2 mol of `N_(2)` is mixed with 6 mol of `H_(2)` in a closed vessel of 1L capacity. If `50%` of `N_(2)` is converted into `NH_(3)` at equilibrium, the value of `K_(C )` for the reaction
`N_(2)(g0+3H_(2)(g)hArr 2NH_(3)(g)` is

Options

A. `4//27`
B. `27//4`
C. `1//27`
D. `27`

Explanation

To solve the problem step by step, we will analyze the reaction and the equilibrium conditions given in the question.

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

### Step 2: Determine the initial moles of reactants
From the problem, we have:
- Initial moles of \( N_2 = 2 \) mol
- Initial moles of \( H_2 = 6 \) mol
- Initial moles of \( NH_3 = 0 \) mol (since it is not mentioned)

### Step 3: Define the change in moles at equilibrium
Let \( x \) be the amount of \( N_2 \) that reacts. According to the problem, 50% of \( N_2 \) is converted into \( NH_3 \). Therefore:
- \( x = 50\% \text{ of } 2 \text{ mol} = 1 \text{ mol} \)

### Step 4: Calculate the moles at equilibrium
Using \( x = 1 \):
- Moles of \( N_2 \) left = \( 2 - x = 2 - 1 = 1 \) mol
- Moles of \( H_2 \) consumed = \( 3x = 3 \times 1 = 3 \) mol
- Moles of \( H_2 \) left = \( 6 - 3 = 3 \) mol
- Moles of \( NH_3 \) formed = \( 2x = 2 \times 1 = 2 \) mol

### Step 5: Write the equilibrium concentrations
Since the volume of the vessel is 1 L, the concentrations (in mol/L) are:
- Concentration of \( N_2 = \frac{1 \text{ mol}}{1 \text{ L}} = 1 \text{ M} \)
- Concentration of \( H_2 = \frac{3 \text{ mol}}{1 \text{ L}} = 3 \text{ M} \)
- Concentration of \( NH_3 = \frac{2 \text{ mol}}{1 \text{ L}} = 2 \text{ M} \)

### Step 6: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(2)^2}{(1)(3)^3} \]

### Step 7: Calculate \( K_c \)
Calculating the values:
\[ K_c = \frac{4}{1 \times 27} = \frac{4}{27} \]

### Final Answer
Thus, the value of \( K_c \) is:
\[ K_c = \frac{4}{27} \]