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

Options

A. `4//27`
B. `27//4`
C. `2//27`
D. `20`

Explanation

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

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

### Step 2: Determine initial moles
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

### Step 3: Calculate moles at equilibrium
Given that 50% of \( N_2 \) is converted into \( NH_3 \):
- Moles of \( N_2 \) converted = \( 50\% \times 2 \) mol = \( 1 \) mol
- Remaining moles of \( N_2 \) = \( 2 - 1 = 1 \) mol

For \( H_2 \):
- According to the stoichiometry of the reaction, 3 moles of \( H_2 \) are required for every mole of \( N_2 \) reacted.
- Moles of \( H_2 \) consumed = \( 3 \times 1 = 3 \) mol
- Remaining moles of \( H_2 \) = \( 6 - 3 = 3 \) mol

For \( NH_3 \):
- Moles of \( NH_3 \) produced = \( 2 \times 1 = 2 \) mol

### Step 4: Summarize moles at equilibrium
At equilibrium, we have:
- Moles of \( N_2 = 1 \) mol
- Moles of \( H_2 = 3 \) mol
- Moles of \( NH_3 = 2 \) mol

### Step 5: Calculate concentrations
Since the volume of the vessel is 1 liter, the concentrations (in mol/L) will be the same as the number of moles:
- Concentration of \( N_2 = 1 \) M
- Concentration of \( H_2 = 3 \) M
- Concentration of \( NH_3 = 2 \) M

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

### Step 7: Substitute the equilibrium concentrations into the \( K_c \) expression
Substituting the concentrations we found:
\[ K_c = \frac{(2)^2}{(1)(3)^3} = \frac{4}{27} \]

### Step 8: Final answer
Thus, the value of \( K_c \) is:
\[ K_c = \frac{4}{27} \]