At equilibrium degree of dissociation of `SO_(3)` is `50%` for the reaction `2SO_(3)(g)hArr 2SO_(2)(g)+O_(2)(g)`. The equilibrium constant for the reaction is

Options

A. 0.15
B. 0.25
C. 2.5
D. 0.025

Answer

To find the equilibrium constant for the reaction \(2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g)\) with a degree of dissociation of \(50\%\), we can follow these steps:

### Step 1: Define the initial conditions
Assume we start with 1 mole of \(SO_3\) and no \(SO_2\) or \(O_2\) present initially.
- Initial moles of \(SO_3 = 1\)
- Initial moles of \(SO_2 = 0\)
- Initial moles of \(O_2 = 0\)

### Step 2: Determine the degree of dissociation
Given that the degree of dissociation of \(SO_3\) is \(50\%\), this means that half of the \(SO_3\) will dissociate.
- Degree of dissociation = \(50\%\) implies that \(0.5 \times 1 = 0.5\) moles of \(SO_3\) will dissociate.

### Step 3: Set up the changes in moles at equilibrium
From the stoichiometry of the reaction:
- For every 2 moles of \(SO_3\) that dissociate, 2 moles of \(SO_2\) and 1 mole of \(O_2\) are produced.
- Thus, if \(0.5\) moles of \(SO_3\) dissociate, we can calculate the changes:
- Moles of \(SO_3\) at equilibrium = \(1 - 0.5 = 0.5\)
- Moles of \(SO_2\) produced = \(0.5 \times 2 = 0.5\)
- Moles of \(O_2\) produced = \(0.5 \times 1 = 0.25\)

### Step 4: Write the equilibrium concentrations
At equilibrium, we have:
- \(SO_3\): \(0.5\) moles
- \(SO_2\): \(0.5\) moles
- \(O_2\): \(0.25\) moles

### Step 5: Calculate the equilibrium constant \(K_c\)
The equilibrium constant \(K_c\) is given by the expression:
\[ K_c = \frac{[SO_2]^2 \cdot [O_2]}{[SO_3]^2} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(0.5)^2 \cdot (0.25)}{(0.5)^2} \]
This simplifies to:
\[ K_c = \frac{0.25 \cdot 0.25}{0.25} = 0.25 \]

### Final Answer
The equilibrium constant \(K_c\) for the reaction is \(0.25\).

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