To solve the problem, we need to find the degree of association of selenium in the reaction:
\[ 3 \text{Se}_2(g) \rightleftharpoons \text{Se}_6(g) \]
### Step 1: Calculate the number of moles of selenium vapor
We start by calculating the number of moles of selenium vapor using the ideal gas law:
\[
PV = nRT
\]
Where:
- \( P = 1 \, \text{atm} \)
- \( V = 2.463 \, \text{mL} = 2.463 \times 10^{-3} \, \text{L} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \) (ideal gas constant)
- \( T = 27^\circ C = 300 \, \text{K} \)
Rearranging the ideal gas law to find \( n \):
\[
n = \frac{PV}{RT}
\]
Substituting the values:
\[
n = \frac{(1 \, \text{atm}) \times (2.463 \times 10^{-3} \, \text{L})}{(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}) \times (300 \, \text{K})}
\]
Calculating \( n \):
\[
n = \frac{2.463 \times 10^{-3}}{24.63} \approx 1.000 \times 10^{-4} \, \text{mol}
\]
### Step 2: Calculate the initial moles of selenium
Given the mass of selenium is \( 0.020 \, \text{g} \) and the molar mass of selenium (\( \text{Se} \)) is \( 79 \, \text{g/mol} \):
\[
\text{Moles of } \text{Se} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.020 \, \text{g}}{79 \, \text{g/mol}} \approx 2.53165 \times 10^{-4} \, \text{mol}
\]
### Step 3: Set up the equilibrium expression
Let \( \alpha \) be the degree of association. Initially, we have:
- Moles of \( \text{Se}_2 \) = \( A = 2.53165 \times 10^{-4} \, \text{mol} \)
At equilibrium, the moles of \( \text{Se}_2 \) that dissociate are \( \frac{A \alpha}{3} \), and the moles of \( \text{Se}_6 \) formed are \( \frac{A \alpha}{3} \).
Thus, the moles of \( \text{Se}_2 \) at equilibrium:
\[
\text{Moles of } \text{Se}_2 = A - \frac{A \alpha}{3} = A(1 - \frac{\alpha}{3})
\]
And the moles of \( \text{Se}_6 \) at equilibrium:
\[
\text{Moles of } \text{Se}_6 = \frac{A \alpha}{3}
\]
### Step 4: Calculate the average molar mass of the mixture
The average molar mass \( M \) of the mixture can be calculated as:
\[
M = \frac{\text{Total mass}}{\text{Total moles}} = \frac{0.020 \, \text{g}}{n}
\]
Using the total moles calculated earlier:
\[
M = \frac{0.020 \, \text{g}}{1.000 \times 10^{-4} \, \text{mol}} \approx 200 \, \text{g/mol}
\]
### Step 5: Relate the molar masses to find the degree of association
The molar mass of \( \text{Se}_2 \):
\[
\text{Molar mass of } \text{Se}_2 = 2 \times 79 = 158 \, \text{g/mol}
\]
Using the relationship:
\[
\frac{M_{\text{reactant}}}{M_{\text{mixture}}} = \frac{1}{1 - \alpha}
\]
Substituting the values:
\[
\frac{158}{200} = \frac{1}{1 - \alpha}
\]
Cross-multiplying gives:
\[
158(1 - \alpha) = 200
\]
Solving for \( \alpha \):
\[
158 - 158\alpha = 200 \\
-158\alpha = 200 - 158 \\
-158\alpha = 42 \\
\alpha = \frac{42}{158} \approx 0.2658
\]
### Final Result
The degree of association \( \alpha \) of selenium is approximately \( 0.2658 \) or \( 26.58\% \).
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