For the equlibrium `SO_(3)(g)hArrSO_(2)(g)+1/2O(2)(g)` the molar mass at equlibrium was observed to be 60. then the degree of dissociation of `SO_(3)` would be

Options

A. `0.33`
B. `0.66`
C. `0.25`
D. `0.50`

Answer

To find the degree of dissociation of \( SO_3 \) in the equilibrium reaction

\[ SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2} O_2(g) \]

given that the molar mass at equilibrium is 60 g/mol, we can follow these steps:

### Step 1: Define Initial Conditions
Assume we start with 1 mole of \( SO_3 \) at time \( t = 0 \).

- Initial moles of \( SO_3 \) = 1
- Initial moles of \( SO_2 \) = 0
- Initial moles of \( O_2 \) = 0

### Step 2: Define Changes at Equilibrium
Let \( \alpha \) be the degree of dissociation of \( SO_3 \). At equilibrium:

- Moles of \( SO_3 \) = \( 1 - \alpha \)
- Moles of \( SO_2 \) = \( \alpha \)
- Moles of \( O_2 \) = \( \frac{\alpha}{2} \)

### Step 3: Calculate Total Moles at Equilibrium
Total moles at equilibrium can be expressed as:

\[ \text{Total moles} = (1 - \alpha) + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2} \]

### Step 4: Calculate Molar Mass at Equilibrium
The molar mass of the equilibrium mixture is given as 60 g/mol.

To find the molar mass of the equilibrium mixture, we need to calculate the average molar mass based on the moles present:

\[ \text{Average Molar Mass} = \frac{\text{Total Mass}}{\text{Total Moles}} \]

The molar mass of \( SO_3 \) is:

\[ \text{Molar mass of } SO_3 = 32 + (16 \times 3) = 80 \text{ g/mol} \]

The total mass at equilibrium is:

\[ \text{Total Mass} = (1 - \alpha) \times 80 + \alpha \times 64 + \frac{\alpha}{2} \times 32 \]

Calculating the mass contributions:

- Mass from \( SO_3 \): \( (1 - \alpha) \times 80 \)
- Mass from \( SO_2 \): \( \alpha \times 64 \)
- Mass from \( O_2 \): \( \frac{\alpha}{2} \times 32 \)

Thus, the total mass is:

\[ \text{Total Mass} = 80 - 80\alpha + 64\alpha + 16\alpha = 80 - 80\alpha + 80\alpha = 80 \]

### Step 5: Relate Molar Mass to Total Moles
We know that the average molar mass at equilibrium is given as 60 g/mol:

\[ \frac{80}{1 + \frac{\alpha}{2}} = 60 \]

### Step 6: Solve for \( \alpha \)
Cross-multiplying gives:

\[ 80 = 60 \left(1 + \frac{\alpha}{2}\right) \]

Expanding this:

\[ 80 = 60 + 30\alpha \]

Rearranging gives:

\[ 20 = 30\alpha \implies \alpha = \frac{20}{30} = \frac{2}{3} \]

### Conclusion
The degree of dissociation \( \alpha \) of \( SO_3 \) is:

\[ \alpha = \frac{2}{3} \approx 0.66 \]

### Final Answer
The degree of dissociation of \( SO_3 \) is \( \frac{2}{3} \) or approximately 0.66.

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