Two moles of `NH_3` gas are introduced into a previously evacuated one litre vesel in which it partially dissociates at high temperature as `2NH_3(g) hArr N_2 (g) +3H_2(g)` .at equilibrium , one mole of `NH_3(g)` remain . The value of `K_c` is

Options

A. 3
B. `27//16`
C. `3//2`
D. `27//64`

Answer

To find the equilibrium constant \( K_c \) for the reaction

\[ 2 \text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3 \text{H}_2(g) \]

we can follow these steps:

### Step 1: Set up the initial conditions
We start with 2 moles of \( \text{NH}_3 \) in a 1-liter vessel. Therefore, the initial concentration of \( \text{NH}_3 \) is:

\[ [\text{NH}_3] = \frac{2 \text{ moles}}{1 \text{ L}} = 2 \text{ M} \]

### Step 2: Determine the change in concentration at equilibrium
At equilibrium, it is given that 1 mole of \( \text{NH}_3 \) remains. Therefore, the change in concentration of \( \text{NH}_3 \) is:

\[ \text{Change in } [\text{NH}_3] = 2 \text{ M} - 1 \text{ M} = 1 \text{ M} \]

Since the stoichiometry of the reaction shows that 2 moles of \( \text{NH}_3 \) produce 1 mole of \( \text{N}_2 \) and 3 moles of \( \text{H}_2 \), we can express the changes in concentration of the products as follows:

- For \( \text{N}_2 \): \( +\frac{1}{2} \text{ M} \) (since 2 moles of \( \text{NH}_3 \) produce 1 mole of \( \text{N}_2 \))
- For \( \text{H}_2 \): \( +\frac{3}{2} \text{ M} \) (since 2 moles of \( \text{NH}_3 \) produce 3 moles of \( \text{H}_2 \))

### Step 3: Calculate equilibrium concentrations
Now we can calculate the equilibrium concentrations:

- \( [\text{NH}_3] = 1 \text{ M} \)
- \( [\text{N}_2] = \frac{1}{2} \text{ M} = 0.5 \text{ M} \)
- \( [\text{H}_2] = \frac{3}{2} \text{ M} = 1.5 \text{ M} \)

### Step 4: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:

\[ K_c = \frac{[\text{N}_2][\text{H}_2]^3}{[\text{NH}_3]^2} \]

### Step 5: Substitute the equilibrium concentrations into the \( K_c \) expression
Substituting the equilibrium concentrations into the expression for \( K_c \):

\[ K_c = \frac{(0.5)(1.5)^3}{(1)^2} \]

Calculating \( (1.5)^3 \):

\[ (1.5)^3 = 3.375 \]

Now substituting this back into the equation:

\[ K_c = \frac{(0.5)(3.375)}{1} = 1.6875 \]

### Final Answer
Thus, the value of \( K_c \) is:

\[ K_c = 1.6875 \]

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