`SO_(3)(g)hArrSO_(2)(g)+(1)/(2)O_(2)(g)`
If observed vapour density of mixture at equilibrium is `35` then find out value of `alpha`

Options

A. `0.28`
B. `0.38`
C. `0.48`
D. `0.58`

Explanation

To solve the problem, we need to find the value of alpha (α) in the equilibrium reaction:

\[ \text{SO}_3(g) \rightleftharpoons \text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \]

Given that the observed vapor density (d) of the mixture at equilibrium is 35, we can follow these steps:

### Step 1: Calculate the molar mass of SO3
The molar mass of SO3 can be calculated as follows:
- Molar mass of S = 32 g/mol
- Molar mass of O = 16 g/mol

Thus, the molar mass of SO3 is:
\[ \text{Molar mass of SO}_3 = 32 + (16 \times 3) = 32 + 48 = 80 \, \text{g/mol} \]

### Step 2: Calculate the vapor density (D) of SO3
The vapor density (D) is defined as half of the molar mass:
\[ D = \frac{\text{Molar mass of SO}_3}{2} = \frac{80}{2} = 40 \]

### Step 3: Determine the number of moles of products
From the reaction, we can see:
- 1 mole of SO2
- 0.5 moles of O2

Thus, the total number of moles of products (n) is:
\[ n = 1 + \frac{1}{2} = 1.5 \]

### Step 4: Use the formula for alpha (α)
The formula to calculate alpha is given by:
\[ \alpha = \frac{D - d}{(n - 1) \cdot d} \]

Substituting the values we have:
- D = 40
- d = 35
- n = 1.5

We can now substitute these values into the formula:
\[ \alpha = \frac{40 - 35}{(1.5 - 1) \cdot 35} \]

### Step 5: Calculate α
Now, we calculate the numerator and denominator:
- Numerator: \( 40 - 35 = 5 \)
- Denominator: \( (1.5 - 1) \cdot 35 = 0.5 \cdot 35 = 17.5 \)

Thus,
\[ \alpha = \frac{5}{17.5} \]

Calculating this gives:
\[ \alpha = 0.2857 \approx 0.28 \]

### Final Answer
The value of alpha (α) is approximately **0.28**.

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