For the reaction `N_2O_4 hArr 2NO_2` , if degrees of dissociation of `N_2O_4` are 25%,50%,80% and 100% , the gradation of observed vapour densities is

Options

A. `d_1 gt d_2 gt d_3 gt d_4`
B. `d_4 gt d_3 gt d_2 d_1`
C. `d_1 =d_2 =d_3 =d_4`
D. none of the above

Answer

To solve the problem regarding the reaction \( N_2O_4 \rightleftharpoons 2NO_2 \) and the degrees of dissociation (25%, 50%, 80%, and 100%), we need to analyze how the degree of dissociation affects the observed vapor density.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction given is:
\[ N_2O_4 \rightleftharpoons 2NO_2 \]
This means that one mole of \( N_2O_4 \) dissociates to form two moles of \( NO_2 \).

2. **Degree of Dissociation**:
The degree of dissociation (\( \alpha \)) is defined as the fraction of the original substance that has dissociated. For example:
- \( \alpha = 0.25 \) (25% dissociation)
- \( \alpha = 0.50 \) (50% dissociation)
- \( \alpha = 0.80 \) (80% dissociation)
- \( \alpha = 1.00 \) (100% dissociation)

3. **Calculating Moles at Equilibrium**:
Let's assume we start with 1 mole of \( N_2O_4 \):
- At 25% dissociation (\( \alpha = 0.25 \)):
- Moles of \( N_2O_4 \) left = \( 1 - 0.25 = 0.75 \)
- Moles of \( NO_2 \) formed = \( 2 \times 0.25 = 0.50 \)
- Total moles = \( 0.75 + 0.50 = 1.25 \)

- At 50% dissociation (\( \alpha = 0.50 \)):
- Moles of \( N_2O_4 \) left = \( 1 - 0.50 = 0.50 \)
- Moles of \( NO_2 \) formed = \( 2 \times 0.50 = 1.00 \)
- Total moles = \( 0.50 + 1.00 = 1.50 \)

- At 80% dissociation (\( \alpha = 0.80 \)):
- Moles of \( N_2O_4 \) left = \( 1 - 0.80 = 0.20 \)
- Moles of \( NO_2 \) formed = \( 2 \times 0.80 = 1.60 \)
- Total moles = \( 0.20 + 1.60 = 1.80 \)

- At 100% dissociation (\( \alpha = 1.00 \)):
- Moles of \( N_2O_4 \) left = \( 0 \)
- Moles of \( NO_2 \) formed = \( 2 \times 1.00 = 2.00 \)
- Total moles = \( 2.00 \)

4. **Vapor Density Calculation**:
The vapor density (\( D \)) is related to the number of moles of gas present. The formula for vapor density is:
\[ D \propto \frac{M}{n} \]
where \( M \) is the molar mass and \( n \) is the number of moles.

Since the molar mass of \( N_2O_4 \) is greater than that of \( NO_2 \), as the degree of dissociation increases, the total number of moles increases, leading to a decrease in vapor density.

5. **Order of Vapor Densities**:
- For \( \alpha = 0.25 \): Total moles = 1.25, \( D_1 \)
- For \( \alpha = 0.50 \): Total moles = 1.50, \( D_2 \)
- For \( \alpha = 0.80 \): Total moles = 1.80, \( D_3 \)
- For \( \alpha = 1.00 \): Total moles = 2.00, \( D_4 \)

Therefore, the observed vapor densities will be in the order:
\[ D_1 > D_2 > D_3 > D_4 \]

### Conclusion:
The correct gradation of observed vapor densities is:
\[ D_1 > D_2 > D_3 > D_4 \]