To calculate the equilibrium constant $K$ for the reaction:

$2 \, \text{SO}_3 (g) \rightleftharpoons 2 \, \text{SO}_2 (g) + \text{O}_2 (g)$

we use the expression for the equilibrium constant in terms of concentrations:

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

Given:

• $[\text{SO}_3] = 1.32 \times 10^{-2} \, \text{M}$
• $[\text{SO}_2] = 2.53 \times 10^{-2} \, \text{M}$
• $[\text{O}_2] = 4.41 \times 10^{-2} \, \text{M}$

Now, let's calculate $K_c$ by substituting these values into the formula.

$K_c = \frac{(2.53 \times 10^{-2})^2 \times (4.41 \times 10^{-2})}{(1.32 \times 10^{-2})^2}$

I'll compute this for you.The equilibrium constant $K_c$ for the reaction at 1260 K is approximately $0.162$.

Would you like any further details or clarifications? Here are 5 related questions to deepen your understanding:

1. How is the equilibrium constant affected by temperature?
2. What is the difference between $K_c$ and $K_p$ in equilibrium expressions?
3. How does Le Chatelier's Principle apply to this reaction?
4. Why do we square the concentrations of $\text{SO}_2$ and $\text{SO}_3$ in the $K_c$ expression?
5. How can the equilibrium constant be used to predict the extent of a reaction?

Tip: The value of the equilibrium constant $K_c$ indicates whether reactants or products are favored at equilibrium; a small value means reactants are favored, while a large value means products are favored.