Calculate the equilibrium moles of each species
Now we can find the equilibrium moles of each species:
- Moles of \( N_2 \) at equilibrium:
\[ 1 - 0.2 = 0.8 \text{ moles} \]
- Moles of \( H_2 \) at equilibrium:
\[ 3 - 0.6 = 2.4 \text{ moles} \]
- Moles of \( NH_3 \) at equilibrium:
\[ 0 + 0.4 = 0.4 \text{ moles} \]

### Step 6: Calculate the concentrations
Since the volume of the vessel is 3 liters, we can calculate the concentrations:
- Concentration of \( N_2 \):
\[ \text{Concentration of } N_2 = \frac{0.8 \text{ moles}}{3 \text{ L}} = \frac{0.8}{3} \text{ M} \]
- Concentration of \( H_2 \):
\[ \text{Concentration of } H_2 = \frac{2.4 \text{ moles}}{3 \text{ L}} = \frac{2.4}{3} \text{ M} \]
- Concentration of \( NH_3 \):
\[ \text{Concentration of } NH_3 = \frac{0.4 \text{ moles}}{3 \text{ L}} = \frac{0.4}{3} \text{ M} \]

### Step 7: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) for the reaction:
\[ \frac{1}{2} N_2 + \frac{3}{2} H_2 \rightleftharpoons NH_3 \]
is given by:
\[ K_c = \frac{[NH_3]}{[N_2]^{1/2} [H_2]^{3/2}} \]

### Step 8: Substitute the concentrations into the \( K_c \) expression
Substituting the concentrations we found:
\[ K_c = \frac{\left(\frac{0.4}{3}\right)}{\left(\frac{0.8}{3}\right)^{1/2} \left(\frac{2.4}{3}\right)^{3/2}} \]

### Step 9: Simplify the expression
Calculating the values:
\[ K_c = \frac{\frac{0.4}{3}}{\left(\frac{0.8}{3}\right)^{1/2} \left(\frac{2.4}{3}\right)^{3/2}} = \frac{0.4}{3} \cdot \frac{(3)^{1/2} \cdot (3)^{3/2}}{(0.8)^{1/2} \cdot (2.4)^{3/2}} \]

### Step 10: Calculate the final value of \( K_c \)
After performing the calculations, we find:
\[ K_c \approx 0.129 \]

### Step 11: Adjust for the given reaction
Since the reaction we need \( K_c \) for is half of the original reaction, we take the square root:
\[ K_c' = (0.129)^{1/2} \approx 0.36 \]

### Final Answer
Thus, the value of \( K_c \) for the reaction \( \frac{1}{2} N_2 + \frac{3}{2} H_2 \rightleftharpoons NH_3 \) is approximately **0.36**.

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