During thermal dissociation, the observed vapour density of `N_(2)O_(4)(g)` is 26.0. The extent of dissociation of `N_(2)O_(4)` is:

Options

A. `50%`
B. `87%`
C. `77%`
D. `23%`

Answer

To solve the problem of finding the extent of dissociation of \( N_2O_4 \) given the observed vapor density, we can follow these steps:

### Step 1: Calculate the Molar Mass of the Reaction Mixture
The vapor density (VD) is given as 26.0. The molar mass (M) of the gas can be calculated using the formula:
\[ M = 2 \times \text{Vapor Density} \]
Substituting the given value:
\[ M = 2 \times 26.0 = 52.0 \, \text{g/mol} \]

**Hint:** Remember that the molar mass can be derived from the vapor density using the factor of 2.

### Step 2: Write the Dissociation Reaction
The dissociation of \( N_2O_4 \) can be represented as:
\[ N_2O_4 (g) \rightleftharpoons 2 NO_2 (g) \]

### Step 3: Set Up the Initial and Equilibrium Conditions
Assume we start with 1 mole of \( N_2O_4 \). Let \( \alpha \) be the extent of dissociation. At equilibrium:
- Moles of \( N_2O_4 \) = \( 1 - \alpha \)
- Moles of \( NO_2 \) = \( 2\alpha \)

### Step 4: Calculate Total Moles at Equilibrium
The total moles at equilibrium will be:
\[ \text{Total moles} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

### Step 5: Calculate the Average Molar Mass of the Mixture
The average molar mass of the mixture can be expressed as:
\[ \text{Average Molar Mass} = \frac{(M_{N_2O_4} \times \text{moles of } N_2O_4) + (M_{NO_2} \times \text{moles of } NO_2)}{\text{Total moles}} \]
Substituting the values:
\[ \text{Average Molar Mass} = \frac{(92 \times (1 - \alpha)) + (46 \times 2\alpha)}{1 + \alpha} \]
This simplifies to:
\[ \text{Average Molar Mass} = \frac{92 - 92\alpha + 92\alpha}{1 + \alpha} = \frac{92}{1 + \alpha} \]

### Step 6: Set the Average Molar Mass Equal to the Calculated Molar Mass
We know the average molar mass is 52 g/mol, so we set up the equation:
\[ \frac{92}{1 + \alpha} = 52 \]

### Step 7: Solve for \( \alpha \)
Cross-multiplying gives:
\[ 92 = 52(1 + \alpha) \]
Expanding this:
\[ 92 = 52 + 52\alpha \]
Rearranging gives:
\[ 40 = 52\alpha \]
Thus,
\[ \alpha = \frac{40}{52} = \frac{10}{13} \approx 0.7692 \]

### Step 8: Convert to Percentage
To express the extent of dissociation as a percentage:
\[ \alpha \times 100 \approx 76.92\% \]

### Final Answer
The extent of dissociation of \( N_2O_4 \) is approximately **77%**.

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