3 mole of `SO_(3)(g)` are taken in 8.0 L container at `800^(@)C`. At equilibrium, 0.58 mol of `O_(2)(g)` are formed. Find the value of `K_(c)` for `2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)`

Answer

To solve the problem step by step, we will follow the process of determining the equilibrium constant \( K_c \) for the reaction:

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

### Step 1: Write the balanced equation and identify initial conditions
The balanced equation for the reaction is:

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

Initially, we have:
- Moles of \( SO_3 \) = 3 moles
- Moles of \( SO_2 \) = 0 moles
- Moles of \( O_2 \) = 0 moles

### Step 2: Determine the change in moles at equilibrium
At equilibrium, we know that 0.58 moles of \( O_2 \) are formed. Let \( x \) be the amount of \( O_2 \) formed, which is given as \( x = 0.58 \) moles.

From the stoichiometry of the reaction:
- For every 1 mole of \( O_2 \) formed, 2 moles of \( SO_2 \) are produced.
- Therefore, moles of \( SO_2 \) formed = \( 2x = 2 \times 0.58 = 1.16 \) moles.
- The moles of \( SO_3 \) that dissociate to produce these moles of \( SO_2 \) and \( O_2 \) can be calculated as follows:
- Initial moles of \( SO_3 \) = 3 moles
- Moles of \( SO_3 \) at equilibrium = \( 3 - x \) (where \( x \) is the total moles of \( SO_3 \) that dissociate)
- Therefore, \( 3 - 0.58 = 1.84 \) moles of \( SO_3 \) remain at equilibrium.

### Step 3: Set up the equilibrium concentrations
The equilibrium concentrations can be calculated using the formula:

\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume (L)}} \]

Given that the volume of the container is 8.0 L, we can calculate the concentrations:

- Concentration of \( SO_3 \):
\[ [SO_3] = \frac{1.84 \text{ moles}}{8.0 \text{ L}} = 0.23 \text{ M} \]

- Concentration of \( SO_2 \):
\[ [SO_2] = \frac{1.16 \text{ moles}}{8.0 \text{ L}} = 0.145 \text{ M} \]

- Concentration of \( O_2 \):
\[ [O_2] = \frac{0.58 \text{ moles}}{8.0 \text{ L}} = 0.0725 \text{ M} \]

### Step 4: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:

\[ K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]} \]

### Step 5: Substitute the equilibrium concentrations into the \( K_c \) expression
Substituting the calculated concentrations into the \( K_c \) expression:

\[ K_c = \frac{(0.23)^2}{(0.145)^2 \times (0.0725)} \]

### Step 6: Calculate \( K_c \)
Calculating the values:

- \( (0.23)^2 = 0.0529 \)
- \( (0.145)^2 = 0.021025 \)
- Therefore,

\[ K_c = \frac{0.0529}{0.021025 \times 0.0725} = \frac{0.0529}{0.001523} \approx 34.71 \]

### Final Answer
Thus, the value of \( K_c \) for the reaction is approximately:

\[ K_c \approx 34.71 \]