For the reaction , `N_2O_4(g) hArr2NO_2(g)` , if percentage dissociation of `N_2O` are 25% , 50% , 75% and 100, then the sequence of observed vapour densities will be

Options

A. `d_(25) gt d_(50) gt d_(75)gt d_(100)`
B. `d_(100) gt d_(75) gt d_(50)gt d_(25)`
C. `d_(25) = d_(50) = d_(75)= d_(100)`
D. `(d_(25) = d_(50)) gt (d_(75)= d_(100))`

Answer

To solve the problem regarding the reaction \( N_2O_4(g) \rightleftharpoons 2NO_2(g) \) and the effect of percentage dissociation on observed vapor densities, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**:
The reaction given is \( N_2O_4(g) \rightleftharpoons 2NO_2(g) \). The dissociation of \( N_2O_4 \) produces \( NO_2 \) gas.

2. **Define Degree of Dissociation**:
Let \( \alpha \) be the degree of dissociation of \( N_2O_4 \). The percentage dissociation values given are 25%, 50%, 75%, and 100%. These correspond to \( \alpha = 0.25, 0.50, 0.75, \) and \( 1.00 \), respectively.

3. **Calculate the Initial and Final Moles**:
- Initially, we have 1 mole of \( N_2O_4 \).
- At equilibrium, if \( \alpha \) is the degree of dissociation:
- Moles of \( N_2O_4 \) remaining = \( 1 - \alpha \)
- Moles of \( NO_2 \) formed = \( 2\alpha \)
- Total moles at equilibrium = \( (1 - \alpha) + 2\alpha = 1 + \alpha \)

4. **Relate Vapor Density to Moles**:
The vapor density \( D \) is calculated using the formula:
\[ D = \frac{\text{mass of gas}}{\text{volume of gas}} \times \frac{1}{\text{molar mass of hydrogen}} \]
Since the molar mass of \( N_2O_4 \) is 92 g/mol and \( NO_2 \) is 46 g/mol, we can express the vapor density in terms of the total moles.

5. **Vapor Density Calculation**:
The vapor density \( D \) at equilibrium can be expressed as:
\[ D \propto \frac{\text{mass}}{\text{total moles}} \propto \frac{92(1 - \alpha)}{1 + \alpha} \]
Thus, we can see that vapor density is inversely proportional to the total moles at equilibrium.

6. **Determine the Order of Vapor Densities**:
Since vapor density is inversely proportional to \( 1 + \alpha \):
- For \( \alpha = 0.25 \): \( D \propto \frac{92(1 - 0.25)}{1 + 0.25} = \frac{69}{1.25} \)
- For \( \alpha = 0.50 \): \( D \propto \frac{92(1 - 0.50)}{1 + 0.50} = \frac{46}{1.50} \)
- For \( \alpha = 0.75 \): \( D \propto \frac{92(1 - 0.75)}{1 + 0.75} = \frac{23}{1.75} \)
- For \( \alpha = 1.00 \): \( D \propto \frac{92(1 - 1)}{1 + 1} = 0 \)

7. **Sequence of Observed Vapor Densities**:
The sequence of observed vapor densities from highest to lowest will be:
- \( \alpha = 0.25 \) (highest density)
- \( \alpha = 0.50 \)
- \( \alpha = 0.75 \)
- \( \alpha = 1.00 \) (lowest density)

### Final Order:
Thus, the sequence of observed vapor densities is:
\[ D_{25\%} > D_{50\%} > D_{75\%} > D_{100\%} \]