1.0 mol sulfur dioxide and 1.00 mol of oxygen are sealed in a 1.00 L rigid container at 1000K. When equilibrium is established, 0.925 mol of sulfur trioxide is formed. Calculate Ke... 1.0 mol sulfur dioxide and 1.00 mol of oxygen are sealed in a 1.00 L rigid container at 1000K. When equilibrium is established, 0.925 mol of sulfur trioxide is formed. Calculate Keq for this reaction.
Understand the Problem
The question describes a chemical reaction at equilibrium involving sulfur dioxide, oxygen, and sulfur trioxide. Given initial amounts of reactants and the amount of product formed at equilibrium, the task is to calculate the equilibrium constant (Keq) for the reaction.
Answer
$K_{eq} \approx 283.0$
Answer for screen readers
$K_{eq} \approx 283.0$
Steps to Solve
- Write the balanced chemical equation
The reaction between sulfur dioxide ($SO_2$) and oxygen ($O_2$) to form sulfur trioxide ($SO_3$) is:
$$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$$
- Set up the ICE table
ICE stands for Initial, Change, and Equilibrium. We use this table to track the concentrations of the reactants and products. Since the volume is 1.00 L, the number of moles is equal to the molar concentration.
| $2SO_2$ | $O_2$ | $2SO_3$ | |
|---|---|---|---|
| Initial (I) | 1.00 | 1.00 | 0 |
| Change (C) | -2x | -x | +2x |
| Equil (E) | 1.00-2x | 1.00-x | 2x |
- Determine the change (x)
The problem states that at equilibrium, 0.925 mol of $SO_3$ is formed. From the ICE table, we know that the equilibrium concentration of $SO_3$ is $2x$. $2x = 0.925$, so $x = 0.4625$
- Calculate equilibrium concentrations
Substitute the value of $x$ back into the Equilibrium row of the ICE table:
$[SO_2] = 1.00 - 2x = 1.00 - 2(0.4625) = 1.00 - 0.925 = 0.075 , M$
$[O_2] = 1.00 - x = 1.00 - 0.4625 = 0.5375 , M$
$[SO_3] = 2x = 0.925 , M$
- Write the expression for Keq
$K_{eq} = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$
- Calculate Keq
Substitute the equilibrium concentrations into the Keq expression:
$K_{eq} = \frac{(0.925)^2}{(0.075)^2(0.5375)} = \frac{0.855625}{0.005625 * 0.5375} = \frac{0.855625}{0.0030234375} \approx 283.0$
$K_{eq} \approx 283.0$
More Information
The equilibrium constant, $K_{eq}$, indicates the ratio of products to reactants at equilibrium. A large $K_{eq}$ value, such as 283.0, suggests that the products are highly favored at equilibrium.
Tips
- Forgetting to square the concentrations of $SO_2$ and $SO_3$ in the Keq expression, as dictated by the balanced chemical equation.
- Incorrectly calculating the change in concentrations based on the stoichiometry of the reaction. For example, the change in $SO_2$ is $-2x$ while the change in $O_2$ is $-x$.
- Using the initial concentrations directly in the Keq expression instead of calculating the equilibrium concentrations.
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