What should be the value of `K_(c)` for the reaction `2SO_(2_((g)))+O_(2_((g))hArr2SO_(3_((g)))`. If the amount are `SO_(3)=48g`. `SO_(2)=12.8 "and" O_(2)=9.6` at equilibrium and the volume of the container is one litre?

Options

A. `64`
B. `30`
C. `42`
D. `8.5`

Explanation

To find the value of \( K_c \) for the reaction

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

given the amounts of substances at equilibrium, we will follow these steps:

### Step 1: Calculate the number of moles of each substance

1. **For \( SO_2 \)**:
- Given mass = 12.8 g
- Molar mass of \( SO_2 \) = \( 32 + (16 \times 2) = 64 \) g/mol
- Moles of \( SO_2 \) = \( \frac{12.8 \text{ g}}{64 \text{ g/mol}} = 0.2 \text{ mol} \)

2. **For \( SO_3 \)**:
- Given mass = 48 g
- Molar mass of \( SO_3 \) = \( 32 + (16 \times 3) = 80 \) g/mol
- Moles of \( SO_3 \) = \( \frac{48 \text{ g}}{80 \text{ g/mol}} = 0.6 \text{ mol} \)

3. **For \( O_2 \)**:
- Given mass = 9.6 g
- Molar mass of \( O_2 \) = \( 16 \times 2 = 32 \) g/mol
- Moles of \( O_2 \) = \( \frac{9.6 \text{ g}}{32 \text{ g/mol}} = 0.3 \text{ mol} \)

### Step 2: 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 3: Substitute the values into the \( K_c \) expression

Since the volume of the container is 1 liter, the concentration is equal to the number of moles:

- \([SO_3] = 0.6 \, \text{mol/L}\)
- \([SO_2] = 0.2 \, \text{mol/L}\)
- \([O_2] = 0.3 \, \text{mol/L}\)

Substituting these values into the \( K_c \) expression:

\[ K_c = \frac{(0.6)^2}{(0.2)^2 \times (0.3)} \]

### Step 4: Calculate \( K_c \)

Calculating the values:

1. \( (0.6)^2 = 0.36 \)
2. \( (0.2)^2 = 0.04 \)
3. Therefore, \( K_c = \frac{0.36}{0.04 \times 0.3} = \frac{0.36}{0.012} = 30 \)

### Final Answer

The value of \( K_c \) is \( 30 \).

---