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# 19.2: Le Châtelier’s Principle

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## Lesson Objectives

• Explain the factors that stress a system at equilibrium, and use Le Châtelier’s principle to predict how the system will respond to each stress.
• Explain how a reaction is able to go to completion.
• Analyze how a change in the concentration of a reaction component can change the equilibrium position but not the value of the equilibrium constant.

## Lesson Vocabulary

• Le Châtelier’s principle

### Recalling Prior Knowledge

• How is equilibrium defined?
• What is an equilibrium constant and how is it calculated?

A chemical system that is at equilibrium can be disrupted by changes in concentrations, temperature, or pressure. In this lesson, you will learn how to analyze stresses to a system at equilibrium and predict how the system will respond to that stress.

## Factors Affecting an Equilibrium System

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: When a chemical system that is at equilibrium 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 increases.

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 removes 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.

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.

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.

An equilibrium can also be disrupted by the removal of one of the substances. 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, 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, the exothermic direction. For the Haber-Bosch process, a decrease in temperature favors the forward reaction. 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 three molecules of N2 combine with nine molecules of H2 to form six molecules 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, the reaction that produces fewer total moles of gas becomes favored. 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.

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

### 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 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.

### Going to Completion

When one of the products of a reaction is removed from the chemical equilibrium system as soon as it is produced, the reverse reaction cannot establish itself and equilibrium is never reached. Reactions such as these are said to go to completion. Reactions which go to completion tend to produce one of three types of products: (1) an insoluble precipitate, (2) a gas, (3) a molecular compound such as water. Examples of these reactions are shown below.

1. Formation of a precipitate: $\mathrm{AgNO_3}(aq) \mathrm{+NaCl}(aq) \mathrm{\rightarrow NaNO_3}(aq) \mathrm{+AgCl}(s)$
2. Formation of a gas: $\mathrm{Mg}(s) \mathrm{+2HCl}(aq) \mathrm{\rightarrow MgCl_2}(aq) \mathrm{+H_2}(g)$
3. Formation of water: $\mathrm{HCl}(aq) \mathrm{+NaOH}(aq) \mathrm{\rightarrow NaCl}(aq) \mathrm{+H_2O}(l)$

## Response of Keq

Occasionally, when students apply LeChatelier’s principle to an equilibrium problem involving a change in concentration, they assume that Keq must change. This seems logical, since we talk about “shifting” the equilibrium in one direction or the other. However, Keq is a constant, for a given equilibrium at a given temperature, so it must not change. Here is an example of how this works. Consider the simplified equilibrium below:

$\mathrm{A \rightleftharpoons B}$

Let’s say we have a 1.0 liter container. At equilibrium, the following amounts are measured:

A = 0.50 mol
B = 1.0 mol

The value of Keq is given by:

$\mathrm{K_{eq}=\dfrac{[B]}{[A]}=\dfrac{1.0 \ M}{0.50 \ M}=2.0}$

Now we will disturb the equilibrium by adding 0.50 mole of A to the mixture. The equilibrium will shift towards the right, forming more B. Immediately after the addition of A and before any response, we now have 1.0 mol of A and 1.0 mol of B. The equilibrium then shifts in the forward direction. We will introduce a variable (x), which will represent the change in concentrations as the reaction proceeds. Since the mole ratio of A:B is 1:1, if [A] decreases by the x moles, [B] increases by the same amount. We can now set up an analysis called ICE, which stands for Initial, Change, and Equilibrium. The values in Table below represent molar concentrations.

A B
Initial 1.0 1.0
Change -x +x
Equilibrium 1.0-x 1.0+x

At the new equilibrium position, the values for A and B as a function of x can be set equal to the value of the Keq. Then, one can solve for x.

$\mathrm{K_{eq}=2.0=\dfrac{[B]}{[A]}=\dfrac{1.0+x}{1.0-x}}$

Solving for x:

2.0(1.0 - x) = 1.0 + x
2.0 - 2.0x = 1.0 + x
3.0x = 1.0
x = 0.33

This value for x is now plugged back in to the Equilibrium line of the table, and the final concentrations of A and B after the reaction are calculated.

[A] = 1.0 – x = 0.67 M
[B] = 1.0 + x = 1.33 M

The value of Keq has been maintained since 1.33/0.67 = 2.0. This shows that even though a change in concentration of one of the substances in an equilibrium causes a shift in the equilibrium position, the value of the equilibrium constant does not change.

## Lesson Summary

• A system at equilibrium can be disrupted by a change in concentration of one of the substances or by a change in temperature or pressure. Le Châtelier’s principle states that such a system will respond by attempting to counteract the stress. Either the forward or reverse reaction will temporarily be favored until equilibrium is reestablished.
• A catalyst increases the rate of both the forward and reverse reactions equally and does not change the equilibrium position.
• Reactions in which a product is continually removed from the system do not attain equilibrium and instead proceed to completion.
• A change in concentration or pressure does not change the value of the equilibrium constant for a reaction. Only a temperature change alters the equilibrium constant.

## Lesson Review Questions

### Reviewing Concepts

1. What are three stresses that can upset the equilibrium of a chemical system?
2. Which stress or stresses change the value of the equilibrium constant?
3. What conditions can drive a reaction to completion?
4. What must be true of the reaction in order for pressure to have an effect on the equilibrium position?
5. Does the use of a catalyst influence the position of an equilibrium? Explain.

### Problems

1. Given the following equilibrium equation: $\mathrm{N_2}(g) \mathrm{+2O_2}(g) \mathrm{+66.2\:kJ \rightleftharpoons 2NO_2}(g)$. Predict the direction of equilibrium that will be favored (forward, reverse, or neither) for each of the following changes.
2. O2 is removed.
3. The temperature is increased.
4. The pressure is increased.
5. A catalyst is used.
6. NO2 is removed.
7. The temperature is decreased.
8. The system volume is increased.
2. For the system in question 6, how would the concentration of NO2 at equilibrium be affected by each change?
3. For the system in question 6, how would the value of Keq be affected by each change?
4. Given the following reaction for the formation of sulfur trioxide from sulfur dioxide and oxygen: $\mathrm{2SO_2}(g) \mathrm{+O_2}(g) \mathrm{\rightleftharpoons 2SO_3}(g) \mathrm{+198\:kJ}$. What conditions of temperature and pressure would maximize the concentration of SO3 at equilibrium?

## Points to Consider

In a saturated solution with excess solute present, an equilibrium exists between dissolved and undissolved solute.

• How is the equilibrium constant for a solution equilibrium determined?
• What is the relationship between the equilibrium constant and the solubility of various ionic compounds?

Mar 01, 2013

Sep 09, 2014