One mole of pure ammonia was injected into a one litre flask at a certain temperature. The equilibrium mixture was then analysed and found to contain 0.30 mole of `H_(2)` . Calculate (i) the concentration of of `N_(2)` and (ii) the concentration of `NH_(3)` at equilibrium.

Explanation

To solve the problem, we first need to write the balanced chemical equation for the reaction involving ammonia (NH₃), nitrogen (N₂), and hydrogen (H₂). The reaction can be represented as follows:

\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

Given that 1 mole of pure ammonia (NH₃) was injected into a 1-liter flask, we can denote the initial concentration of NH₃ as:

- Initial moles of NH₃ = 1 mole
- Initial concentration of NH₃ = 1 M (since volume = 1 L)

At equilibrium, we are given that the concentration of H₂ is 0.30 moles.

### Step 1: Determine the change in moles of H₂

From the balanced equation, we see that 3 moles of H₂ are produced for every 2 moles of NH₃ that are consumed. Therefore, if we denote the change in moles of NH₃ as \( -x \), the change in moles of H₂ will be \( +\frac{3}{2}x \).

### Step 2: Set up the equilibrium expression

Let’s denote:
- The initial moles of NH₃ = 1 mole
- The change in moles of NH₃ = \( -x \)
- The moles of H₂ at equilibrium = 0.30 moles

From the balanced equation, we can express the moles of NH₃ at equilibrium:

\[ \text{Moles of } NH_3 = 1 - x \]

And for H₂, we can express it as:

\[ \text{Moles of } H_2 = \frac{3}{2}x \]

Setting the moles of H₂ equal to the given value:

\[ \frac{3}{2}x = 0.30 \]

### Step 3: Solve for x

Now we can solve for \( x \):

\[ x = \frac{0.30 \times 2}{3} = 0.20 \]

### Step 4: Calculate the concentration of NH₃ at equilibrium

Now substituting \( x \) back into the expression for moles of NH₃ at equilibrium:

\[ \text{Moles of } NH_3 = 1 - 0.20 = 0.80 \text{ moles} \]

Since the volume of the flask is 1 L, the concentration of NH₃ at equilibrium is:

\[ \text{Concentration of } NH_3 = \frac{0.80 \text{ moles}}{1 \text{ L}} = 0.80 \text{ M} \]

### Step 5: Calculate the concentration of N₂ at equilibrium

From the balanced equation, we see that for every 2 moles of NH₃ produced, 1 mole of N₂ is consumed. Thus, the change in moles of N₂ will be:

\[ \text{Moles of } N_2 = \frac{1}{2} \times (1 - (1 - x)) = \frac{1}{2} \times 0.20 = 0.10 \text{ moles} \]

Since the volume of the flask is 1 L, the concentration of N₂ at equilibrium is:

\[ \text{Concentration of } N_2 = \frac{0.10 \text{ moles}}{1 \text{ L}} = 0.10 \text{ M} \]

### Final Answers

1. Concentration of \( N_2 \) at equilibrium = 0.10 M
2. Concentration of \( NH_3 \) at equilibrium = 0.80 M