One mole of `SO_(3)` was placed in a litre reaction vessel at a certain temperature. The following equilibrium was established `2SO_(3)(g)hArr 2SO_(2)(g)+O_(2)(g)` At equilibrium 0.6 moles of `SO_(2)` were formed. The equilibrium constant of the reaction will be

Options

A. 0.36
B. 0.45
C. 0.54
D. 0.675

Answer

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

\[ 2 \text{SO}_3(g) \rightleftharpoons 2 \text{SO}_2(g) + \text{O}_2(g) \]

### Step 1: Write the Initial and Equilibrium Concentrations

Initially, we have 1 mole of \( \text{SO}_3 \) in a 1-litre vessel, so the initial concentration of \( \text{SO}_3 \) is:

\[ [\text{SO}_3]_{initial} = 1 \, \text{mol/L} \]

At equilibrium, we are told that 0.6 moles of \( \text{SO}_2 \) are formed. Since the stoichiometry of the reaction shows that 2 moles of \( \text{SO}_2 \) are produced for every 2 moles of \( \text{SO}_3 \) that react, we can find the change in concentration of \( \text{SO}_3 \).

### Step 2: Determine the Change in Concentrations

Let \( x \) be the amount of \( \text{SO}_3 \) that dissociates. According to the stoichiometry of the reaction:

- For every 2 moles of \( \text{SO}_3 \) that react, 2 moles of \( \text{SO}_2 \) and 1 mole of \( \text{O}_2 \) are produced.
- If 0.6 moles of \( \text{SO}_2 \) are formed, then:

\[ \text{From the stoichiometry: } 2 \text{SO}_3 \rightarrow 2 \text{SO}_2 \Rightarrow 0.6 \text{ moles of } \text{SO}_2 \text{ means } 0.6 \text{ moles of } \text{SO}_3 \text{ reacted.} \]

Thus, we can say:

\[ x = 0.6 \text{ moles of } \text{SO}_3 \text{ reacted.} \]

### Step 3: Calculate the Equilibrium Concentrations

Now we can calculate the equilibrium concentrations:

- Concentration of \( \text{SO}_3 \) at equilibrium:

\[ [\text{SO}_3]_{eq} = [\text{SO}_3]_{initial} - x = 1 - 0.6 = 0.4 \, \text{mol/L} \]

- Concentration of \( \text{SO}_2 \) at equilibrium:

\[ [\text{SO}_2]_{eq} = 0.6 \, \text{mol/L} \]

- Concentration of \( \text{O}_2 \) at equilibrium (since 0.6 moles of \( \text{SO}_3 \) produce 0.3 moles of \( \text{O}_2 \)):

\[ [\text{O}_2]_{eq} = \frac{0.6}{2} = 0.3 \, \text{mol/L} \]

### Step 4: Write the Expression for the Equilibrium Constant \( K_c \)

The equilibrium constant \( K_c \) for the reaction is given by:

\[ K_c = \frac{[\text{SO}_2]^2 [\text{O}_2]}{[\text{SO}_3]^2} \]

### Step 5: Substitute the Equilibrium Concentrations into the Expression

Now substituting the equilibrium concentrations:

\[ K_c = \frac{(0.6)^2 \cdot (0.3)}{(0.4)^2} \]

Calculating this gives:

\[ K_c = \frac{0.36 \cdot 0.3}{0.16} = \frac{0.108}{0.16} = 0.675 \]

### Final Answer

Thus, the equilibrium constant \( K_c \) for the reaction is:

\[ K_c = 0.675 \]

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