Problem 111 - General Chemistry: Principles and Modern Applications

Question

An equilibrium mixture of \(\mathrm{SO}_{2}, \mathrm{SO}_{3},\) and \(\mathrm{O}_{2}\) gases is maintained in a \(2.05 \mathrm{L}\) flask at a temperature at which \(K_{\mathrm{c}}=35.5\) for the reaction $$2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})$$ (a) If the numbers of moles of \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3}\) in the flask are equal, how many moles of \(\mathrm{O}_{2}\) are present? (b) If the number of moles of \(\mathrm{SO}_{3}\) in the flask is twice the number of moles of \(\mathrm{SO}_{2}\), how many moles of \(\mathrm{O}_{2}\) are present?

Step-by-Step Solution

Answer

The number of moles of \(O_2\) are approximately 0.028 for case (a) and approximately 0.113 for case (b).

Step 1: Analyzing Case (a)

In case (a), the number of moles of \(\mathrm{SO_{2}}\) and \(\mathrm{SO_{3}}\) are equal. This implies that \(n(\mathrm{SO_{2}}) = n(\mathrm{SO_{3}})\). Let's assume this to be 'x' moles. Additionally, let's assume the number of moles of \(O_2\) to be 'y'. The expression for the equilibrium constant for the given reaction is \[K_c = \frac{(n(\mathrm{SO_{3}})/V)^2}{(n(\mathrm{SO_{2}})/V)^2\times (n(\mathrm{O_{2}})/V)}\] Since the volume, V, is the same for all the gases, it cancels out from the equation and we have: \[K_c = \frac{(n(\mathrm{SO_{3}}))^2}{(n(\mathrm{SO_{2}}))^2\times n(\mathrm{O_{2}})}\] Using this formula, we insert the known values and solve for y.

Step 2: Calculating moles of \(O_2\) for Case (a)

For case (a), since \(n(\mathrm{SO_{2}}) = n(\mathrm{SO_{3}}) =x\), the formula becomes: \[35.5 = \frac{x^2}{x^2\times y} = \frac{1}{y}\] Solving for y, we get \[y = \frac{1}{35.5} = 0.028 moles\] Thus, there are approximately 0.028 moles of \(O_2\) in the flask.

Step 3: Analyzing Case (b)

In case (b), the number of moles of \(\mathrm{SO_{3}}\) is twice the number of moles of \(\mathrm{SO_{2}}\). This implies that \(n(\mathrm{SO_{2}}) =x\) and \(n(\mathrm{SO_{3}}) =2x\). We use the same expression for equilibrium and adjust the values according to these quantities to find \(y\) the moles of \(O_2\).

Step 4: Calculating moles of \(O_2\) for Case (b)

To solve for the moles of \(O_2\) in case (b), we insert the given values into the equilibrium formula: \[35.5 = \frac{(2x)^2}{x^2 \times y}\] Solving for y, we get \[y = \frac{4}{35.5} = 0.113 moles\] Thus, there are approximately 0.113 moles of \(O_2\) when the number of moles of \(\mathrm{SO_{3}}\) is twice the number of moles of \(\mathrm{SO_{2}}\).

Key Concepts

Chemical Equilibrium Constant

The chemical equilibrium constant, denoted by \(K_c\), is a fundamental concept in understanding equilibrium in chemical reactions. For the reaction provided, where \(2 \mathrm{SO}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\), \(K_c\) offers a quantitative measure of the concentration ratio of products to reactants when the system has reached equilibrium. This number is specific to a given temperature. In the provided problem, \(K_c\) is 35.5.
The equilibrium constant expression for a reaction is formulated as the concentration of products raised to their stoichiometric coefficients, divided by the concentration of reactants again raised to their stoichiometric coefficients. This can be illustrated by:

• Products over Reactants: \(K_c = \frac{[\mathrm{SO}_{3}]^2}{[\mathrm{SO}_{2}]^2[\mathrm{O}_{2}]}\)

Understanding \(K_c\) is critical because it describes the extent to which a reaction proceeds to completion under a given set of conditions. A high \(K_c\) value indicates that, at equilibrium, the reaction mixture contains a larger concentration of products relative to reactants, as seen in this problem's context.

Reactant and Product Concentration

The concentration of reactants and products is at the heart of understanding and calculating chemical equilibria. In chemical reactions at equilibrium, the concentrations of the reactants and products remain constant over time. This balance results from the forward and reverse reactions occurring at the same rate.
When solving for different variables, such as the number of moles of \(O_2\) at equilibrium, we use the concentrations of other species in the mixture. In scenarios (a) and (b) of the problem, we know:

• Case (a): Equal moles of \(\mathrm{SO_{2}}\) and \(\mathrm{SO_{3}}\) lead to solving for \(O_2\) using the equilibrium expression.
• Case (b): \(\mathrm{SO_{3}}\) is twice \(\mathrm{SO_{2}}\) indicated a different set-up in the equilibrium expression.

By substituting these relationships into the equilibrium expression, the students can solve for unknown concentrations, such as \(y = 0.028 \text{ moles of } O_2\) when concentrations are equal, and \(y = 0.113 \text{ moles of } O_2\) when \(\mathrm{SO_{3}}\) is double that of \(\mathrm{SO_{2}}\). These calculations emphasize the importance of knowing the concentrations to use the equilibrium constant effectively.

Stoichiometry in Chemical Reactions

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It provides a fundamental approach to solving chemical equilibrium problems, where knowing the stoichiometric coefficients allows for accurate setting up of equilibrium expressions like those in this exercise.
In the given reaction, \(2 \mathrm{SO}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\), the coefficients 2, 1, and 2 reflect the stoichiometric ratio among the compounds. This ratio is crucial in forming the equilibrium expression \(K_c = \frac{[\mathrm{SO}_{3}]^2}{[\mathrm{SO}_{2}]^2[\mathrm{O}_{2}]}\). Each substance's exponent in the \(K_c\) expression corresponds to its stoichiometric coefficient.
By knowing the stoichiometric principles, students can determine how changes in one component's concentration affect the others. They can also predict moles of gases involved when provided with partial information, as seen where the stoichiometric relationship between \(\mathrm{SO_{3}}\) and \(\mathrm{SO_{2}}\) varied in cases (a) and (b). This understanding is key to mastering chemical equilibrium and solving similar problems efficiently.