`2SO_(2)+O_(2)hArr 2SO_(3)`. Starting with 2 moles `SO_(2)` and 1 mole `O_(2)` in 1L flask, mixture required 0.4 mole `MnO_(4)^(-)` in acidic medium. Hence, `K_9c)` is

Options

A. 2
B. 0.4
C. 1.6
D. 2.6

Explanation

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

\[ 2SO_2 + O_2 \rightleftharpoons 2SO_3 \]

### Step 1: Determine the moles of \( SO_2 \) reacted
We start with 2 moles of \( SO_2 \) and 1 mole of \( O_2 \) in a 1L flask. The mixture required 0.4 moles of \( MnO_4^- \) in acidic medium to fully react with the \( SO_2 \).

From the video transcript, we know that the reaction of \( SO_2 \) with \( MnO_4^- \) in acidic medium is:

\[ 2KMnO_4 + 5SO_2 + 6H^+ \rightarrow 2K_2SO_4 + 2MnSO_4 + 3H_2O + 2O_2 \]

This indicates that 2 moles of \( KMnO_4 \) react with 5 moles of \( SO_2 \).

### Step 2: Calculate moles of \( SO_2 \) that reacted
Since 0.4 moles of \( KMnO_4 \) were used, we can find out how many moles of \( SO_2 \) reacted:

\[ \text{Moles of } SO_2 \text{ reacted} = \left( \frac{5}{2} \right) \times 0.4 = 1 \text{ mole of } SO_2 \]

### Step 3: Determine moles of \( SO_2 \) at equilibrium
Initially, we had 2 moles of \( SO_2 \). After reacting 1 mole of \( SO_2 \), the moles left at equilibrium are:

\[ \text{Moles of } SO_2 \text{ at equilibrium} = 2 - 1 = 1 \text{ mole} \]

### Step 4: Determine moles of \( O_2 \) at equilibrium
Initially, we had 1 mole of \( O_2 \). The stoichiometry of the reaction shows that 1 mole of \( O_2 \) is consumed for every 2 moles of \( SO_2 \) reacted. Since we reacted 1 mole of \( SO_2 \):

\[ \text{Moles of } O_2 \text{ reacted} = \frac{1}{2} \text{ mole} \]

Thus, the moles of \( O_2 \) at equilibrium are:

\[ \text{Moles of } O_2 \text{ at equilibrium} = 1 - 0.5 = 0.5 \text{ moles} \]

### Step 5: Determine moles of \( SO_3 \) at equilibrium
From the reaction, 2 moles of \( SO_2 \) produce 2 moles of \( SO_3 \). Since we have reacted 1 mole of \( SO_2 \), we will produce 1 mole of \( SO_3 \):

\[ \text{Moles of } SO_3 \text{ at equilibrium} = 1 \text{ mole} \]

### Step 6: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:

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

### Step 7: Substitute the equilibrium concentrations
Substituting the equilibrium values into the expression:

- Concentration of \( SO_3 = 1 \text{ mole/L} \)
- Concentration of \( SO_2 = 1 \text{ mole/L} \)
- Concentration of \( O_2 = 0.5 \text{ mole/L} \)

Thus,

\[ K_c = \frac{(1)^2}{(1)^2 \cdot (0.5)} = \frac{1}{0.5} = 2 \]

### Final Answer
The value of \( K_c \) is:

\[ K_c = 2 \]

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