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 will follow the reaction and the changes in moles of the reactants and products, and then calculate the equilibrium constant \( K_c \).

### Step 1: Write the balanced chemical equation.
The balanced equation for the formation of ammonia from nitrogen and hydrogen is:
\[ \frac{1}{2} N_2(g) + \frac{3}{2} H_2(g) \rightleftharpoons NH_3(g) \]

### Step 2: Determine the initial moles of reactants.
We have:
- Nitrogen (\( N_2 \)): 1 mole
- Hydrogen (\( H_2 \)): 3 moles

### Step 3: Calculate the amount of nitrogen converted to ammonia.
Given that 20% of nitrogen is converted into ammonia:
\[ \text{Moles of } N_2 \text{ converted} = 20\% \text{ of } 1 \text{ mole} = 0.2 \text{ moles} \]

### Step 4: Determine the change in moles of reactants and products.
From the balanced equation:
- For every 0.2 moles of \( N_2 \) that reacts, \( NH_3 \) produced will be:
\[ \text{Moles of } NH_3 = 0.2 \text{ moles} \times 2 = 0.4 \text{ moles} \]
- The change in moles of \( N_2 \) and \( H_2 \):
\[ \text{Moles of } H_2 \text{ consumed} = 0.2 \text{ moles} \times 3 = 0.6 \text{ moles} \]

### Step 5: Calculate 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 of each species.
The volume of the vessel is 3 liters, so the concentrations are:
- Concentration of \( N_2 \):
\[ [C_{N_2}] = \frac{0.8 \text{ moles}}{3 \text{ L}} = \frac{0.8}{3} \text{ M} \]
- Concentration of \( H_2 \):
\[ [C_{H_2}] = \frac{2.4 \text{ moles}}{3 \text{ L}} = \frac{2.4}{3} \text{ M} \]
- Concentration of \( NH_3 \):
\[ [C_{NH_3}] = \frac{0.4 \text{ moles}}{3 \text{ L}} = \frac{0.4}{3} \text{ M} \]

### Step 7: Write the expression for \( K_c \).
The equilibrium constant \( K_c \) for the reaction 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 values:
\[ K_c = \frac{\left(\frac{0.4}{3}\right)}{\left(\frac{0.8}{3}\right)^{1/2} \left(\frac{2.4}{3}\right)^{3/2}} \]

### Step 9: Simplify the expression.
Calculating the denominator:
\[ \left(\frac{0.8}{3}\right)^{1/2} = \frac{\sqrt{0.8}}{\sqrt{3}} \quad \text{and} \quad \left(\frac{2.4}{3}\right)^{3/2} = \frac{(2.4)^{3/2}}{(3)^{3/2}} \]
Now, substituting these back into the \( K_c \) expression and simplifying will yield the final value.

### Step 10: Calculate the final value of \( K_c \).
After performing the calculations, we find:
\[ K_c \approx 0.36 \]