Problem 29 - General Chemistry: Principles and Modern Applications

Question

Can a mixture of \(2.2 \mathrm{mol} \mathrm{O}_{2}, 3.6 \mathrm{mol} \mathrm{SO}_{2},\) and \(1.8 \mathrm{mol}\) \(\mathrm{SO}_{3}\) be maintained indefinitely in a \(7.2 \mathrm{L}\) flask at a temperature at which \(K_{\mathrm{c}}=100\) in this reaction? Explain. $$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) $$

Step-by-Step Solution

Answer

No, the given mixture cannot be maintained indefinitely in the flask because the initial conditions do not correspond to the state of equilibrium for this reaction at the given temperature. The reaction will shift towards the products to reach equilibrium.

Step 1: Understand the Reaction

The reaction under consideration is: \[2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\] At equilibrium, the concentrations of the constituents obey the law of mass action, which is given by \(K_c\), the equilibrium constant.

Step 2: Calculate the Initial Molar Concentration of Reactants and Products

Before any reaction occurs, the molar concentration of each substance can be calculated by dividing the number of moles by the volume of the flask. For \(SO_2\), \(O_2\), and \(SO_3\), the initial molar concentrations are: \[ [SO_2] = \frac{3.6 \mathrm{mol}}{7.2 \mathrm{L}} = 0.5 \mathrm{M} \] \[ [O_2] = \frac{2.2 \mathrm{mol}}{7.2 \mathrm{L}} = 0.306 \mathrm{M} \] \[ [SO_3] = \frac{1.8 \mathrm{mol}}{7.2 \mathrm{L}} = 0.25 \mathrm{M} \]

Step 3: Calculate the Reaction Quotient, Q

The next step is determining the reaction quotient \(Q\), which is similar to the equilibrium constant but involves the initial concentrations, before equilibrium is reached. Its calculation is based on the law of mass action as well: \[ Q = \frac{([SO_3]^2)}{([SO_2]^2 [O_2])} = \frac{(0.25 \mathrm{M})^2}{(0.5 \mathrm{M})^2 \times 0.306 \mathrm{M}} = 0.326 \]

Step 4: Comparing Q and Kc to Predict the Reaction Shift

Let's compare \(Q\) with \(K_c\). If \(Q < K_c\), the reaction will shift to the right, i.e., towards the products, to achieve equilibrium. If \(Q > K_c\), it will shift to the left, i.e., towards the reactants. Here, \(Q = 0.326\) and \(K_c = 100\). Since \(Q < K_c\), the reaction will shift to the right to attain equilibrium, converting some \(SO_2\) and \(O_2\) into \(SO_3\). Thus, the given mixture cannot be maintained indefinitely, as the reaction will proceed until equilibrium is reached.

Key Concepts

Equilibrium Constant

In the world of chemical reactions, the equilibrium constant, denoted as \(K_c\), plays a pivotal role. It provides insight into the ratio of concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. When we talk about chemical equilibrium, we're referring to a state where the rates of the forward and reverse reactions are equal, meaning the concentrations of all reactants and products remain constant over time.

For the reaction \(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\), the equilibrium constant expression based on the law of mass action is:

• \(K_c = \frac{[SO_3]^2}{[SO_2]^2 [O_2]}\)

This expression is derived from the balanced chemical equation and provides a quantifiable measure of the equilibrium position. A large \(K_c\) value suggests that the equilibrium lies towards the products, indicating the reaction proceeds mostly towards forming \(SO_3\). Conversely, a small \(K_c\) value indicates that reactants are favored, and only a small amount of products are formed at equilibrium.

Reaction Quotient

The reaction quotient, \(Q\), serves as a snapshot of a reaction's current state before it reaches equilibrium. Calculating \(Q\) involves the same formula as \(K_c\), but \(Q\) uses the initial concentrations of reactants and products rather than those at equilibrium.

In the exercise given, before reaching equilibrium, we determined \(Q\) with the initial concentrations as follows:

• \(Q = \frac{([SO_3]^2)}{([SO_2]^2 [O_2])} = \frac{(0.25 \text{ M})^2}{(0.5 \text{ M})^2 \times 0.306 \text{ M}} = 0.326\)

By comparing \(Q\) to \(K_c\), we can predict the direction in which the reaction will proceed to reach equilibrium. If \(Q < K_c\), the reaction will move towards the products, attempting to increase \(Q\) until it equals \(K_c\). Conversely, if \(Q > K_c\), the reaction will shift towards the reactants.

Law of Mass Action

The law of mass action is a foundational principle in chemical equilibrium, serving as the backbone for both the equilibrium constant and the reaction quotient. This law states that at a given temperature, the rate of a chemical reaction is proportional to the product of the concentrations of the reactants, each raised to the power of their stoichiometric coefficients.

For the reaction \(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\), the law of mass action gives rise to the expression:

• \(K_c = \frac{[SO_3]^2}{[SO_2]^2 [O_2]}\)

This inherently defines how concentrations of chemical species relate to each other at equilibrium. Moreover, it explains why, during a reaction, changes in concentrations lead to shifts towards becoming either more product or reactant-favored until equilibrium is reached. The law of mass action underscores the dynamic nature of equilibrium, illustrating that while concentrations remain constant at equilibrium, reactions are continuously occurring in both directions.