Two moles of `NH_(3)` when put into a previously evacuated vessel ( one litre), partially dossociate into `N_(2) and H_(2).` If at equlib rium one mole of `NH_(3)` is present, the equlibrium constant is

Options

A. `3//4 mol^(2) litre^(-2)`
B. `27//64 mol^(2) litre^(-2)`
C. `27//32 mol^(2) litre^(-2)`
D. `27//16 mol^(2) litre^(-2)`

Explanation

To solve the problem, we need to analyze the dissociation of ammonia (NH₃) into nitrogen (N₂) and hydrogen (H₂) and calculate the equilibrium constant (K) based on the given conditions.

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
The dissociation of ammonia can be represented as:
\[ 2 \text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3 \text{H}_2(g) \]

2. **Set up the initial concentrations:**
Initially, we have 2 moles of NH₃ in a 1-liter vessel, so:
- \([\text{NH}_3] = 2 \, \text{mol/L}\)
- \([\text{N}_2] = 0 \, \text{mol/L}\)
- \([\text{H}_2] = 0 \, \text{mol/L}\)

3. **Define the change in concentration:**
Let \(x\) be the amount of NH₃ that dissociates at equilibrium. According to the stoichiometry of the reaction:
- Change in NH₃: \( -2x \)
- Change in N₂: \( +x \)
- Change in H₂: \( +3x \)

4. **Write the equilibrium concentrations:**
At equilibrium, the concentrations will be:
- \([\text{NH}_3] = 2 - 2x\)
- \([\text{N}_2] = x\)
- \([\text{H}_2] = 3x\)

5. **Use the information given:**
We know that at equilibrium, 1 mole of NH₃ is present:
\[ 2 - 2x = 1 \]
Solving for \(x\):
\[ 2x = 2 - 1 \\ 2x = 1 \\ x = 0.5 \]

6. **Calculate the equilibrium concentrations:**
Now substituting \(x = 0.5\):
- \([\text{NH}_3] = 2 - 2(0.5) = 1 \, \text{mol/L}\)
- \([\text{N}_2] = 0.5 \, \text{mol/L}\)
- \([\text{H}_2] = 3(0.5) = 1.5 \, \text{mol/L}\)

7. **Write the expression for the equilibrium constant (K):**
The equilibrium constant \(K\) for the reaction is given by:
\[ K = \frac{[\text{N}_2][\text{H}_2]^3}{[\text{NH}_3]^2} \]

8. **Substitute the equilibrium concentrations into the expression:**
\[ K = \frac{(0.5)(1.5)^3}{(1)^2} \]
Calculate \( (1.5)^3 = 3.375 \):
\[ K = \frac{(0.5)(3.375)}{1} = 1.6875 \]

9. **Convert to fraction form if needed:**
\[ K = \frac{27}{16} \]

### Final Answer:
Thus, the equilibrium constant \(K\) is:
\[ K = \frac{27}{16} \]