Starting with one moles of `O_(2)` and two moles of `SO_(2)`, the equilibrium for the formation of `SO_(3)` was established at a certain temperature. If V is the volume of the vessel and 2 x is the number of moles of `SO_(3)` present, the equilibrium cosntant will be

Options

A. `(x^(2)V)/((1-x)^(3))`
B. `(4x^(2))/((2-x)(1-x))`
C. `((1-x)^(3))/(2V)`
D. `(x^(2))/((2-x)(1-x))`

Answer

To solve the problem, we need to establish the equilibrium constant \( K_c \) for the reaction:

\[ \text{O}_2 + 2\text{SO}_2 \rightleftharpoons 2\text{SO}_3 \]

### Step 1: Write the initial concentrations
Initially, we have:
- \( [\text{O}_2] = 1 \) mole
- \( [\text{SO}_2] = 2 \) moles
- \( [\text{SO}_3] = 0 \) moles

Since the volume of the vessel is \( V \), the initial concentrations are:
- \( [\text{O}_2] = \frac{1}{V} \)
- \( [\text{SO}_2] = \frac{2}{V} \)
- \( [\text{SO}_3] = 0 \)

### Step 2: Define the change in concentration at equilibrium
Let \( x \) be the amount of \( \text{O}_2 \) that reacts. At equilibrium:
- The concentration of \( \text{O}_2 \) will be \( \frac{1 - x}{V} \)
- The concentration of \( \text{SO}_2 \) will be \( \frac{2 - 2x}{V} \)
- The concentration of \( \text{SO}_3 \) will be \( \frac{2x}{V} \)

### Step 3: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant expression for the reaction is given by:

\[ K_c = \frac{[\text{SO}_3]^2}{[\text{O}_2][\text{SO}_2]^2} \]

Substituting the equilibrium concentrations, we get:

\[ K_c = \frac{\left(\frac{2x}{V}\right)^2}{\left(\frac{1 - x}{V}\right)\left(\frac{2 - 2x}{V}\right)^2} \]

### Step 4: Simplify the expression
This simplifies to:

\[ K_c = \frac{\frac{4x^2}{V^2}}{\frac{(1 - x)(2 - 2x)^2}{V^5}} \]

Now, simplifying further:

\[ K_c = \frac{4x^2 \cdot V^3}{(1 - x)(2 - 2x)^2} \]

### Step 5: Substitute \( (2 - 2x)^2 \)
We can rewrite \( (2 - 2x)^2 \) as \( 4(1 - x)^2 \):

\[ K_c = \frac{4x^2 \cdot V^3}{(1 - x) \cdot 4(1 - x)^2} \]

### Step 6: Cancel out the common terms
The \( 4 \) cancels out:

\[ K_c = \frac{x^2 \cdot V^3}{(1 - x)^3} \]

### Final Answer
Thus, the equilibrium constant \( K_c \) is:

\[ K_c = \frac{x^2 \cdot V^3}{(1 - x)^3} \]