For a reaction, `2SO_(2(g))+O_(2(g))hArr2SO_(3(g))`, 1.5 moles of `SO_(2)` and 1 mole of `O_(2)` are taken in a 2 L vessel. At equilibrium the concentration of `SO_(3)` was found to be 0.35 mol `L^(-1)` The `K_(c)` for the reaction would be

Options

A. `5.1"L mol"^(-1)`
B. `1.4"L mol"^(-1)`
C. `0.6"L mol"^(-1)`
D. `2.95"L mol"^(-1)`

Explanation

To find the equilibrium constant \( K_c \) for the reaction

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

given the initial moles of \( SO_2 \) and \( O_2 \) and the equilibrium concentration of \( SO_3 \), we can follow these steps:

### Step 1: Calculate Initial Concentrations
We start with the initial moles of the reactants and convert them into concentrations using the formula:

\[ \text{Concentration} = \frac{\text{Moles}}{\text{Volume (L)}} \]

- For \( SO_2 \):
\[ \text{Initial moles of } SO_2 = 1.5 \text{ moles} \]
\[ \text{Concentration of } SO_2 = \frac{1.5 \text{ moles}}{2 \text{ L}} = 0.75 \text{ mol/L} \]

- For \( O_2 \):
\[ \text{Initial moles of } O_2 = 1 \text{ mole} \]
\[ \text{Concentration of } O_2 = \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ mol/L} \]

- For \( SO_3 \):
\[ \text{Initial concentration of } SO_3 = 0 \text{ mol/L} \]

### Step 2: Set Up the Change in Concentrations
At equilibrium, we know the concentration of \( SO_3 \) is \( 0.35 \text{ mol/L} \). We can denote the change in concentration of \( SO_3 \) as \( +0.35 \text{ mol/L} \).

From the stoichiometry of the reaction, for every 2 moles of \( SO_3 \) produced, 2 moles of \( SO_2 \) are consumed and 1 mole of \( O_2 \) is consumed. Therefore, the changes in concentrations will be:

- Change in \( SO_2 \): \( -0.35 \times \frac{2}{2} = -0.35 \)
- Change in \( O_2 \): \( -0.35 \times \frac{1}{2} = -0.175 \)

### Step 3: Calculate Equilibrium Concentrations
Now we can calculate the equilibrium concentrations:

- For \( SO_2 \):
\[ \text{Equilibrium concentration of } SO_2 = 0.75 - 0.35 = 0.40 \text{ mol/L} \]

- For \( O_2 \):
\[ \text{Equilibrium concentration of } O_2 = 0.5 - 0.175 = 0.325 \text{ mol/L} \]

- For \( SO_3 \):
\[ \text{Equilibrium concentration of } SO_3 = 0.35 \text{ mol/L} \]

### Step 4: Write the Expression for \( K_c \)
The equilibrium constant \( K_c \) is expressed as:

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

### Step 5: Substitute the Equilibrium Concentrations
Now we substitute the equilibrium concentrations into the \( K_c \) expression:

\[ K_c = \frac{(0.35)^2}{(0.40)^2 \cdot (0.325)} \]

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

\[ K_c = \frac{0.1225}{0.16 \cdot 0.325} \]
\[ K_c = \frac{0.1225}{0.052} \]
\[ K_c \approx 2.35 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction is approximately \( 2.35 \, \text{mol/L} \).