Question

Consider the reaction: \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g) \quad \Delta H^{\circ}=-198.2 \mathrm{~kJ} / \mathrm{mol}\) Comment on the changes in the concentrations of \(\mathrm{SO}_{2}\), \(\mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at equilibrium if we were to (a) increase the temperature, (b) increase the pressure, (c) increase \(\mathrm{SO}_{2}\), (d) add a catalyst, (e) add helium at constant volume.

Short answer

Temperature increase shifts equilibrium left; pressure increase shifts it right; more \(\text{SO}_{2}\) shifts it right; catalysts have no effect; inert gas has no effect at constant volume.

Step-by-step solution

Understand the Reaction and Equilibrium

The reaction given is an exothermic equilibrium, meaning it releases heat. The equilibrium can be shifted by changing conditions such as temperature, pressure, and concentration. Le Chatelier's principle helps predict the direction of the shift. Given the enthalpy change \(\Delta H^{\circ} = -198.2 \, \text{kJ/mol}\), the reaction tends to shift to the left if temperature increases as the system wants to absorb the added heat.

Temperature Change

(a) Increasing the temperature adds heat to the system, thus shifting the equilibrium to the left (towards the reactants, \(\text{SO}_{2}\) and \(\text{O}_{2}\)) to absorb this heat. As a result, the concentration of \(\text{SO}_{2}\) and \(\text{O}_{2}\) will increase, while \(\text{SO}_{3}\) will decrease.

Pressure Change

(b) Increasing the pressure by decreasing the volume will shift the equilibrium towards the side with fewer moles of gas. Here, 3 moles of reactants (\(2 \, \text{SO}_{2} + \text{O}_{2}\)) become 2 moles of \(\text{SO}_{3}\). Thus, equilibrium shifts right, increasing \(\text{SO}_{3}\) and decreasing \(\text{SO}_{2}\) and \(\text{O}_{2}\).

Concentration Change

(c) Increasing \(\text{SO}_{2}\) increases its concentration, causing the equilibrium to shift right to consume the added \(\text{SO}_{2}\). \(\text{SO}_{2}\) concentration decreases as \(\text{SO}_{3}\) is formed. \(\text{O}_{2}\) also decreases.

Catalysts

(d) Adding a catalyst does not change the concentrations of \(\text{SO}_{2}\), \(\text{O}_{2}\), or \(\text{SO}_{3}\) at equilibrium. It only speeds up the rate at which equilibrium is achieved.

Inert Gas at Constant Volume

(e) Adding helium at constant volume does not affect the total pressure involved in the species' partial pressures that influence the reaction. Therefore, the concentrations of \(\text{SO}_{2}\), \(\text{O}_{2}\), and \(\text{SO}_{3}\) remain unchanged.

Key concepts

Equilibrium Shift

Le Chatelier's principle explores how a system at equilibrium responds to changes in conditions. In essence, if a system in equilibrium experiences a change in concentration, temperature, or pressure, it will adjust in such a way as to counteract the imposed change. This adjustment is known as an equilibrium shift. For instance, if you increase the concentration of a reactant, the system will shift towards the products to reduce this change. Conversely, increasing a product's concentration will cause a shift back towards the reactants. Equilibrium shifts are nature's way of maintaining balance and are essential for understanding how the various factors can influence chemical reactions.

Exothermic Reaction

An exothermic reaction releases heat into its surroundings. In the exercise, the reaction between sulfur dioxide (\(2 \mathrm{SO}_{2}\)) and oxygen (\(\mathrm{O}_{2}\)) to form sulfur trioxide (\(2 \mathrm{SO}_{3}\)) is exothermic, as indicated by the negative enthalpy change (\(\Delta H^{\circ} = -198.2 \, \text{kJ/mol}\)). This means that increasing the temperature will add more heat to an already warm system, encouraging the reaction to shift towards the reactants (\(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\)) to absorb some of this heat. For exothermic reactions, temperature changes can significantly alter product formation, showcasing the delicate balance between heat and reaction progress.

Pressure Change

Pressure plays a crucial role in gaseous reactions. According to Le Chatelier's principle, an increase in pressure will favor the direction which results in fewer moles of gas. For the given reaction, starting with 3 moles of reactants (\(2 \, \text{SO}_{2} + \text{O}_{2}\)) and forming 2 moles of \(\text{SO}_{3}\), increasing the pressure will shift the equilibrium to the right. This shift results because the system strives to minimize pressure by reducing the number of gas particles. Consequently, the concentration of \(\text{SO}_{3}\) increases, while the concentrations of \(\text{SO}_{2}\) and \(\text{O}_{2}\) decrease.

Catalyst Effect

Catalysts are essential in chemistry as they speed up chemical reactions without being consumed. In the context of equilibrium, a catalyst will lower the activation energy required for both the forward and reverse reactions equally. However, it does not alter the concentrations of the reactants or products at equilibrium. This means that while a catalyst can help the system reach equilibrium faster, it doesn't shift the position of equilibrium itself. For the reaction given in the exercise, adding a catalyst would make the establishment of equilibrium quicker, but the concentrations of \(\text{SO}_{2}\), \(\text{O}_{2}\), and \(\text{SO}_{3}\) would remain as they were predetermined by the original equilibrium conditions.