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 \( K_p \) for the reaction

\[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \]

given that the degree of dissociation of \( SO_3 \) is 50%, we can follow these steps:

### Step 1: Define the degree of dissociation
Let the degree of dissociation \( \alpha \) be 50%. This means that 50% of the initial \( SO_3 \) will dissociate. In decimal form, \( \alpha = 0.5 \).

### Step 2: Set up the initial and equilibrium concentrations
Assume we start with 2 moles of \( SO_3 \):
- Initial moles of \( SO_3 = 2 \)
- Initial moles of \( SO_2 = 0 \)
- Initial moles of \( O_2 = 0 \)

At equilibrium, the moles of each species will be:
- Moles of \( SO_3 \) at equilibrium = \( 2 - 2\alpha = 2 - 2(0.5) = 1 \)
- Moles of \( SO_2 \) at equilibrium = \( 2\alpha = 2(0.5) = 1 \)
- Moles of \( O_2 \) at equilibrium = \( \alpha = 0.5 \)

### Step 3: Calculate total moles at equilibrium
Total moles at equilibrium = Moles of \( SO_3 + SO_2 + O_2 \)
\[ = 1 + 1 + 0.5 = 2.5 \]

### Step 4: Calculate partial pressures
Assuming that the total pressure is \( P \), the partial pressures can be expressed as:
- Partial pressure of \( SO_3 \) = \( \frac{1}{2.5} P \)
- Partial pressure of \( SO_2 \) = \( \frac{1}{2.5} P \)
- Partial pressure of \( O_2 \) = \( \frac{0.5}{2.5} P \)

### Step 5: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{SO_2})^2 \cdot (P_{O_2})}{(P_{SO_3})^2} \]

Substituting the partial pressures:
\[ K_p = \frac{\left(\frac{1}{2.5} P\right)^2 \cdot \left(\frac{0.5}{2.5} P\right)}{\left(\frac{1}{2.5} P\right)^2} \]

### Step 6: Simplify the expression
\[ K_p = \frac{\left(\frac{1}{2.5}\right)^2 \cdot \left(\frac{0.5}{2.5}\right) P^3}{\left(\frac{1}{2.5}\right)^2 P^2} \]
\[ = \frac{0.5}{2.5} \]

### Step 7: Calculate \( K_p \)
\[ K_p = \frac{0.5}{2.5} = 0.2 \]

Thus, the equilibrium constant \( K_p \) for the reaction is **0.2**.

---