Problem 12 - Chemistry and Chemical Reactivity

Question

You place 3.00 mol of pure \(\mathrm{SO}_{3}\) in an \(8.00-\mathrm{L}\). flask at \(1150 \mathrm{K}\). At equilibrium, 0.58 mol of \(\mathrm{O}_{2}\) has been formed. Calculate \(K\) for the reaction at \(1150 \mathrm{K}\) $$ 2 \mathrm{SO}_{3}(\mathrm{g}) \rightleftarrows 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$

Step-by-Step Solution

Answer

The equilibrium constant \(K\) is \(0.0312\).

Step 1: Write the balanced equation

The balanced chemical equation is given as \(2 \text{SO}_3(\text{g}) \rightleftharpoons 2 \text{SO}_2(\text{g}) + \text{O}_2(\text{g})\). It expresses the decomposition of \(\text{SO}_3\) into \(\text{SO}_2\) and \(\text{O}_2\).

Step 2: Initial moles and change in moles

Initially, we have 3.00 moles of \(\text{SO}_3\) and 0 moles of \(\text{SO}_2\) and \(\text{O}_2\). At equilibrium, 0.58 moles of \(\text{O}_2\) are formed. According to the stoichiometry of the reaction, the formation of 0.58 moles of \(\text{O}_2\) corresponds to the consumption of 2 * 0.58 = 1.16 moles of \(\text{SO}_3\) and formation of 1.16 moles of \(\text{SO}_2\).

Step 3: Equilibrium moles calculation

At equilibrium, the moles of \(\text{SO}_3\) remain \(3.00 - 1.16 = 1.84\) moles, \(\text{SO}_2\) is \(1.16\) moles, and \(\text{O}_2\) is \(0.58\) moles.

Step 4: Calculate concentrations at equilibrium

The equilibrium concentrations in the 8.00 L flask are calculated as:\(\text{[SO}_3\text{]} = \frac{1.84}{8.00} = 0.230 \, \text{M}\), \(\text{[SO}_2\text{]} = \frac{1.16}{8.00} = 0.145 \, \text{M}\), and \(\text{[O}_2\text{]} = \frac{0.58}{8.00} = 0.0725 \, \text{M}\).

Step 5: Expression for the equilibrium constant \(K\)

The equilibrium constant expression \(K\) for the reaction \(2\text{SO}_3 \rightleftharpoons 2\text{SO}_2 + \text{O}_2\) is given by \(K = \frac{[\text{SO}_2]^2 [\text{O}_2]}{[\text{SO}_3]^2}\).

Step 6: Calculate the equilibrium constant \(K\)

Substitute the equilibrium concentrations into the expression:\[K = \frac{(0.145)^2 \times 0.0725}{(0.230)^2}\]. This evaluates to \(K \approx 0.0312\).

Key Concepts

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the backward reaction. In the case of our decomposition reaction involving sulfur trioxide
(\(2 \text{SO}_3(\text{g}) \rightleftharpoons 2 \text{SO}_2(\text{g}) + \text{O}_2(\text{g})\)), equilibrium is achieved when the concentration of reactants and products remains constant over time.
At equilibrium, reactions continue occurring, but the concentrations of the substances do not change significantly. This balance is crucial because it allows us to calculate the equilibrium constant, \(K\).

• It gives insight into the position of equilibrium — whether the reactants or products are favored.
• The larger \(K\) is, the more the products are favored at equilibrium.
• If \(K\) is small, the reactants are favored.

Understanding chemical equilibrium helps predict the composition of the mixture at any given time during a reaction under specified conditions.

Reaction Stoichiometry

Stoichiometry refers to the calculation of reactants and products in chemical reactions. It reflects the balanced nature of chemical equations, emphasizing the concept that matter is neither created nor destroyed.
In our reaction:\(2\text{SO}_3 \rightleftharpoons 2\text{SO}_2 + \text{O}_2\), stoichiometry tells us that two moles of \(\text{SO}_3\) decompose into two moles of \(\text{SO}_2\) and one mole of \(\text{O}_2\).
Understanding stoichiometry is crucial for:

• Predicting the amount of products formed in a reaction.
• Determining the required amounts of reactants to achieve a desired product yield.
• Calculating changes in moles when a reaction proceeds to equilibrium.

In this case, for each mole of \(\text{O}_2\) formed, 2 moles of \(\text{SO}_3\) are used, consistently maintaining the stoichiometric ratio outlined in the balanced equation.

Concentration Calculation

Concentration refers to the amount of a substance in a given volume. In equilibrium reactions, calculating concentrations helps us understand the quantities of reactants and products present at equilibrium.
In our sulfur trioxide decomposition, concentrations are calculated using the formula:\(\text{Concentration} = \frac{\text{Number of moles}}{\text{Volume in liters}}.\)
For instance, initially, there were 3 moles of \(\text{SO}_3\) in an 8-liter container, but at equilibrium, this number decreased, leading to new concentrations.

• The concentration of \(\text{SO}_3\) became 0.230 M.
• The concentration of \(\text{SO}_2\) was 0.145 M.
• For \(\text{O}_2\), it reached 0.0725 M.

Concentration calculations are foundational for determining the equilibrium constant, and ensure that we have accurate data to analyze chemical reactions effectively.