In reaction `N_(2)O_(4)(g)rarr 2NO_(2)(g),` The observed molecular weight `"80 gmol"^(-1)` at 350 K. The percentage dissociation of `N_(2)O_(4)(g)` at 350 K is

Options

A. `10%`
B. `15%`
C. `20%`
D. `18%`

Answer

To find the percentage dissociation of \( N_2O_4 \) at 350 K, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction is:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

### Step 2: Calculate the molar mass of the products
The molar mass of \( NO_2 \) is calculated as follows:
- Molar mass of Nitrogen (N) = 14 g/mol
- Molar mass of Oxygen (O) = 16 g/mol
- Therefore, molar mass of \( NO_2 = 14 + (2 \times 16) = 14 + 32 = 46 \) g/mol

Since the reaction produces 2 moles of \( NO_2 \):
\[ \text{Molar mass of } 2NO_2 = 2 \times 46 = 92 \text{ g/mol} \]

### Step 3: Use the observed molecular weight
The observed molecular weight is given as 80 g/mol.

### Step 4: Calculate the van't Hoff factor (i)
The van't Hoff factor \( i \) can be calculated using the formula:
\[ i = \frac{\text{Calculated molar mass}}{\text{Observed molar mass}} = \frac{92}{80} = 1.15 \]

### Step 5: Relate the van't Hoff factor to dissociation
For the dissociation of \( N_2O_4 \):
- Initial moles of \( N_2O_4 \) = 1
- Change in moles = \( -\alpha \) for \( N_2O_4 \) and \( +2\alpha \) for \( NO_2 \)

At equilibrium:
- Moles of \( N_2O_4 = 1 - \alpha \)
- Moles of \( NO_2 = 2\alpha \)

Total moles at equilibrium:
\[ \text{Total moles} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

### Step 6: Relate i to alpha
The van't Hoff factor \( i \) is also related to the degree of dissociation \( \alpha \):
\[ i = 1 + \alpha \]
Substituting the value of \( i \):
\[ 1.15 = 1 + \alpha \]

### Step 7: Solve for alpha
\[ \alpha = 1.15 - 1 = 0.15 \]

### Step 8: Calculate the percentage dissociation
Percentage dissociation is given by:
\[ \text{Percentage dissociation} = \alpha \times 100 = 0.15 \times 100 = 15\% \]

### Final Answer
The percentage dissociation of \( N_2O_4 \) at 350 K is **15%**.

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