For the reaction `N_2O_4 hArr 2NO_2(g)`, if percentage dissociation of `N_2O_4` are 20%, 45%, 65%, 80% then the sequence of observed vapour densities wil be :

Options

A. `d_(20)gtd_(45)gtd_(65)gtd_(80)`
B. `d_(80)gtd_(63)gtd_(45)gtd_(20)`
C. `d_(20)=d_(45)=d_(65)=d_(80)`
D. `d_(2)=d_(45)gt d_(65)=d_(80)`

Answer

To solve the problem regarding the reaction \( N_2O_4 \rightleftharpoons 2NO_2(g) \) and the sequence of observed vapor densities for different percentages of dissociation of \( N_2O_4 \), we can follow these steps:

### Step 1: Understand the Reaction and Vapor Density Concept
The reaction involves the dissociation of \( N_2O_4 \) into \( 2NO_2 \). The vapor density of a gas is related to the molar mass of the gas and the number of moles present. As \( N_2O_4 \) dissociates into \( NO_2 \), the total number of moles of gas increases, which affects the vapor density.

### Step 2: Calculate Molar Masses
- Molar mass of \( N_2O_4 \) = \( 2 \times 14 + 4 \times 16 = 28 + 64 = 92 \, g/mol \)
- Molar mass of \( NO_2 \) = \( 14 + 2 \times 16 = 14 + 32 = 46 \, g/mol \)

### Step 3: Determine the Effect of Dissociation on Vapor Density
When \( N_2O_4 \) dissociates:
- For 0% dissociation: All \( N_2O_4 \) remains, vapor density \( D = \frac{M}{V} = \frac{92}{1} = 92 \, g/L \).
- For 20% dissociation: 20% of \( N_2O_4 \) dissociates to form \( NO_2 \).
- Moles of \( N_2O_4 \) remaining = \( 0.8 \) moles
- Moles of \( NO_2 \) formed = \( 0.4 \) moles
- Total moles = \( 0.8 + 0.4 = 1.2 \)
- Average molar mass = \( \frac{(0.8 \times 92) + (0.4 \times 46)}{1.2} = \frac{73.6 + 18.4}{1.2} = \frac{92}{1.2} = 76.67 \, g/mol \)

- For 45% dissociation:
- Moles of \( N_2O_4 \) remaining = \( 0.55 \)
- Moles of \( NO_2 \) formed = \( 0.9 \)
- Total moles = \( 0.55 + 0.9 = 1.45 \)
- Average molar mass = \( \frac{(0.55 \times 92) + (0.9 \times 46)}{1.45} = \frac{50.6 + 41.4}{1.45} = \frac{92}{1.45} = 63.45 \, g/mol \)

- For 65% dissociation:
- Moles of \( N_2O_4 \) remaining = \( 0.35 \)
- Moles of \( NO_2 \) formed = \( 1.3 \)
- Total moles = \( 0.35 + 1.3 = 1.65 \)
- Average molar mass = \( \frac{(0.35 \times 92) + (1.3 \times 46)}{1.65} = \frac{32.2 + 59.8}{1.65} = \frac{92}{1.65} = 55.76 \, g/mol \)

- For 80% dissociation:
- Moles of \( N_2O_4 \) remaining = \( 0.2 \)
- Moles of \( NO_2 \) formed = \( 1.6 \)
- Total moles = \( 0.2 + 1.6 = 1.8 \)
- Average molar mass = \( \frac{(0.2 \times 92) + (1.6 \times 46)}{1.8} = \frac{18.4 + 73.6}{1.8} = \frac{92}{1.8} = 51.11 \, g/mol \)

### Step 4: Order the Vapor Densities
Now we can summarize the average molar masses (which correspond to vapor densities):
- 20% dissociation: \( 76.67 \, g/mol \)
- 45% dissociation: \( 63.45 \, g/mol \)
- 65% dissociation: \( 55.76 \, g/mol \)
- 80% dissociation: \( 51.11 \, g/mol \)

### Step 5: Sequence of Observed Vapor Densities
The sequence of observed vapor densities from highest to lowest is:
1. 20% dissociation: \( 76.67 \, g/mol \)
2. 45% dissociation: \( 63.45 \, g/mol \)
3. 65% dissociation: \( 55.76 \, g/mol \)
4. 80% dissociation: \( 51.11 \, g/mol \)

Thus, the final sequence of observed vapor densities is:
**76.67 > 63.45 > 55.76 > 51.11**