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According to Le Châtelier's principle, does the equilibrium shift in the direction of products or reactants when \(\mathrm{O}_{2}\) is added to the equilibrium mixture of each of the following reactions? (13.5) a. \(3 \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{O}_{3}(g)\) b. \(2 \mathrm{CO}_{2}(g) \rightleftarrows 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)\) c. \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g)\) d. \(2 \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{~S}(g)+3 \mathrm{O}_{2}(g)\)

Short Answer

Expert verified
a. To productsb. To reactantsc. To productsd. To reactants

Step by step solution

01

Understand Le Châtelier's Principle

Le Châtelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
02

Analyze Reaction (a)

For the reaction \(3 \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{O}_{3}(g)\)adding \( \mathrm{O}_{2} \) will increase the concentration of one of the reactants. According to Le Châtelier's principle, the equilibrium will shift to the right to counteract the change by forming more \( \mathrm{O}_{3} \).
03

Analyze Reaction (b)

For the reaction \(2 \mathrm{CO}_{2}(g) \rightleftarrows 2 \mathrm{CO}(g) + \mathrm{O}_{2}(g)\)adding \( \mathrm{O}_{2} \) will increase the concentration of one of the products. According to Le Châtelier's principle, the equilibrium will shift to the left to counteract the change by forming more \( \mathrm{CO}_{2} \).
04

Analyze Reaction (c)

For the reaction \(2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g)\)adding \( \mathrm{O}_{2} \) will increase the concentration of one of the reactants. According to Le Châtelier's principle, the equilibrium will shift to the right to counteract the change by forming more \( \mathrm{SO}_{3} \).
05

Analyze Reaction (d)

For the reaction \(2 \mathrm{SO}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{~S}(g) + 3 \mathrm{O}_{2}(g)\)adding \( \mathrm{O}_{2} \) will increase the concentration of one of the products. According to Le Châtelier's principle, the equilibrium will shift to the left to counteract the change by forming more \( \mathrm{SO}_{2} \) and \( \mathrm{H}_{2} \mathrm{O} \).

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Key Concepts

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

equilibrium shift
When a chemical system at equilibrium is disturbed by a change in concentration, pressure, or temperature, the equilibrium position shifts to counteract the disturbance. This is known as the equilibrium shift. Le Châtelier's principle is the guiding rule.

To understand how it works, consider the reaction: 3 \, \mathrm{O}_{2}(g) \rightleftarrows 2 \, \mathrm{O}_{3}(g).

Adding more \( \mathrm{O}_{2} \) (a reactant) will cause the system to shift toward forming more products, in this case, \( \mathrm{O}_{3} \). This happens because the system seeks to restore equilibrium by using up the added \( \mathrm{O}_{2} \). Hence, the equilibrium shifts to the right.

Conversely, if we add more \( \mathrm{O}_{2} \) to the reaction: \[ 2 \, \mathrm{CO}_{2}(g) \rightleftarrows 2 \, \mathrm{CO}(g) + \mathrm{O}_{2}(g) \] we increase the concentration of a product. Here, the equilibrium will shift to the left to counteract the change, producing more \( \mathrm{CO}_{2} \).
chemical equilibrium
Chemical equilibrium is a state in which both reactants and products are present at concentrations that have no further tendency to change over time. In other words, the forward and reverse reactions occur at equal rates.

For example, in the reaction: \[ 2 \, \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightleftarrows 2 \, \mathrm{SO}_{3}(g) \] both \( \mathrm{SO}_{2} \) and \( \mathrm{O}_{2} \) are converting to \( \mathrm{SO}_{3} \) and vice versa at the same rate. When the system is at equilibrium and more \( \mathrm{O}_{2} \) is added, the system reacts to the disturbance by shifting the equilibrium to the right, producing more \( \mathrm{SO}_{3} \), in order to re-establish equilibrium.

Understanding equilibrium helps in predicting how a change in conditions affects the reaction mixture.
reaction analysis
Analyzing reactions based on Le Châtelier's principle is crucial in anticipating shifts in equilibrium. Consider the reaction:

\( 2 \, \mathrm{SO}_{2}(g) + 2 \, \mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows 2 \, \mathrm{H}_{2} \mathrm{S}(g) + 3 \, \mathrm{O}_{2}(g) \).

When we add \( \mathrm{O}_{2} \) to this equilibrium system, we increase the concentration of a product. According to Le Châtelier's principle, the equilibrium will shift to the left. As a result, more \( \mathrm{SO}_{2} \) and \( \mathrm{H}_{2} \mathrm{O} \) will form to counteract the addition of \( \mathrm{O}_{2} \).

This knowledge enables chemists to manipulate reaction conditions to favor the formation of desired products, making it invaluable in industrial processes and laboratory settings.

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Most popular questions from this chapter

A person who has been a smoker and has a low oxygen blood saturation uses an oxygen tank for supplemental oxygen. When oxygen is first supplied, a. What is the stress on the hemoglobin equilibrium? b. In what direction does the hemoglobin equilibrium shift?

What is meant by the term reversible reaction?

Indicate if you would increase or decrease the temperature of the reaction to increase the yield of the products in each of the following: \((13.5)\) a. \(\operatorname{COBr}_{2}(g)+\) Heat \(\rightleftarrows \mathrm{CO}(g)+\mathrm{Br}_{2}(g)\) b. \(2 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 4 \mathrm{NO}(g)+\) Heat c. \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(\mathrm{~g})+\) Heat \(\rightleftarrows 2 \mathrm{HBr}(\mathrm{g})\)

Would decreasing the volume of the container for each of the following reactions cause the equilibrium to shift in the direction of products, reactants, or not change? \((13.5)\) a. \(3 \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{O}_{3}(g)\) b. \(2 \mathrm{CO}_{2}(g) \rightleftarrows 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)\) c. \(\mathrm{P}_{4}(g)+5 \mathrm{O}_{2}(g) \rightleftarrows \mathrm{P}_{4} \mathrm{O}_{10}(s)\) d. \(2 \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{~S}(g)+3 \mathrm{O}_{2}(g)\)

Consider the following reaction: \((13.3,13.4,13.5)\) $$ \mathrm{PCl}_{5}(g) \rightleftarrows \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ a. Write the equilibrium expression for the reaction. b. Initially, \(0.60 \mathrm{~mol}\) of \(\mathrm{PCl}_{5}\) is placed in a \(1.0\) - \(\mathrm{L}\) flask. At equilibrium, there is \(0.16 \mathrm{~mol}\) of \(\mathrm{PCl}_{3}\) in the flask. What are the equilibrium concentrations of \(\mathrm{PCl}_{5}\) and \(\mathrm{Cl}_{2}\) ? c. What is the numerical value of the equilibrium constant, \(K_{\mathrm{c}}\), for the reaction? d. If \(0.20 \mathrm{~mol}\) of \(\mathrm{Cl}_{2}\) is added to the equilibrium mixture, will the concentration of \(\mathrm{PCl}_{5}\) increase or decrease?

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