Problem 53 - Chemistry

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 $K$ for the reaction $2 \mathrm{SO_{3}(g)} \rightleftharpoons 2 \mathrm{SO_{2}(g)} + \mathrm{O_{2}(g)}$ under the given conditions is 4.5.

Step 1: Write the reaction and determine the initial concentrations

The given reaction is: \[2 \mathrm{SO_{3}(g)} \rightleftharpoons 2 \mathrm{SO_{2}(g)} + \mathrm{O_{2}(g)}\] We are given initial moles of \(\mathrm{SO_{3}}\) and volume, therefore, we can calculate the initial concentration of \(\mathrm{SO_{3}}\) using the formula: Concentration = Moles/Volume \[[\mathrm{SO_{3}}]_0 = \frac{12.0\: \text{moles}}{3.0\: \text{L}} = 4.0\: M\] As no other amounts have been specified, we assume that initially there are no \(\mathrm{SO_{2}}\) and \(\mathrm{O_{2}}\).

Step 2: Write an expression for the changes in concentrations

Assume that x moles of \(\mathrm{SO_{3}}\) dissociates. Since the stoichiometry is 2:2:1, we see that the change in concentration of \(\mathrm{SO_{2}}\) will be 2x and that of \(\mathrm{O_{2}}\) will be x. So the changes in concentrations for the reaction are: \[\begin{cases} \Delta [\mathrm{SO_{3}}] = -2x \\ \Delta [\mathrm{SO_{2}}] = 2x \\ \Delta [\mathrm{O_{2}}] = x \end{cases}\]

Step 3: Form the equilibrium expressions

At equilibrium, we can write the concentrations as follows: \[[\mathrm{SO_{3}}]_{eq} = 4.0\: M - 2x\] \[[\mathrm{SO_{2}}]_{eq} = 2x\] \[[\mathrm{O_{2}}]_{eq} = x\] We are given that there are 3.0 moles of \(\mathrm{SO_{2}}\) at equilibrium. We will use this information to find the value of x. \[2x = 3.0\: \text{moles}\] \[x = 1.5\: \text{moles}\] Now, we can find the equilibrium concentrations by plugging in the value of x to the equilibrium expressions.

Step 4: Calculate the equilibrium concentrations

Using the value of x we just found, we can calculate the equilibrium concentrations: \[[\mathrm{SO_{3}}]_{eq} = 4.0\: M - 2(1.5\: \text{moles}) = 1.0\: M\] \[[\mathrm{SO_{2}}]_{eq} = 2(1.5\: \text{moles}) = 3.0\: M\] \[[\mathrm{O_{2}}]_{eq} = 1.5\: \text{moles} = 0.5\: M\]

Step 5: Determine the equilibrium constant (K)

We can now determine the equilibrium constant K for the reaction by forming the expression for K and plugging in the equilibrium concentrations we found in the previous step: \[K = \frac{[\mathrm{SO_{2}}]^2 \times [\mathrm{O_{2}}]}{[\mathrm{SO_{3}}]^2}\] Substituting the equilibrium concentrations: \[K = \frac{(3.0\: M)^2 \times (0.5\: M)}{(1.0\: M)^2}\] \[K = \frac{9.0 \times 0.5}{1.0}\] \[K = 4.5\] So, the equilibrium constant K for this reaction is 4.5.

Key Concepts

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction achieves a state where the concentrations of reactants and products remain constant over time. This doesn't mean that the reactions have stopped. Instead, the rates of the forward and reverse reactions are equal, leading to a stable ratio of product to reactant concentrations. Understanding chemical equilibrium is essential because it allows us to predict the concentrations of various species in a reaction mixture under equilibrium conditions.

In the case of our dissociation reaction of sulfur trioxide (\(\mathrm{SO}_3\)) the reaction has reached an equilibrium when the amounts of \(\mathrm{SO}_3\), \(\mathrm{SO}_2\), and \(\mathrm{O}_2\) remain unchanged. At this point, the system has balanced the rate at which \(\mathrm{SO}_3\) dissociates into \(\mathrm{SO}_2\) and \(\mathrm{O}_2\) with the rate at which \(\mathrm{SO}_2\) and \(\mathrm{O}_2\) react to form \(\mathrm{SO}_3\) again. The equilibrium constant, denoted as \(K\), helps quantify this equilibrium by representing the ratio of the products' concentrations to the reactants' concentrations raised to the power of their stoichiometric coefficients.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It's like a recipe that dictates how much of each ingredient is needed and what will be produced. In our reaction of interest, 2 moles of \(\mathrm{SO}_3\) dissociate to produce 2 moles of \(\mathrm{SO}_2\) and 1 mole of \(\mathrm{O}_2\).

Using stoichiometry, we can determine how changes in the quantity of one compound will affect the others. For example, in this reaction, if some \(\mathrm{SO}_3\) is used up, twice the amount of \(\mathrm{SO}_2\) is produced, and for every 2 moles of \(\mathrm{SO}_3\) that react, 1 mole of \(\mathrm{O}_2\) forms.

• Initial concentrations can be determined from the moles and volume provided.
• Stoichiometry helps us define the 'change' variable \(x\) in the expression which correlates how the reactant \(\mathrm{SO}_3\) decreases as the products \(\mathrm{SO}_2\) and \(\mathrm{O}_2\) increase.

This stoichiometric balance is crucial in finding the equilibrium concentrations of all participating substances.

Dissociation Reaction

A dissociation reaction is a chemical process in which a compound breaks down into smaller molecules or ions. The dissociation we have here involves sulfur trioxide \((\mathrm{SO}_3)\) breaking down into sulfur dioxide \((\mathrm{SO}_2)\) and oxygen gas \((\mathrm{O}_2)\). The balanced equation, \(2 \mathrm{SO}_3 (g) \rightleftharpoons 2 \mathrm{SO}_2 (g) + \mathrm{O}_2 (g)\), shows how each molecule of \(\mathrm{SO}_3\) splits into these products.

• In equilibrium, the reaction doesn’t fully shift to either side but has both reactants and products present.
• Dissociation reactions are characterized by a specific stoichiometry, which dictates how substances change during the reaction.

Understanding how dissociation happens in terms of molecular interaction and collision theories helps predict how different conditions affect the extent of a reaction under equilibrium.

Concentration Calculation

Calculating concentrations accurately allows us to determine the equilibrium condition and the equilibrium constant \(K\). Concentration is a measure of how much of a substance is present in a given volume and is typically expressed in molarity (M), which is moles per liter.

In our exercise, initial concentrations are calculated using the formula \(\text{Concentration} = \frac{\text{Moles}}{\text{Volume}}\). Given the moles of \(\mathrm{SO}_3\) and an initial volume of the container, we find the initial concentration of \(\mathrm{SO}_3\). As no initial concentrations for \(\mathrm{SO}_2\) and \(\mathrm{O}_2\) are given, they start at zero.

Determining how concentrations change requires understanding the stoichiometry of the reaction. By defining the changes as \(x\), we can use the balanced reaction to express the equilibrium concentrations of all species. By using known equilibrium conditions, such as the presence of 3 moles of \(\mathrm{SO}_2\), we back-calculate \(x\) and determine concentrations of all components at equilibrium, which ultimately allows us to calculate \(K\). This step-by-step approach ensures the accuracy of our calculations throughout the process.