Find the percentage dissociation of ammonia into `N_(2) "and' H_(2)` if the dissociation is carried out at constant pressure and the volume at equilibrium is `20%` greater than initial volume. (Initially. Equal moles of `NH_(3) "and" N_(2)` are present with no hydrogen)

Explanation

To find the percentage dissociation of ammonia into nitrogen and hydrogen, we can follow these steps:

### Step 1: Write the dissociation reaction
The dissociation of ammonia can be represented as:
\[ 2 \text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3 \text{H}_2(g) \]

### Step 2: Set up initial conditions
Let’s assume we start with:
- 2 moles of \( \text{NH}_3 \)
- 2 moles of \( \text{N}_2 \)
- 0 moles of \( \text{H}_2 \)

### Step 3: Define the degree of dissociation
Let \( \alpha \) be the degree of dissociation of ammonia. At equilibrium, the moles will change as follows:
- Moles of \( \text{NH}_3 \) at equilibrium: \( 2 - 2\alpha \)
- Moles of \( \text{N}_2 \) at equilibrium: \( 2 + \alpha \)
- Moles of \( \text{H}_2 \) at equilibrium: \( 3\alpha \)

### Step 4: Calculate total moles at equilibrium
The total number of moles at equilibrium (\( N_t \)) can be calculated as:
\[ N_t = (2 - 2\alpha) + (2 + \alpha) + (3\alpha) = 4 + 2\alpha \]

### Step 5: Relate volume change to moles
It is given that the volume at equilibrium is 20% greater than the initial volume. The initial volume corresponds to the initial moles, which is \( 4 \) moles (2 moles of \( \text{NH}_3 \) and 2 moles of \( \text{N}_2 \)). Therefore, the volume at equilibrium is:
\[ \text{Volume at equilibrium} = 4 + 0.2 \times 4 = 4 + 0.8 = 4.8 \text{ moles} \]

### Step 6: Set up the equation
Since the total number of moles at equilibrium is also equal to the volume at equilibrium:
\[ 4 + 2\alpha = 4.8 \]

### Step 7: Solve for \( \alpha \)
Rearranging the equation gives:
\[ 2\alpha = 4.8 - 4 = 0.8 \]
\[ \alpha = \frac{0.8}{2} = 0.4 \]

### Step 8: Calculate percentage dissociation
To find the percentage dissociation, we multiply \( \alpha \) by 100:
\[ \text{Percentage dissociation} = \alpha \times 100 = 0.4 \times 100 = 40\% \]

### Final Answer
The percentage dissociation of ammonia into nitrogen and hydrogen is **40%**.

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