The reaction given is:

$2 \text{SO}_2 + \text{O}_2 \rightleftharpoons 2 \text{SO}_3$

with an equilibrium constant ($K_\text{eq}$) of $2.80 \times 10^{-4}$.

Understanding the Equilibrium Constant

The equilibrium constant, $K_\text{eq}$, indicates the ratio of the concentrations of products to reactants at equilibrium. Specifically:

[ K_\text{eq} = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 \cdot [\text{O}_2]} ]

What $K_\text{eq}$ Value Implies

• A small $K_\text{eq}$ value (less than 1), like $2.80 \times 10^{-4}$, indicates that the equilibrium position favors the reactants.
• This means that, at equilibrium, the concentration of $\text{SO}_2$ and $\text{O}_2$ is higher than the concentration of $\text{SO}_3$.

Answer

Since $K_\text{eq} = 2.80 \times 10^{-4}$, the reactants ($\text{SO}_2$ and $\text{O}_2$) are favored at equilibrium.

Would you like more details on interpreting equilibrium constants?

---

Here are five related questions to deepen your understanding:

1. What does it mean if $K_\text{eq}$ is greater than 1 in a reaction?
2. How does temperature affect the value of $K_\text{eq}$ for this reaction?
3. How would adding more $\text{O}_2$ affect the equilibrium position?
4. What would the effect of a catalyst be on the value of $K_\text{eq}$?
5. How do you calculate $K_\text{eq}$ if given the equilibrium concentrations of all species?

Tip: For reversible reactions, a small $K_\text{eq}$ suggests that the reaction does not proceed far towards the products, while a large $K_\text{eq}$ suggests the reaction heavily favors products at equilibrium.