Problem 47 - Chemistry An Atoms First Approach

Question

At a particular temperature, 12.0 moles of \(\mathrm{SO}_{3}\) is placed into a 3.0-L rigid container, and the \(\mathrm{SO}_{3}\) dissociates by the reaction $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$. At equilibrium, 3.0 moles of \(\mathrm{SO}_{2}\) is present. Calculate \(K\) for this reaction.

Step-by-Step Solution

Answer

The equilibrium constant for the given reaction is \(K = 0.056\).

Step 1: Determine initial and equilibrium moles

At the beginning, we have 12.0 moles of \(\mathrm{SO}_3\) and 0 moles of \(\mathrm{SO}_2\) and \(\mathrm{O}_2\). When the reaction reaches equilibrium, we know there are 3.0 moles of \(\mathrm{SO}_2\). Since the stoichiometry of the reaction is \(2\mathrm{SO}_3 \rightleftharpoons 2\mathrm{SO}_2 + 1\mathrm{O}_2\), let's find out how many moles of \(\mathrm{SO}_3\) have reacted and also the moles of \(\mathrm{O}_2\) at equilibrium: 1. Moles of \(\mathrm{SO}_{2}\) formed = 3.0 moles 2. Moles of \(\mathrm{SO}_{3}\) reacted = Moles of \(\mathrm{SO}_{2}\) formed = 3.0 moles 3. Moles of \(\mathrm{SO}_{3}\) remaining = 12.0 (initial) - 3.0 (reacted) = 9.0 moles 4. Moles of \(\mathrm{O}_{2}\) formed = 0.5 × Moles of \(\mathrm{SO}_{2}\) formed = 0.5 × 3.0 = 1.5 moles

Step 2: Determine initial and equilibrium concentrations

Now, we can convert moles to concentrations by dividing with the volume of the container (3.0 liters). Initial concentrations: 1. [\(\mathrm{SO}_3\)] = 12.0 moles / 3.0 L = 4.0 M 2. [\(\mathrm{SO}_2\)] = 0 moles / 3.0 L = 0 M 3. [\(\mathrm{O}_2\)] = 0 moles / 3.0 L = 0 M Equilibrium concentrations: 1. [\(\mathrm{SO}_3\)] = 9.0 moles / 3.0 L = 3.0 M 2. [\(\mathrm{SO}_2\)] = 3.0 moles / 3.0 L = 1.0 M 3. [\(\mathrm{O}_2\)] = 1.5 moles / 3.0 L = 0.5 M

Step 3: Calculate the equilibrium constant K using the reaction quotient formula

Knowing the equilibrium concentrations, we can use the reaction quotient formula: $$K = \frac{[\mathrm{SO}_{2}]^2[\mathrm{O}_{2}]}{[\mathrm{SO}_{3}]^2}$$ Substitute the equilibrium concentrations: $$K = \frac{(1.0 \,\mathrm{M})^2 (0.5\, \mathrm{M})}{(3.0\, \mathrm{M})^2}$$ Simplify the expression: $$K = \frac{0.5}{9}$$ Calculate the value: $$K = 0.056$$ The equilibrium constant for the given reaction is 0.056.

Key Concepts

Equilibrium Constant (K)

In the realm of chemical equilibrium, the Equilibrium Constant, denoted as \( K \), plays a crucial role. It quantifies the balance between the products and reactants in a reversible chemical reaction at equilibrium. Understanding \( K \) requires us to dive into a specific equation, often referred to as the reaction quotient formula. For this particular reaction, where sulfur trioxide \( \mathrm{SO}_{3} \) dissociates into sulfur dioxide \( \mathrm{SO}_{2} \) and oxygen \( \mathrm{O}_{2} \), the equilibrium constant expression is described by:

• \( K = \frac{[\mathrm{SO}_{2}]^2[\mathrm{O}_{2}]}{[\mathrm{SO}_{3}]^2} \)

Here, the square brackets denote the equilibrium concentrations of the substances involved. By calculating \( K \), you are essentially measuring the extent to which the forward reaction occurs relative to the reverse reaction at a given temperature. A \( K \) value greater than 1 suggests that the products are favored at equilibrium, while a \( K \) less than 1 indicates a favor towards the reactants. In this exercise, the calculated \( K \) value of 0.056 indicates the presence of more reactants than products at equilibrium.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. Each coefficient in a balanced chemical equation directly corresponds to the number of moles of a substance. Understanding this concept is fundamental as it allows us to predict how much product will form from a given amount of reactant.
In the dissociation reaction of sulfur trioxide \( 2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \), stoichiometry tells us:

• Two moles of \( \mathrm{SO}_{3} \) break down to form two moles of \( \mathrm{SO}_{2} \).
• Additionally, for every two moles of \( \mathrm{SO}_{3} \) that dissociate, one mole of \( \mathrm{O}_{2} \) is produced.

This stoichiometric relationship is crucial in determining how the concentration of each species changes as the system reaches equilibrium. Given that 3.0 moles of \( \mathrm{SO}_{2} \) are formed, 3.0 moles of \( \mathrm{SO}_{3} \) must have reacted based on a 1:1 stoichiometry with \( \mathrm{SO}_{2} \), and consequently, 1.5 moles of \( \mathrm{O}_{2} \) were produced, following a 2:1 stoichiometry with \( \mathrm{SO}_{3} \).

Equilibrium Concentration

Equilibrium concentration refers to the amount of each reactant and product in a reaction mixture when the rates of the forward and reverse reactions have balanced. To find equilibrium concentrations, start with known initial concentrations and apply stoichiometry to determine the shifts.
In this exercise, sulfur trioxide is initially present at 4.0 M concentration. As the reaction progresses to equilibrium, 3.0 moles of \( \mathrm{SO}_{2} \) form, indicating that the concentration of \( \mathrm{SO}_{3} \) has decreased. At equilibrium, we compute:

• \( \text{[SO}_3\text{]} = 3.0 \, ext{M } \) since 9.0 moles remain in a 3.0 L container.
• \( \text{[SO}_2\text{]} = 1.0 \, ext{M } \) because 3.0 moles have formed in 3.0 L of space.
• \( \text{[O}_2\text{]} = 0.5 \, ext{M } \) as 1.5 moles exist over the same volume.

These equilibrium concentrations are pivotal when substituting into the equilibrium constant expression to solve for \( K \). Understanding equilibrium concentration assists in predicting reaction behavior under different conditions, which is critical for both academic and practical chemistry applications.