The density of a sample of `SO_(3)` gas is 2.5 g/L at `0^(@)`C and 1 atm . It's degree of dissociation into `SO_(2)` and `O_(2)` gases is :

Options

A. `(6)/(7)`
B. `(1)/(7)`
C. `(3)/(7)`
D. `(5)/(7)`

Answer

To find the degree of dissociation of \( SO_3 \) gas into \( SO_2 \) and \( O_2 \), we can follow these steps:

### Step 1: Understand the Reaction
The dissociation of sulfur trioxide can be represented as:
\[ SO_3 \rightleftharpoons SO_2 + \frac{1}{2} O_2 \]
From this reaction, we can see that 1 mole of \( SO_3 \) produces 1 mole of \( SO_2 \) and 0.5 moles of \( O_2 \), totaling 1.5 moles of products.

### Step 2: Use the Density to Find Molecular Weight
We are given the density of the gas mixture at equilibrium:
\[ \text{Density} = 2.5 \, \text{g/L} \]
Using the ideal gas law, we can relate density to molecular weight:
\[ PV = nRT \implies P = \frac{n}{V}RT \implies \text{Density} = \frac{n}{V} \cdot \text{Molecular Weight} \]
Thus, we can express it as:
\[ \text{Density} = \frac{P \cdot \text{Molecular Weight}}{RT} \]
Where:
- \( P = 1 \, \text{atm} \)
- \( R = 0.081 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 273 \, \text{K} \)

Rearranging gives:
\[ \text{Molecular Weight} = \frac{\text{Density} \cdot RT}{P} \]

### Step 3: Calculate the Molecular Weight of the Mixture
Substituting the values:
\[ \text{Molecular Weight} = \frac{2.5 \, \text{g/L} \cdot 0.081 \, \text{L atm K}^{-1} \text{mol}^{-1} \cdot 273 \, \text{K}}{1 \, \text{atm}} \]
Calculating this gives:
\[ \text{Molecular Weight} = \frac{2.5 \cdot 0.081 \cdot 273}{1} \approx 56 \, \text{g/mol} \]

### Step 4: Calculate the Vapor Density at Equilibrium
The vapor density \( d \) of the mixture can be calculated as:
\[ d = \frac{\text{Molecular Weight}}{2} = \frac{56}{2} = 28 \, \text{g/mol} \]

### Step 5: Calculate the Initial Vapor Density of \( SO_3 \)
The molecular weight of \( SO_3 \) is 80 g/mol, so its vapor density is:
\[ D = \frac{80}{2} = 40 \, \text{g/mol} \]

### Step 6: Use the Degree of Dissociation Formula
The degree of dissociation \( \alpha \) can be calculated using the formula:
\[ \alpha = \frac{D - d}{n - 1} \cdot d \]
Where \( n \) is the total number of moles of gas after dissociation. Here, \( n = \frac{3}{2} \) (1 mole of \( SO_3 \) gives 1 mole of \( SO_2 \) and 0.5 moles of \( O_2 \)).

Substituting the values:
\[ \alpha = \frac{40 - 28}{\frac{3}{2} - 1} \cdot 28 \]
Calculating the numerator:
\[ 40 - 28 = 12 \]
Calculating the denominator:
\[ \frac{3}{2} - 1 = \frac{1}{2} \]
Thus:
\[ \alpha = \frac{12}{\frac{1}{2}} \cdot 28 = 12 \cdot 2 \cdot 28 = 24 \cdot 28 = 6 \]
Finally, we find:
\[ \alpha = \frac{12}{14} = \frac{6}{7} \]

### Conclusion
The degree of dissociation of \( SO_3 \) into \( SO_2 \) and \( O_2 \) is:
\[ \alpha = \frac{6}{7} \]