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 of finding the equilibrium constant for the reaction \( O_2 + 2SO_2 \rightleftharpoons 2SO_3 \), we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the formation of sulfur trioxide (\( SO_3 \)) from oxygen (\( O_2 \)) and sulfur dioxide (\( SO_2 \)) is:
\[ O_2 + 2SO_2 \rightleftharpoons 2SO_3 \]

### Step 2: Define initial moles and changes at equilibrium
- Initial moles:
- \( O_2 = 1 \) mole
- \( SO_2 = 2 \) moles
- \( SO_3 = 0 \) moles

- Change in moles at equilibrium:
- Let \( 2x \) be the number of moles of \( SO_3 \) formed at equilibrium.
- Therefore, \( O_2 \) will decrease by \( x \) moles (since 1 mole of \( O_2 \) is required for 2 moles of \( SO_3 \)).
- \( SO_2 \) will decrease by \( 2x \) moles (since 2 moles of \( SO_2 \) are required for 2 moles of \( SO_3 \)).

- At equilibrium:
- Moles of \( O_2 = 1 - x \)
- Moles of \( SO_2 = 2 - 2x \)
- Moles of \( SO_3 = 2x \)

### Step 3: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by the formula:
\[ K_c = \frac{[\text{Products}]}{[\text{Reactants}]} \]
For our reaction:
\[ K_c = \frac{[SO_3]^2}{[O_2][SO_2]^2} \]

### Step 4: Substitute the equilibrium concentrations into the expression
The concentrations can be expressed in terms of moles and volume \( V \):
- Concentration of \( SO_3 = \frac{2x}{V} \)
- Concentration of \( O_2 = \frac{1 - x}{V} \)
- Concentration of \( SO_2 = \frac{2 - 2x}{V} \)

Substituting these into the \( K_c \) expression:
\[ K_c = \frac{\left(\frac{2x}{V}\right)^2}{\left(\frac{1 - x}{V}\right)\left(\frac{2 - 2x}{V}\right)^2} \]

### Step 5: Simplify the expression
\[ K_c = \frac{\frac{4x^2}{V^2}}{\frac{(1 - x)(2 - 2x)^2}{V^3}} \]
\[ K_c = \frac{4x^2 \cdot V}{(1 - x)(2 - 2x)^2} \]

### Step 6: Final expression for \( K_c \)
Thus, the equilibrium constant \( K_c \) can be expressed as:
\[ K_c = \frac{4x^2 \cdot V}{(1 - x)(2 - 2x)^2} \]

### Summary
The equilibrium constant \( K_c \) for the reaction is:
\[ K_c = \frac{4x^2 \cdot V}{(1 - x)(2 - 2x)^2} \]