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# 19.3: Factors Affecting Chemical Equilibria

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

• Determine whether various changes would alter the position of an equilibrium.
• Describe Le Châtelier’s Principle, and use it to predict the direction that a reaction would shift in response to various changes.

## Vocabulary

• Le Châtelier’s Principle: States that when a chemical system is at equilibrium and is disturbed by a stress, the system will respond by attempting to counteract that stress until a new equilibrium is established.

• Write the equilibrium constant expression for the following chemical reaction:

$2NOBr(g)\rightleftarrows2NO(s)+Br_2(g)$

• If Kc for this process is 0.0142, would you expect the reaction to be reactant-favored or product-favored?

## Introduction

So far you have learned about what chemical equilibrium means and how to calculate an equilibrium constant for a reaction based on reaction rate and reactant component concentrations. In order to establish chemical equilibrium, the rates of the forward and reverse reactions change until they become equal. In this lesson, you will learn how a chemical reaction maintains equilibrium when a stress is put on the system.

## Le Châtelier's Principle

Equilibrium represents a balance between the reactants and the products of a chemical reaction. Changes to the conditions of the system can disturb that equilibrium. When this occurs, the system reacts in such a way as to restore the equilibrium. However, the position of equilibrium will be changed following the disturbance. In other words, the response of the system involves a change in the amounts of the reactants and products. Some will increase and some will decrease until equilibrium is reestablished.

Chemical equilibrium was studied by French chemist Henri Le Châtelier (1850-1936), and his description of how a system at equilibrium responds to a change in conditions has become known as Le Châtelier’s principle. This principle states that when a chemical system is at equilibrium and is disturbed by a stress, the system will respond by attempting to counteract that stress until a new equilibrium is established. Stresses to a chemical system include changes in the concentrations of reactants or products, changes in the temperature of the system, or changes in the pressure of the system. We will discuss each of these stresses separately. In each case, the change to the equilibrium position will cause either the forward or reverse reaction to be favored over the opposite process. When the forward reaction is favored, the concentrations of the products increase, and the concentrations of the reactants decrease. When the reverse reaction is favored, the concentrations of the products decrease, and the concentrations of the reactants increase.

Original Equilibrium Favored Reaction Result
$\mathrm{A \rightleftharpoons B}$ forward: $\mathrm{A \rightarrow B}$ [A] decreases; [B] increases
$\mathrm{A \rightleftharpoons B}$ reverse: $\mathrm{A \leftarrow B}$ [A] increases; [B] decreases

### Concentration

A change in the concentration of one of the substances in an equilibrium system typically involves either the addition or the removal of one of the reactants or products. Consider the Haber-Bosch process for the industrial production of ammonia from nitrogen and hydrogen gases:

$\mathrm{N_2}(g) \mathrm{+3H_2}(g) \mathrm{\rightleftharpoons 2NH_3}(g)$

If the concentration of one substance in a system is increased, the system will respond by favoring the reaction that consumes that substance. When more N2 is added, the forward reaction will be favored because the forward reaction uses up N2 and converts it to NH3. Initially, the forward reaction speeds up because one of the reactants is present at a higher concentration, but the rate of the reverse reaction is unaffected. Since the two rates are no longer equal, the system is no longer at equilibrium, and there will be a net shift to the right (producing more NH3) until the two rates are once again balanced. The concentration of NH3 increases, while the concentrations of N2 and H2 decrease. After some time passes, equilibrium is reestablished with new concentrations of all three substances. As can be seen in Figure below, the new concentration of NH3 is higher than it was originally, because the forward reaction became temporarily favored due to the stress. The new concentration of H2 is lower. The final concentration of N2 is higher than it was in the original equilibrium, but lower than it was immediately after the addition of N2 that disturbed the original equilibrium. By responding in this way, the value of the equilibrium constant for the reaction, Keq, does not change as a result of the stress to the system.

Conversely, if more NH3 were added, the reverse reaction would be favored. This “favoring” of a reaction means temporarily speeding up the reaction in that direction until equilibrium is reestablished. Recall that once equilibrium is reestablished, the rates of the forward and reverse reactions are again equal. The addition of NH3 would result in a net increase in the formation of the reactants, N2 and H2.

The Haber-Bosch process is an equilibrium between the reactants (N2 and H2) and the product (NH3). When more N2 is added, the system favors the forward reaction until equilibrium is reestablished.

An equilibrium can also be disrupted by the full or partial removal of a reactant or product. If the concentration of a substance is decreased, the system will respond by favoring the reaction that replaces that substance. In the industrial Haber-Bosch process, NH3 is removed from the equilibrium system as the reaction proceeds. As a result, the forward reaction is favored so that more NH3 will be produced. The concentrations of N2 and H2 decrease. Continued removal of NH3 will eventually force the reaction to go to completion until all of the reactants are used up. If either N2 or H2 were removed from the equilibrium system, the reverse reaction would be favored, and the concentration of NH3 would decrease.

The effects of changes in concentration on a system at equilibrium are summarized in Table below.

Stress Response
addition of reactant forward reaction favored
addition of product reverse reaction favored
removal of reactant reverse reaction favored
removal of product forward reaction favored

### Temperature

Increasing or decreasing the temperature of a system at equilibrium is also a stress to the system. The equation for the Haber-Bosch process is written again below, this time as a thermochemical equation.

$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) + 91 \ \text{kJ}$

The forward reaction is the exothermic direction; the formation of NH3 releases heat. The reverse reaction is the endothermic direction; as NH3 decomposes to N2 and H2, heat is absorbed. An increase in the temperature of a system favors the direction of the reaction that absorbs heat, or the endothermic direction. Absorption of heat in this case is a relief of the stress provided by the temperature increase. For the Haber-Bosch process, an increase in temperature favors the reverse reaction. The concentration of NH3 in the system decreases, while the concentrations of N2 and H2 increase.

Conversely, a decrease in the temperature of a system favors the direction of the reaction that releases heat, or the exothermic direction. For the Haber-Bosch process, a decrease in temperature favors the forward reaction. Therefore the concentration of NH3 in the system increases, while the concentrations of N2 and H2 decrease.

For changes in concentration, the system responds in such a way that the value of the equilibrium constant, Keq, is unchanged. However, a change in temperature shifts the equilibrium and changes the value of Keq. As discussed in the previous section, values of Keq are dependent on the temperature. When the temperature of the system for the Haber-Bosch process is increased, the resultant shift in equilibrium towards the reactants means that the Keq value decreases. When the temperature is decreased, the shift in equilibrium towards the products means that the Keq value increases.

Le Châtelier’s principle as related to temperature changes can be illustrated easily by the equilibrium between dinitrogen tetroxide and nitrogen dioxide.

$\text{N}_2\text{O}_4(g) + \text{heat} \rightleftharpoons 2\text{NO}_2(g)$

Dinitrogen tetroxide (N2O4) is colorless, while nitrogen dioxide (NO2) is dark brown in color. When N2O4 breaks down into NO2, heat is absorbed according to the forward reaction above. Therefore, an increase in the temperature of the system will favor the forward reaction, while a decrease in temperature will favor the reverse reaction. By changing the temperature, the equilibrium between colorless N2O4 and brown NO2 can be manipulated, resulting in a visible color change.

The video below shows three sealed glass tubes containing N2O4 and NO2. When one tube is placed in hot water, the equilibrium favors the brown NO2. When another tube is placed in ice cold water, the equilibrium lies in favor of the colorless N2O4.

### Pressure

Changing the pressure of an equilibrium system in which gases are involved is also a stress to the system. A change in the pressure on a liquid or a solid has a negligible effect. We will return again to the equilibrium for the Haber-Bosch process. Imagine the gases are contained in a closed system in which the volume of the system is controlled by an adjustable piston as shown in Figure below.

(A) A mixture of nitrogen, hydrogen, and ammonia in equilibrium. (B) When the pressure is increased on the equilibrium mixture, the forward reaction is favored because that results in a reduction of the total moles of gas present. (C) Fewer moles of gas will exert a lower total pressure, so the stress is partially relieved by such a shift.

On the far left, the reaction system contains primarily N2 and H2, with only one molecule of NH3 present. As the piston is pushed inwards, the pressure of the system increases according to Boyle’s Law. This is a stress to the equilibrium. In the middle image, the same number of molecules are now confined to a smaller space, so the pressure has increased. According to Le Châtelier’s principle, the system responds in order to relieve the stress. In the image on the right, the forward reaction has been favored, in which each one molecule of N2 combines with three molecules of H2 to form one molecule of NH3. The overall result is a decrease in the number of gas molecules in the entire system. This decreases the pressure and counteracts the original stress of a pressure increase. When the pressure is increased by decreasing the available volume, system responds by favoring the reaction direction that produces fewer gas molecules. In this case, it is the forward reaction that is favored.

A decrease in pressure on the above system could be achieved by pulling the piston outward, increasing the container volume. The equilibrium would respond by favoring the reverse reaction, in which NH3 decomposes to N2 and H2. This is because the overall number of gas molecules would increase, and so would the pressure. When the pressure of a system at equilibrium is decreased by providing more total volume, the reaction that produces more total moles of gas becomes favored. This is summarized in the Table below.

Stress Response
pressure increase reaction produces fewer gas molecules
pressure decrease reaction produces more gas molecules

Like changes in concentration, the Keq value for a given reaction is unchanged by a change in pressure.

It is important to remember when analyzing the effect of a pressure change on equilibrium that only gases are affected. If a certain reaction involves liquids or solids, they should be ignored. Calcium carbonate decomposes according to the equilibrium reaction:

$\mathrm{CaCO_3}(s) \mathrm{\rightleftharpoons CaO}(s) \mathrm{+O_2}(g)$

Oxygen is the only gas in the system. An increase in the pressure of the system has no effect on the rate of decomposition of CaCO3, but it speeds the reverse reaction by forcing the oxygen molecules closer together, causing a net shift to the left. When a system contains equal moles of gas on both sides of the equation, pressure has no effect on the equilibrium position, as in the formation of HCl from H2 and Cl2.

$\mathrm{H_2}(g) \mathrm{+Cl_2}(g) \mathrm{\rightleftharpoons 2HCl}(g)$

It should also be noted that increasing the pressure by the addition of an inert gas (nonreactive gas) has no effect on the equilibrium. This can be thought of in terms of the pressure-dependent equilibrium constant (Kp). Unlike an increase or decrease in the volume of the container, the addition of an inert gas does not affect the partial pressures of any reaction components. Because the ratio of partial pressures is still equal to the equilibrium constant, no change is necessary to maintain equilibrium.

### Use of a Catalyst

Since a catalyst speeds up the rate of a reaction, you might think that it would have an effect on the equilibrium position. However, catalysts have equal effects on the forward and reverse rates, so for a system that is already at equilibrium, these two rates remain equal. A system will reach equilibrium more quickly in the presence of a catalyst, but the equilibrium position itself is unaffected.

## Lesson Summary

• Le Chatelier’s principle describes how a chemical system responds to stresses in order to reestablish equilibrium.
• Changes in the concentrations of reactants or products will cause a reaction to shift left or right in order to reestablish an equilibrium position. These changes do not change the value of Keq.
• Changes in temperature will change the value of Keq. The direction of the change depends on whether the reaction is endothermic or exothermic.
• When the pressure of a system is changed by changing the volume of the container, reactions in which some components are in the gas phase may shift to reestablish equilibrium. Changing the pressure by adding an inert gas does not affect a system at equilibrium.
• Adding a catalyst speeds up the progression of the reaction towards equilibrium, but does not change the equilibrium position itself.

## Review Questions

1. Describe the basis of Le Chatlier’s principle.
2. Is the value of the equilibrium constant changed with the addition or removal of reactant or product? If so, how? If not, why not?
3. Is the value of the equilibrium constant changed when the temperature is increased or decreased? If so, how? If not, why not?
4. How does the addition of an inert gas affect the equilibrium of a system?
5. Why does the addition of a catalyst have no effect on the equilibrium position?
1. The following reaction is allowed to reach equilibrium: $CaCO_3(s)+ heat\rightleftarrows CaO(s)+CO_2(g)$ Would the reaction shift left, shift right, or remain unchanged in response to each of the following changes:
1. Increasing the pressure by decreasing the volume at a constant temperature
2. Increasing the pressure by adding an inert gas at a constant temperature and volume
3. Increasing the temperature at a constant pressure
2. The following reaction is allowed to reach equilibrium: $CO(g)+H_2O(g)+heat\rightarrow CO_2(g)+H_2(g)$ Would the reaction shift left, shift right, or remain unchanged in response to each of the following changes:
1. Increasing [CO]
2. Decreasing [H2]
3. Decreasing [H2O]
4. Increasing [CO2]

## Points to Consider

• When chemical equilibrium is achieved for a given chemical process, the rate at which the forward process occurs is the same as the rate at which the reverse process occurs. What factors do you suppose affect the rate at which equilibrium is achieved?
• Some chemical processes are spontaneous, while others are not. How do you suppose chemical equilibrium is related to whether or not a chemical process is spontaneous or not?
• Can you think of examples where a particular process is spontaneous under certain conditions (say temperature) and nonspontaneous under other conditions?

Sep 09, 2013

May 06, 2015