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Chapter 15: Problem 60

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

Expert verified
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

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01

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.
02

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.
03

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 \(\text{O}_{2}\).
04

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.
05

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.
06

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

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Ammonium carbamate \(\left(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}\right)\) decomposes as follows: $$ \mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}(s) \rightleftarrows 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) $$ Starting with only the solid, it is found that when the system reaches equilibrium at \(40^{\circ} \mathrm{C},\) the total gas pressure \(\left(\mathrm{NH}_{3}\right.\) and \(\mathrm{CO}_{2}\) ) is 0.363 atm. Calculate the equilibrium constant \(K_{P}\).

At \(20^{\circ} \mathrm{C},\) the vapor pressure of water is \(0.0231 \mathrm{~atm} .\) Calculate \(K_{P}\) and \(K_{\mathrm{c}}\) for the process: $$\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{H}_{2} \mathrm{O}(g)$$

Consider the following reaction at a certain temperature: $$ \mathrm{A}_{2}+\mathrm{B}_{2} \rightleftarrows 2 \mathrm{AB} $$ The mixing of 1 mole of \(\mathrm{A}_{2}\) with 3 moles of \(\mathrm{B}_{2}\) gives rise to \(x\) mole of \(\mathrm{AB}\) at equilibrium. The addition of 2 more moles of \(\mathrm{A}_{2}\) produces another \(x\) mole of \(\mathrm{AB}\). What is the equilibrium constant for the reaction?

About 75 percent of hydrogen for industrial use is produced by the steam- reforming process. This process is carried out in two stages called primary and secondary reforming. In the primary stage, a mixture of steam and methane at about 30 atm is heated over a nickel catalyst at \(800^{\circ} \mathrm{C}\) to give hydrogen and carbon monoxide: \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \quad \Delta H^{\circ}=206 \mathrm{~kJ} / \mathrm{mol}\) The secondary stage is carried out at about \(1000^{\circ} \mathrm{C},\) in the presence of air, to convert the remaining methane to hydrogen: \(\mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftarrows \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \quad \Delta H^{\circ}=35.7 \mathrm{~kJ} / \mathrm{mol}\) (a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondary stages? (b) The equilibrium constant \(K_{\mathrm{c}}\) for the primary stage is 18 at \(800^{\circ} \mathrm{C}\). (i) Calculate \(K_{P}\) for the reaction. (ii) If the partial pressures of methane and steam were both 15 atm at the start, what are the pressures of all the gases at equilibrium?

The dissociation of molecular iodine into iodine atoms is represented as: $$ \mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{I}(g) $$ At \(1000 \mathrm{~K},\) the equilibrium constant \(K_{\mathrm{c}}\) for the reaction is \(3.80 \times 10^{-5}\). Suppose you start with 0.0456 mole of \(\mathrm{I}_{2}\) in a 2.30-L flask at \(1000 \mathrm{~K}\). What are the concentrations of the gases at equilibrium?
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