A mixture consisting of \(1.1 \mathrm{mmol} \mathrm{SO}_{2}\) and \(2.2 \mathrm{mmol} \mathrm{O}_{2}\) in a \(250-\mathrm{mL}\) container was heated to \(500 \mathrm{~K}\) and allowed to reach equilibrium. Will more sulfur trioxide be formed if that equilibrium mixture is cooled to \(25^{\circ} \mathrm{C}\) ? For the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\) \(\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{3}(\mathrm{~g}), \mathrm{K}=2.5 \times 10^{10}\) at \(500 \mathrm{~K}\) and \(4.0 \times 10^{24}\) at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Yes, more sulfur trioxide will be formed if the equilibrium mixture is cooled to \(25^\circ \text{C}\).

Step by step solution

01

State the Reaction and Given Information

Write down the balanced chemical equation for the reaction: \(2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g)\). Note the quantities of reactants given: \(1.1 \text{ mmol SO}_2\) and \(2.2 \text{ mmol O}_2\), and the equilibrium constants: \(K = 2.5 \times 10^{10}\) at \(500 \text{ K}\) and \(K = 4.0 \times 10^{24}\) at \(25^\circ \text{C}\) (or 298 K).
02

Assess the Effect of Temperature on Equilibrium

Observe the equilibrium constants at different temperatures. A higher value of \(K\) indicates a higher concentration of products at equilibrium. Since \(K\) at \(25^\circ \text{C}\) is much higher than at \(500 \text{ K}\), we expect a greater concentration of sulfur trioxide, \(\text{SO}_3\), at \(25^\circ \text{C}\).
03

Apply Le Chatelier's Principle

Le Chatelier's Principle states that if a system at equilibrium is disturbed by a change in temperature, pressure, or concentration, the system will shift its equilibrium position to counteract the effect of the disturbance. Cooling the mixture from \(500 \text{ K}\) to \(25^\circ \text{C}\) (298 K) will shift the equilibrium to favor the formation of the products (exothermic reaction) in order to release heat.
04

Conclude on the Effect of Cooling the Mixture

Since the reaction is exothermic and cooling the system would decrease the system's temperature, according to Le Chatelier's Principle, the equilibrium will shift to the right, increasing the production of \(\text{SO}_3\) to counteract the cooling. This is corroborated by the significantly higher equilibrium constant at the lower temperature, which indicates a higher concentration of products at equilibrium.

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Key Concepts

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

Le Chatelier's Principle
Le Chatelier's Principle is a foundational concept in chemistry that helps us predict how a chemical reaction at equilibrium responds to changes in conditions such as temperature, pressure, and concentration. This principle asserts that if a dynamic equilibrium is disturbed, the system will adjust to minimize the disturbance and re-establish equilibrium.

For example, consider a reversible chemical reaction where the reactants turn into products at the same rate as the products revert back to reactants. If we add more of one of the reactants to the system, the principle suggests that the equilibrium will shift to produce more products, thereby using up the additional reactant to re-establish the equilibrium. This adjustment allows students to understand and predict the behavior of chemical systems under stress.
Equilibrium Constant
The equilibrium constant, denoted by K, quantifies the balance between reactants and products in a chemical reaction at equilibrium. It is a number that represents the ratio of product concentrations to reactant concentrations at a specific temperature, each raised to the power of their stoichiometric coefficients.

To calculate the equilibrium constant, we use the expression:
\[ K = \frac{{[products]}}{{[reactants]}} \]
where the square brackets indicate concentrations. The magnitude of K tells us about the position of the equilibrium: a large K (much greater than 1) means that the equilibrium position favors the products, while a small K (much less than 1) indicates that reactants are favored. It's important to note that K is constant only at a fixed temperature and will change if the temperature changes.
Exothermic Reactions
In exothermic reactions, energy is released into the surroundings, usually in the form of heat. These reactions can be identified by a negative enthalpy change, \( \Delta H < 0 \), indicating that the system loses energy. When dealing with equilibrium, exothermic reactions are affected by temperature changes. Cooling the system will typically result in an increase in the production of heat through the formation of more products, as this helps counteract the temperature decrease. This forms the basis for understanding why some reactions produce more products at lower temperatures and is essential for grasping the concepts behind Le Chatelier's Principle.
Reaction Quotient
The reaction quotient, Q, serves as a 'snapshot' of a reaction's position at any given time and determines how the reaction will proceed to reach equilibrium. The formula for Q is similar to that of the equilibrium constant:\[ Q = \frac{{[products]}}{{[reactants]}} \]
However, while K is calculated at equilibrium, Q can be calculated under any conditions. It's instrumental in comparing Q to K in order to predict the direction in which a reaction will shift to achieve equilibrium. If \( Q < K \), the reaction will proceed forward, shifting to the right to produce more products. Conversely, if \( Q > K \), the reaction will proceed in the reverse direction, shifting to the left to produce more reactants.
Temperature Effects on Equilibrium
Temperature has a significant impact on the state of chemical equilibrium. When the temperature of a system in equilibrium is changed, the equilibrium constant, K, will also change, indicating a shift in the equilibrium position. For endothermic reactions, where heat is absorbed, an increase in temperature favors the formation of products, increasing the value of K. On the other hand, for exothermic reactions, like the formation of sulfur trioxide from sulfur dioxide and oxygen, lowering the temperature increases the amount of product formed, as seen in the higher K value at lower temperatures.

This concept is applied in industrial processes, where temperature control is crucial for maximizing product yields. Understanding how temperature affects chemical equilibrium is essential for students when predicting the outcomes of reactions under different conditions.

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Most popular questions from this chapter

The temperature dependence of the equilibrium constant of the reaction \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\), which makes an important contribution to atmospheric nitrogen oxides, can be expressed as \(\ln K=2.5-\) \((21700 \mathrm{~K}) \cdot T^{-1}\). What is the standard enthalpy of the forward reaction?

Determine \(K_{c}\) from the value of \(K\) for cach of the following equilibria. (a) \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{3}(\mathrm{~g}), K=3.4\) at \(1000 \mathrm{~K}\) (b) \(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s})=\mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}), K=9.4 \times 10^{-2}\) at \(24^{\circ} \mathrm{C}\)

Analysis of a reaction mixture showed that it had the composition \(0.417 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}\), \(0.524 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\), and \(0.122 \mathrm{~mol}-\mathrm{L}^{-1} \mathrm{NH}_{3}\) at \(800 \mathrm{~K}\), at which temperature \(K_{c}=0.278\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})=2 \mathrm{NH}_{3}(\mathrm{~g}) \cdot(\mathrm{a}) \mathrm{Calculare}\) the reaction quotient. (b) Is the reaction mixture at equilibrium? (c) If not, is there a tendency to form more reactants or more products?

For the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\), \(K_{\mathrm{c}}=160\) at \(500 \mathrm{~K}\). An analysis of a reaction mixture at \(500 \mathrm{~K}\) showed that it had the composition \(4.8 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}, 2.4 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{I}_{2}\), and \(2.4 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{HI}\). (a) Calculate the reaction quoticnt. (b) Is the reaction mixture at equilibrium? (c) If not, is there a tendency to form more reactants or more products?

The equilibrium constant for the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{3}(\mathrm{~g})\) has the value \(K=2.5 \times 10^{10}\) at \(500 \mathrm{~K}\). Find the value of \(K\) for each of the following reactions at the same temperature. (a) \(\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})=\mathrm{SO}_{3}(\mathrm{~g})\) (b) \(\mathrm{SO}_{3}(\mathrm{~g})=\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) (c) \(3 \mathrm{SO}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 3 \mathrm{SO}_{3}(\mathrm{~g})\)

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