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 step by step, we will follow the process of determining the equilibrium concentrations and then calculating the equilibrium constant \( K_c \) for the given reaction.

### Step 1: Initial Concentrations
We start with the initial amounts of \( N_2 \) and \( H_2 \):
- 1 mole of \( N_2 \)
- 2 moles of \( H_2 \)

Since the volume of the vessel is 4 liters, we can calculate the initial concentrations:
\[ \text{Concentration of } N_2 = \frac{1 \text{ mole}}{4 \text{ L}} = 0.25 \text{ M} \]
\[ \text{Concentration of } H_2 = \frac{2 \text{ moles}}{4 \text{ L}} = 0.5 \text{ M} \]
\[ \text{Concentration of } NH_3 = 0 \text{ M (initially)} \]

### Step 2: Change in Concentrations
According to the problem, 50% of \( N_2 \) is converted to \( NH_3 \). Therefore, the amount of \( N_2 \) that reacts is:
\[ \text{Amount of } N_2 \text{ reacted} = 0.5 \times 1 \text{ mole} = 0.5 \text{ moles} \]

Using the stoichiometry of the reaction:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]
For every 1 mole of \( N_2 \) that reacts, 3 moles of \( H_2 \) are consumed and 2 moles of \( NH_3 \) are produced. Therefore, the changes in concentrations will be:
- \( N_2 \): \( 0.25 - 0.125 = 0.125 \) M (since \( 0.5 \text{ moles} \) of \( N_2 \) corresponds to \( 0.125 \text{ M} \))
- \( H_2 \): \( 0.5 - 0.375 = 0.125 \) M (since \( 0.5 \text{ moles} \) of \( N_2 \) reacts with \( 1.5 \text{ moles} \) of \( H_2 \))
- \( NH_3 \): \( 0 + 0.25 = 0.25 \) M (since \( 0.5 \text{ moles} \) of \( N_2 \) produces \( 0.25 \text{ moles} \) of \( NH_3 \))

### Step 3: Equilibrium Concentrations
At equilibrium, we have:
\[ \text{Concentration of } N_2 = 0.125 \text{ M} \]
\[ \text{Concentration of } H_2 = 0.125 \text{ M} \]
\[ \text{Concentration of } NH_3 = 0.25 \text{ M} \]

### Step 4: Write the Expression for \( K_c \)
For the equilibrium reaction:
\[ NH_3 \rightleftharpoons \frac{1}{2} N_2 + \frac{3}{2} H_2 \]
The expression for \( K_c \) is:
\[ K_c = \frac{[N_2]^{1/2} [H_2]^{3/2}}{[NH_3]^2} \]

### Step 5: Substitute the Equilibrium Concentrations
Substituting the equilibrium concentrations into the \( K_c \) expression:
\[ K_c = \frac{(0.125)^{1/2} \cdot (0.125)^{3/2}}{(0.25)^2} \]

Calculating each part:
- \( (0.125)^{1/2} = 0.3536 \)
- \( (0.125)^{3/2} = 0.125 \cdot 0.3536 = 0.0442 \)
- \( (0.25)^2 = 0.0625 \)

Now substituting these values:
\[ K_c = \frac{0.3536 \cdot 0.0442}{0.0625} = \frac{0.0156}{0.0625} = 0.2496 \]

### Step 6: Final Calculation
To find \( K_c \) for the reverse reaction:
\[ K_c' = \frac{1}{K_c^2} = \frac{1}{(0.2496)^2} = 16 \]

Thus, the final value of \( K_c \) for the given equilibrium reaction is:
\[ K_c = \frac{1}{16} \]