One mole of `N_(2)` (g) is mixed with `2` moles of `H_(2)(g)` in a `4` litre vessel If `50%` of `N_(2)`(g) is converted to `NH_(3)`(g) by the following reaction :
`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)`
What will the value of `K_(c)` for the following equilibrium ?
`NH_(3)(g)hArr(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)`

Options

A. `256`
B. `16`
C. `(1)/(16)`
D. None of these

Answer

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

### Step 1: Determine Initial Concentrations
We start with 1 mole of \( N_2 \) and 2 moles of \( H_2 \) in a 4-liter vessel.

- Initial concentration of \( N_2 \):
\[ [N_2] = \frac{1 \text{ mole}}{4 \text{ L}} = 0.25 \text{ M} \]

- Initial concentration of \( H_2 \):
\[ [H_2] = \frac{2 \text{ moles}}{4 \text{ L}} = 0.5 \text{ M} \]

- Initial concentration of \( NH_3 \):
\[ [NH_3] = 0 \text{ M} \]

### Step 2: Calculate Changes at Equilibrium
Given that 50% of \( N_2 \) is converted to \( NH_3 \), we calculate the changes in concentration.

- Change in concentration of \( N_2 \):
\[ \text{Change} = 0.5 \times 0.25 = 0.125 \text{ M} \]
So, at equilibrium:
\[ [N_2] = 0.25 - 0.125 = 0.125 \text{ M} \]

- Change in concentration of \( H_2 \):
According to the stoichiometry of the reaction \( N_2 + 3H_2 \rightarrow 2NH_3 \), for every mole of \( N_2 \), 3 moles of \( H_2 \) are consumed. Therefore:
\[ \text{Change in } H_2 = 3 \times 0.125 = 0.375 \text{ M} \]
So, at equilibrium:
\[ [H_2] = 0.5 - 0.375 = 0.125 \text{ M} \]

- Change in concentration of \( NH_3 \):
For every mole of \( N_2 \) that reacts, 2 moles of \( NH_3 \) are produced. Therefore:
\[ \text{Change in } NH_3 = 2 \times 0.125 = 0.25 \text{ M} \]
So, at equilibrium:
\[ [NH_3] = 0 + 0.25 = 0.25 \text{ M} \]

### Step 3: Write the Equilibrium Expression for the First Reaction
The equilibrium expression for the reaction \( N_2 + 3H_2 \rightleftharpoons 2NH_3 \) is:
\[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(0.25)^2}{(0.125)(0.125)^3} \]

### Step 4: Calculate \( K_c \)
Calculating the values:
\[ K_c = \frac{0.0625}{0.125 \times 0.001953125} = \frac{0.0625}{0.000244140625} \]
\[ K_c = 256.0 \]

### Step 5: Relate \( K_c \) of the First Reaction to the Second Reaction
The second reaction is:
\[ NH_3 \rightleftharpoons \frac{1}{2}N_2 + \frac{3}{2}H_2 \]
To find \( K_c \) for this reaction, we note that it is the reverse of the first reaction divided by 2.

The relationship is:
\[ K' = \frac{1}{K_c^{1/2}} \]
Thus:
\[ K' = \frac{1}{\sqrt{256}} = \frac{1}{16} \]

### Final Answer
The value of \( K_c \) for the equilibrium \( NH_3 \rightleftharpoons \frac{1}{2}N_2 + \frac{3}{2}H_2 \) is:
\[ \boxed{\frac{1}{16}} \]