Consider the following hypothetical equilibrium
`2B(g)hArrB_(2)(g)`
If `d` is observed vapour density and `D` is theoretical vapour density, then degree of association `(alpha)` will be

Options

A. `alpha=((D-d)/(d))`
B. `alpha=(2D-d)/(D)`
C. `alpha=2-(2D)/(d)`
D. `alpha=(2D)/(D-d)`

Answer

To solve the problem regarding the degree of association (α) for the equilibrium reaction \( 2B(g) \rightleftharpoons B_2(g) \), we will follow these steps:

### Step 1: Write the equilibrium reaction
The given equilibrium reaction is:
\[ 2B(g) \rightleftharpoons B_2(g) \]

### Step 2: Define the initial moles
Initially, we have 2 moles of \( B \). When the reaction reaches equilibrium, let \( \alpha \) be the degree of association. The moles of \( B \) that associate to form \( B_2 \) will be \( 2\alpha \) (since 2 moles of \( B \) are needed to form 1 mole of \( B_2 \)).

### Step 3: Calculate the moles at equilibrium
At equilibrium:
- Moles of \( B \) left = Initial moles - Moles that reacted = \( 2 - 2\alpha \)
- Moles of \( B_2 \) formed = \( \alpha \)

### Step 4: Total moles at equilibrium
The total moles at equilibrium will be:
\[ \text{Total moles} = (2 - 2\alpha) + \alpha = 2 - \alpha \]

### Step 5: Relate vapor density to moles
The vapor density (D) is related to the number of moles. The theoretical vapor density \( D \) is calculated based on the assumption that there is no association (i.e., all \( B \) remains as \( B \)):
- Theoretical vapor density \( D = \frac{\text{Molar mass}}{\text{Total moles}} \) when no association occurs.

For the initial 2 moles of \( B \), the theoretical vapor density \( D \) would be:
\[ D = \frac{M}{2} \]
where \( M \) is the molar mass of \( B \).

### Step 6: Observed vapor density
The observed vapor density \( d \) at equilibrium is given by:
\[ d = \frac{M}{\text{Total moles at equilibrium}} = \frac{M}{2 - \alpha} \]

### Step 7: Use the formula for degree of association
The degree of association \( \alpha \) can be calculated using the formula:
\[ \alpha = \frac{D - d}{D} \]

### Step 8: Substitute the values
Substituting \( d \) and \( D \) into the equation:
1. \( D = \frac{M}{2} \)
2. \( d = \frac{M}{2 - \alpha} \)

Now substituting these into the formula:
\[ \alpha = \frac{\frac{M}{2} - \frac{M}{2 - \alpha}}{\frac{M}{2}} \]

### Step 9: Simplify the equation
1. Cancel \( M \) from numerator and denominator:
\[ \alpha = \frac{\frac{1}{2} - \frac{1}{2 - \alpha}}{\frac{1}{2}} \]
2. Simplifying further leads to:
\[ \alpha = 1 - \frac{1}{2 - \alpha} \]

### Step 10: Solve for α
Cross-multiplying and simplifying will yield the final expression for \( \alpha \):
\[ \alpha = \frac{2D - 2d}{d} \]

### Final Result
Thus, the degree of association \( \alpha \) is given by:
\[ \alpha = 2 - \frac{2d}{D} \]