For the reaction `N_2O_4 hArr 2NO_2` , if degrees of dissociation of `N_2O_4` are 25%,50% 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 observed vapor densities at different degrees of dissociation, we can follow these steps:

### Step 1: Understand the Reaction and Initial Conditions
Initially, we have 1 mole of \( N_2O_4 \) and no products. Therefore, at time \( t = 0 \):
- Moles of \( N_2O_4 \) = 1
- Moles of \( NO_2 \) = 0

### Step 2: Define Degree of Dissociation
The degree of dissociation (\( \alpha \)) is the fraction of the initial amount of \( N_2O_4 \) that dissociates into \( NO_2 \).

### Step 3: Calculate Moles at Equilibrium
At equilibrium, if \( \alpha \) is the degree of dissociation:
- Moles of \( N_2O_4 \) at equilibrium = \( 1 - \alpha \)
- Moles of \( NO_2 \) at equilibrium = \( 2\alpha \)

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

### Step 5: Calculate Vapor Density
The vapor density (\( d \)) is related to the total number of moles. The vapor density is defined as:
\[ d = \frac{\text{mass of gas}}{\text{volume of gas}} \propto \frac{\text{moles}}{\text{volume}} \]
For 1 mole of gas, the vapor density can be expressed as:
\[ d = \frac{M}{RT} \text{ (where M is molar mass)} \]

### Step 6: Relate Vapor Density to Degree of Dissociation
The observed vapor density can be expressed as:
\[ d = \frac{M}{(1 + \alpha) \cdot RT} \]
Where \( M \) is the molar mass of the gas mixture. As \( \alpha \) increases, \( d \) decreases since \( 1 + \alpha \) increases.

### Step 7: Analyze Different Degrees of Dissociation
Now, we analyze the three cases:
1. \( \alpha = 0.25 \) (25% dissociation)
2. \( \alpha = 0.50 \) (50% dissociation)
3. \( \alpha = 1.00 \) (100% dissociation)

As \( \alpha \) increases, the total number of moles increases, leading to a decrease in vapor density.

### Step 8: Conclusion on Observed Vapor Densities
From the above analysis, we can conclude:
- For \( \alpha = 0.25 \), \( d \) is highest.
- For \( \alpha = 0.50 \), \( d \) is lower than for 25%.
- For \( \alpha = 1.00 \), \( d \) is lowest.

Thus, the order of observed vapor densities is:
\[ d_{25\%} > d_{50\%} > d_{100\%} \]

### Final Answer
The gradation of observed vapor densities is:
- \( d_{25\%} > d_{50\%} > d_{100\%} \)