In the dissociation of `N_(2)O_(4)` into `NO_(2), (1+ alpha)` values with the vapour densities ratio `((D)/(d))` is given by: [`alpha` degree of dissociation, D-vapour density before dissociation, d-vapour density after dissociation]

Options

A.
B.
C.
D.

Answer

To solve the problem regarding the dissociation of \( N_2O_4 \) into \( NO_2 \) and the relationship between the vapor densities before and after dissociation, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Dissociation Reaction**:
The dissociation of \( N_2O_4 \) can be represented as:
\[ N_2O_4 \rightleftharpoons 2 NO_2 \]

2. **Set Up Initial and Equilibrium Conditions**:
Assume we start with \( C \) moles of \( N_2O_4 \) and no \( NO_2 \):
- Initial moles of \( N_2O_4 \) = \( C \)
- Initial moles of \( NO_2 \) = 0

At equilibrium, if \( \alpha \) is the degree of dissociation:
- Moles of \( N_2O_4 \) at equilibrium = \( C(1 - \alpha) \)
- Moles of \( NO_2 \) at equilibrium = \( 2C\alpha \)

3. **Calculate Vapor Densities**:
Vapor density is defined as the mass of the gas divided by the volume. Since we are dealing with moles, we can relate vapor density to the number of moles:
- Initial vapor density \( D \) is proportional to the initial moles \( C \).
- At equilibrium, the total moles of gas = Moles of \( N_2O_4 \) + Moles of \( NO_2 \):
\[ \text{Total moles at equilibrium} = C(1 - \alpha) + 2C\alpha = C(1 + \alpha) \]
- Therefore, the vapor density at equilibrium \( d \) is proportional to \( C(1 + \alpha) \).

4. **Set Up the Ratio of Vapor Densities**:
The ratio of the vapor densities before and after dissociation can be expressed as:
\[ \frac{D}{d} = \frac{C}{C(1 + \alpha)} = \frac{1}{1 + \alpha} \]

5. **Rearranging the Equation**:
Rearranging the equation gives:
\[ D = \frac{d}{1 + \alpha} \]
This implies that:
\[ \frac{D}{d} = \frac{1}{1 + \alpha} \]

6. **Establishing the Relationship**:
From the equation \( \frac{D}{d} = \frac{1}{1 + \alpha} \), we can see that as \( \alpha \) increases, \( \frac{D}{d} \) decreases. This indicates an inverse relationship.

7. **Identifying the Correct Plot**:
Given the relationship \( \frac{D}{d} = \frac{1}{1 + \alpha} \), we can conclude that the plot of \( \frac{D}{d} \) versus \( \alpha \) will not be a straight line through the origin but rather a hyperbolic curve. However, if we were to plot \( 1 + \alpha \) against \( \frac{D}{d} \), it would yield a straight line.

### Final Answer:
The correct relationship is:
\[ \frac{D}{d} = \frac{1}{1 + \alpha} \]