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 \(K_{c}\) for the reaction at \(1150 \mathrm{K}\). $$ 2 \mathrm{sO}_{3}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$

Step-by-Step Solution

Answer

\(K_c \approx 0.0288\)

Step 1: Understand the Reaction

The balanced chemical reaction is \(2 \mathrm{SO}_{3}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\). This equilibrium involves the decomposition of sulfur trioxide (\(\mathrm{SO}_3\)) into sulfur dioxide (\(\mathrm{SO}_2\)) and oxygen gas (\(\mathrm{O}_2\)).

Step 2: Determine Initial Amounts and Changes

Initially, we have 3.00 mol of \(\mathrm{SO}_3\). No \(\mathrm{SO}_2\) or \(\mathrm{O}_2\) is present initially. If 0.58 mol of \(\mathrm{O}_2\) is formed at equilibrium, then by stoichiometry, 1.16 mol (\(2 \times 0.58\) mol) of \(\mathrm{SO}_3\) is consumed to produce 1.16 mol of \(\mathrm{SO}_2\).

Step 3: Calculate Moles at Equilibrium

At equilibrium, the moles of each substance are:- \(\mathrm{SO}_3\): \(3.00 - 1.16 = 1.84\) mol- \(\mathrm{SO}_2\): \(1.16\) mol- \(\mathrm{O}_2\): \(0.58\) mol

Step 4: Calculate Concentrations

Since the volume of the flask is 8.00 L, we calculate the concentrations as follows:- \([\mathrm{SO}_3]\): \(\frac{1.84}{8.00} = 0.230\,\mathrm{mol/L}\)- \([\mathrm{SO}_2]\): \(\frac{1.16}{8.00} = 0.145\,\mathrm{mol/L}\)- \([\mathrm{O}_2]\): \(\frac{0.58}{8.00} = 0.0725\,\mathrm{mol/L}\)

Step 5: Write the Expression for the Equilibrium Constant, \(K_c\)

The equilibrium constant expression for the reaction is:\[ K_c = \frac{([\mathrm{SO}_2]^2 [\mathrm{O}_2])}{[\mathrm{SO}_3]^2} \]

Step 6: Calculate \(K_c\)

Substitute the equilibrium concentrations into the expression:\[ K_c = \frac{(0.145)^2 \times 0.0725}{(0.230)^2} \]Calculate:1. \((0.145)^2 = 0.021025\)2. \(0.021025 \times 0.0725 = 0.0015253125\)3. \((0.230)^2 = 0.0529\)4. Finally, \(K_c = \frac{0.0015253125}{0.0529} \approx 0.0288\)

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the concentrations of the reactants and products stop changing with time. This happens because the rates of the forward and reverse reactions become equal. In the context of the given exercise, the reaction involves the decomposition of sulfur trioxide (\(\mathrm{SO}_3\)) into sulfur dioxide (\(\mathrm{SO}_2\)) and oxygen (\(\mathrm{O}_2\)).
When the system reaches equilibrium, the quantity of each substance remains constant, though not necessarily equal between reactants and products. Here, 0.58 mol of \(\mathrm{O}_2\) is formed, signaling that equilibrium has been achieved.
It is crucial to understand that equilibrium does not mean the reactants and products are in equal concentrations. It simply means their concentrations are stable over time.
Outcomes of chemical equilibrium are described by equilibrium constants like \(K_c\), which help us understand the relative quantities of products and reactants at equilibrium.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It uses the coefficients from the balanced chemical equation to determine how much of each substance is consumed or produced. In this exercise, the reaction is given as: \[2 \mathrm{SO}_{3}(\text{g}) \rightleftharpoons 2 \mathrm{SO}_2(\text{g}) + \mathrm{O}_2(\text{g})\]. This balanced equation shows us that two moles of \(\mathrm{SO}_3\) decompose to form two moles of \(\mathrm{SO}_2\) and one mole of \(\mathrm{O}_2\).

• For every mole of \(\mathrm{O}_2\) produced, 2 moles of \(\mathrm{SO}_3\) are consumed.
• Similarly, each mole of \(\mathrm{SO}_2\) formed corresponds to one mole of \(\mathrm{SO}_3\) degraded.

By understanding stoichiometry, we calculated that 0.58 mol of \(\mathrm{O}_2\) production corresponds to 1.16 mol of \(\mathrm{SO}_3\) being consumed and an equal amount of \(\mathrm{SO}_2\) being produced. This step is fundamental to proceeding with the calculation of equilibrium concentrations.

Equilibrium Concentration

Equilibrium concentration refers to the amount of each reactant and product present in a system at equilibrium. To find these concentrations, it is crucial to know the initial amounts and any changes that occur as the system reaches equilibrium.
In this exercise, initially, we have 3.00 moles of \(\mathrm{SO}_3\). No \(\mathrm{SO}_2\) or \(\mathrm{O}_2\) is present at the start. Through the reaction stoichiometry, we determine that when equilibrium is reached, the moles of each substance are: 1.84 mol of \(\mathrm{SO}_3\), 1.16 mol of \(\mathrm{SO}_2\), and 0.58 mol of \(\mathrm{O}_2\).

• The concentration \([\mathrm{SO}_3]\) at equilibrium is \(\frac{1.84}{8.00} = 0.230\,\mathrm{mol/L}\)
• The concentration \([\mathrm{SO}_2]\) at equilibrium is \(\frac{1.16}{8.00} = 0.145\,\mathrm{mol/L}\)
• The concentration \([\mathrm{O}_2]\) at equilibrium is \(\frac{0.58}{8.00} = 0.0725\,\mathrm{mol/L}\)

These concentrations are then used to calculate the equilibrium constant \(K_c\), providing insight into the reaction's dynamics at that specific temperature and pressure.