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. (a) `5.1"L mol"^(-1)`
B. (b) `1.4"L mol"^(-1)`
C. (c) `0.6"L mol"^(-1)`
D. (d) `2.35"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: Determine Initial Concentrations
We are given:
- Initial moles of \( SO_2 = 1.5 \) moles
- Initial moles of \( O_2 = 1 \) mole
- Volume of the vessel = 2 L

To find the initial concentrations, we use the formula:

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

Calculating the initial concentrations:
- Initial concentration of \( SO_2 \):
\[ \text{[SO}_2\text{]}_0 = \frac{1.5 \text{ moles}}{2 \text{ L}} = 0.75 \text{ mol L}^{-1} \]
- Initial concentration of \( O_2 \):
\[ \text{[O}_2\text{]}_0 = \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ mol L}^{-1} \]

### Step 2: Set Up the Change in Concentrations
Let \( x \) be the change in concentration of \( SO_3 \) at equilibrium. According to the stoichiometry of the reaction:
- For every 2 moles of \( SO_2 \) that react, 1 mole of \( O_2 \) reacts and 2 moles of \( SO_3 \) are formed.

At equilibrium:
- Concentration of \( SO_2 \):
\[ \text{[SO}_2\text{]} = 0.75 - 2x \]
- Concentration of \( O_2 \):
\[ \text{[O}_2\text{]} = 0.5 - x \]
- Concentration of \( SO_3 \):
\[ \text{[SO}_3\text{]} = 2x \]

### Step 3: Use Given Equilibrium Concentration
We are given that at equilibrium, the concentration of \( SO_3 \) is \( 0.35 \text{ mol L}^{-1} \):
\[ 2x = 0.35 \implies x = 0.175 \]

### Step 4: Calculate Equilibrium Concentrations
Now substituting \( x \) back into the expressions for the equilibrium concentrations:
- Concentration of \( SO_2 \):
\[ \text{[SO}_2\text{]} = 0.75 - 2(0.175) = 0.75 - 0.35 = 0.40 \text{ mol L}^{-1} \]
- Concentration of \( O_2 \):
\[ \text{[O}_2\text{]} = 0.5 - 0.175 = 0.325 \text{ mol L}^{-1} \]

### Step 5: Write the Expression for \( K_c \)
The equilibrium constant \( K_c \) is given by the expression:
\[ K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]} \]

### Step 6: Substitute Equilibrium Concentrations into the Expression
Substituting the equilibrium concentrations into the \( K_c \) expression:
\[ K_c = \frac{(0.35)^2}{(0.40)^2(0.325)} \]

### Step 7: Calculate \( K_c \)
Calculating the values:
\[ K_c = \frac{0.1225}{0.16 \times 0.325} = \frac{0.1225}{0.052} \approx 2.35 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction is approximately \( 2.35 \text{ mol L}^{-1} \).

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