1.0 mol of 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... 1.0 mol of 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 is asking to calculate the equilibrium constant (Keq) for a reaction where sulfur dioxide and oxygen react to form sulfur trioxide. We are given the initial amounts of sulfur dioxide and oxygen, the volume of the container, the temperature, and how much sulfur trioxide is formed at equilibrium. We will need to use an ICE table (Initial, Change, Equilibrium) to determine the concentrations of each species at equilibrium and then use these values to calculate Keq.
Answer
$K_{eq} \approx 283.0$
Answer for screen readers
$K_{eq} \approx 283.0$
Steps to Solve
- Write the balanced chemical equation
The balanced chemical equation for 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
Since the volume of the container is 1.00 L, the number of moles is equal to the concentration in mol/L.
| $2SO_2$ | $O_2$ | $2SO_3$ | |
|---|---|---|---|
| Initial (I) | 1.00 | 1.00 | 0 |
| Change (C) | -2x | -x | +2x |
| Equil (E) | 1.00-2x | 1.00-x | 2x |
- Determine the value of x
We are given that at equilibrium, the amount of $SO_3$ formed is 0.925 mol. From the ICE table, the equilibrium concentration of $SO_3$ is $2x$. Therefore:
$2x = 0.925$ $x = 0.4625$
- Calculate the equilibrium concentrations
Now we can calculate the equilibrium concentrations of all the species:
$[SO_2] = 1.00 - 2x = 1.00 - 2(0.4625) = 1.00 - 0.925 = 0.075 , M$ $[O_2] = 1.00 - x = 1.00 - 0.4625 = 0.5375 , M$ $[SO_3] = 2x = 0.925 , M$
- Write the expression for the equilibrium constant, Keq
$K_{eq} = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$
- Calculate the value of Keq
Plug in the equilibrium concentrations into the expression:
$K_{eq} = \frac{(0.925)^2}{(0.075)^2(0.5375)}$ $K_{eq} = \frac{0.855625}{(0.005625)(0.5375)}$ $K_{eq} = \frac{0.855625}{0.0030234375}$ $K_{eq} \approx 283.0$
$K_{eq} \approx 283.0$
More Information
The equilibrium constant, $K_{eq}$, is a dimensionless quantity. It indicates the ratio of products to reactants at equilibrium. A large $K_{eq}$ value, like the one calculated, signifies that the reaction favors the formation of products at equilibrium.
Tips
- Forgetting to square the concentrations
Remember to raise the concentrations to the power of their stoichiometric coefficients in the equilibrium constant expression. For example, $[SO_3]$ and $[SO_2]$ are squared because their stoichiometric coefficient is 2.
- Incorrectly calculating the change in concentrations
Make sure the 'change' row in your ICE table reflects the stoichiometry of the balanced chemical equation. For every 2 moles of $SO_2$ that react, 1 mole of $O_2$ also reacts, and 2 moles of $SO_3$ are formed.
- Using moles instead of concentrations
The equilibrium constant expression uses concentrations (mol/L), not moles. Since the volume was 1.00 L in this problem, the moles and molarity are numerically equal. If the volume was different, you would need to divide the number of moles by the volume to get the concentration.
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