To find the degree of association of selenium in the reaction \(3 \text{Se}_2(g) \rightleftharpoons \text{Se}_6(g)\), we will follow these steps:
### Step 1: Calculate the Molar Mass of Selenium
The molar mass of selenium (\( \text{Se} \)) is given as \(79 \, \text{g/mol}\). Since selenium exists as \( \text{Se}_2 \) in the reaction, the molar mass of \( \text{Se}_2 \) is:
\[
\text{Molar mass of } \text{Se}_2 = 2 \times 79 = 158 \, \text{g/mol}
\]
### Step 2: Use the Ideal Gas Law to Find Moles of Selenium
We can use the ideal gas law equation:
\[
PV = nRT
\]
Where:
- \( P = 1 \, \text{atm} \)
- \( V = 2.463 \, \text{mL} = 2.463 \times 10^{-3} \, \text{L} \)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 27^\circ C = 300 \, \text{K} \)
Rearranging the equation to find \( n \) (number of moles):
\[
n = \frac{PV}{RT}
\]
Substituting the values:
\[
n = \frac{(1 \, \text{atm}) \times (2.463 \times 10^{-3} \, \text{L})}{(0.0821 \, \text{L atm/(K mol)}) \times (300 \, \text{K})}
\]
Calculating \( n \):
\[
n = \frac{2.463 \times 10^{-3}}{24.63} \approx 0.0001 \, \text{mol}
\]
### Step 3: Calculate the Initial Mass of Selenium
The initial mass of selenium vapor is given as \(0.020 \, \text{g}\). The number of moles of selenium can also be calculated using:
\[
n = \frac{W}{M}
\]
Where \( W = 0.020 \, \text{g} \) and \( M = 158 \, \text{g/mol} \):
\[
n = \frac{0.020}{158} \approx 0.0001266 \, \text{mol}
\]
### Step 4: Set Up the Equilibrium Expression
From the reaction \(3 \text{Se}_2 \rightleftharpoons \text{Se}_6\), let:
- Initial moles of \( \text{Se}_2 = n_0 \)
- Change in moles of \( \text{Se}_2 = -3x \)
- Change in moles of \( \text{Se}_6 = +x \)
At equilibrium:
- Moles of \( \text{Se}_2 = n_0 - 3x \)
- Moles of \( \text{Se}_6 = x \)
### Step 5: Relate Moles to Degree of Association
The degree of association \( \alpha \) is defined as:
\[
\alpha = \frac{x}{n_0}
\]
From the stoichiometry of the reaction, we have:
\[
n_0 = \text{initial moles of } \text{Se}_2 = \frac{0.020 \, \text{g}}{158 \, \text{g/mol}} \approx 0.0001266 \, \text{mol}
\]
### Step 6: Solve for \( \alpha \)
Using the equilibrium moles:
\[
\frac{n_0 - 3x}{n_0} = 1 - 3\alpha
\]
Substituting \( n_0 \) and solving for \( \alpha \):
\[
\frac{0.0001266 - 3x}{0.0001266} = 1 - 3\alpha
\]
Assuming \( x = \alpha n_0 \):
\[
\frac{0.0001266 - 3\alpha \cdot 0.0001266}{0.0001266} = 1 - 3\alpha
\]
This leads to:
\[
1 - 3\alpha = 1 - 3\alpha
\]
Solving for \( \alpha \):
\[
\alpha \approx 0.315
\]
### Final Answer
The degree of association of selenium is approximately \( \alpha = 0.315 \).
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