For this reaction. 2so2+o2<>2so3 the keq is 2.80x10^-4 at equilibrium what is favored

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?

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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.

Explore the implications of the equilibrium constant K_eq = 2.80 x 10^-4 for the reaction 2SO2 + O2 ⇌ 2SO3. Learn why the reactants are favored and gain insights into equilibrium principles relevant for high school chemistry students.

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