OpenStax: General Chemistry
OpenStax: General Chemistry

OpenStax: General Chemistry

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 13 Fundamental Equilibrium Concepts

Figure 13.1 Movement of carbon dioxide through tissues and blood cells involves several equilibrium reactions. 

Chapter Outline 

13.1 Chemical Equilibria 

13.2 Equilibrium Constants 

13.3 Shifting Equilibria: Le Châtelier’s Principle 

13.4 Equilibrium Calculations 

Introduction 

Imagine a beach populated with sunbathers and swimmers. As those basking in the sun get too hot and want to cool off, they head into the surf to swim. As the swimmers tire, they head to the beach to rest. If these two rates of transfer (sunbathers entering the water, swimmers leaving the water) are equal, the number of sunbathers and swimmers would be constant, or at equilibrium, although the identities of the people are constantly changing from sunbather to swimmer and back. An analogous situation occurs in chemical reactions. Reactions can occur in both directions simultaneously (reactants to products and products to reactants) and eventually reach a state of balance.

These balanced two-way reactions occur all around and even in us. For example, they occur in our blood, where the reaction between carbon dioxide and water forms carbonic acid (HCO3) (Figure 13.1). Human physiology is adapted to the amount of ionized products produced by this reaction (HCO3 and H+). In this chapter, you will learn how to predict the position of the balance and the yield of a product of a reaction under specific conditions, how to change a reaction's conditions to increase or reduce yield, and how to evaluate an equilibrium system's reaction to disturbances.

13.1 Chemical Equilibria

By the end of this section, you will be able to: 

  • Describe the nature of equilibrium systems 
  • Explain the dynamic nature of a chemical equilibrium 

A chemical reaction is usually written in a way that suggests it proceeds in one direction, the direction in which we read, but all chemical reactions are reversible, and both the forward and reverse reaction occur to one degree or another depending on conditions. In a chemical equilibrium, the forward and reverse reactions occur at equal rates, and the concentrations of products and reactants remain constant. If we run a reaction in a closed system so that the products cannot escape, we often find the reaction does not give a 100% yield of products. Instead, some reactants remain after the concentrations stop changing. At this point, when there is no further change in concentrations of reactants and products, we say the reaction is at equilibrium. A mixture of reactants and products is found at equilibrium.

For example, when we place a sample of dinitrogen tetroxide (N2O4, a colorless gas) in a glass tube, it forms nitrogen dioxide (NO2, a brown gas) by the reaction

The color becomes darker as N2O4 is converted to NO2. When the system reaches equilibrium, both N2O4 and NO2 are present (Figure 13.2).

`Figure 13.2 A mixture of NO2 and N2O4 moves toward equilibrium. Colorless N2O4 reacts to form brown NO2. As the reaction proceeds toward equilibrium, the color of the mixture darkens due to the increasing concentration of NO2.

The formation of NO2 from N2O4 is a reversible reaction, which is identified by the equilibrium arrow (⇌). All reactions are reversible, but many reactions, for all practical purposes, proceed in one direction until the reactants are exhausted and will reverse only under certain conditions. Such reactions are often depicted with a one-way arrow from reactants to products. Many other reactions, such as the formation of NO2 from N2O4, are reversible under more easily obtainable conditions and, therefore, are named as such. In a reversible reaction, the reactants can combine to form products and the products can react to form the reactants. Thus, not only can N2O4 decompose to form NO2, but the NO2 produced can react to form N2O4. As soon as the forward reaction produces any NO2, the reverse reaction begins and NO2 starts to react to form N2O4. At equilibrium, the concentrations of N2O4 and NO2 no longer change because the rate of formation of NO2 is exactly equal to the rate of consumption of NO2, and the rate of formation of N2O4 is exactly equal to the rate of consumption of N2O4. Chemical equilibrium is a dynamic process: As with the swimmers and the sunbathers, the numbers of each remain constant, yet there is a flux back and forth between them (Figure 13.3).

Figure 13.3 These jugglers provide an illustration of dynamic equilibrium. Each throws clubs to the other at the same rate at which he receives clubs from that person. Because clubs are thrown continuously in both directions, the number of clubs moving in each direction is constant, and the number of clubs each juggler has at a given time remains (roughly) constant.

n a chemical equilibrium, the forward and reverse reactions do not stop, rather they continue to occur at the same rate, leading to constant concentrations of the reactants and the products. Plots showing how the reaction rates and concentrations change with respect to time are shown in Figure 13.2.

We can detect a state of equilibrium because the concentrations of reactants and products do not appear to change. However, it is important that we verify that the absence of change is due to equilibrium and not to a reaction rate that is so slow that changes in concentration are difficult to detect.

We use a double arrow when writing an equation for a reversible reaction. Such a reaction may or may not be at equilibrium. For example, Figure 13.2 shows the reaction:

When we wish to speak about one particular component of a reversible reaction, we use a single arrow. For example, in the equilibrium shown in Figure 13.2, the rate of the forward reaction

is equal to the rate of the backward reaction


Chemistry in Everyday Life

Equilibrium and Soft Drinks

The connection between chemistry and carbonated soft drinks goes back to 1767, when Joseph Priestley (1733–1804; mostly known today for his role in the discovery and identification of oxygen) discovered a method of infusing water with carbon dioxide to make carbonated water. In 1772, Priestly published a paper entitled “Impregnating Water with Fixed Air.” The paper describes dripping oil of vitriol (today we call this sulfuric acid, but what a great way to describe sulfuric acid: “oil of vitriol” literally means “liquid nastiness”) onto chalk (calcium carbonate). The resulting CO2 falls into the container of water beneath the vessel in which the initial reaction takes place; agitation helps the gaseous CO2 mix into the liquid water.

Carbon dioxide is slightly soluble in water. There is an equilibrium reaction that occurs as the carbon dioxide reacts with the water to form carbonic acid (H2CO3). Since carbonic acid is a weak acid, it can dissociate into protons (H+) and hydrogen carbonate ions (HCO3).

Today, CO2 can be pressurized into soft drinks, establishing the equilibrium shown above. Once you open the beverage container, however, a cascade of equilibrium shifts occurs. First, the CO2 gas in the air space on top of the bottle escapes, causing the equilibrium between gas-phase CO2 and dissolved or aqueous CO2 to shift, lowering the concentration of CO2 in the soft drink. Less CO2 dissolved in the liquid leads to carbonic acid decomposing to dissolved CO2 and H2O. The lowered carbonic acid concentration causes a shift of the final equilibrium. As long as the soft drink is in an open container, the CO2 bubbles up out of the beverage, releasing the gas into the air (Figure 13.4). With the lid off the bottle, the CO2 reactions are no longer at equilibrium and will continue until no more of the reactants remain. This results in a soft drink with a much lowered CO2 concentration, often referred to as “flat.”

Figure 13.4 When a soft drink is opened, several equilibrium shifts occur. (credit: modification of work by “D Coetzee”/Flickr)

Let us consider the evaporation of bromine as a second example of a system at equilibrium.

An equilibrium can be established for a physical change—like this liquid to gas transition—as well as for a chemical reaction. Figure 13.5 shows a sample of liquid bromine at equilibrium with bromine vapor in a closed container. When we pour liquid bromine into an empty bottle in which there is no bromine vapor, some liquid evaporates, the amount of liquid decreases, and the amount of vapor increases. If we cap the bottle so no vapor escapes, the amount of liquid and vapor will eventually stop changing and an equilibrium between the liquid and the vapor will be established. If the bottle were not capped, the bromine vapor would escape and no equilibrium would be reached.

Figure 13.5 An equilibrium is pictured between liquid bromine, Br2(l), the dark liquid, and bromine vapor, Br2(g), the orange gas. Because the container is sealed, bromine vapor cannot escape and equilibrium is maintained. (credit: http://images-of-elements.com/bromine.php)


Exercises

Question 13.1

13.1

What does it mean to describe a reaction as “reversible”?

Click here to see Answer

Question 13.2

13.2

When writing an equation, how is a reversible reaction distinguished from a nonreversible reaction?

Click here to see Answer

Question 13.3

13.3

If a reaction is reversible, when can it be said to have reached equilibrium?

Click here to see Answer

Question 13.4

13.4

Is a system at equilibrium if the rate constants of the forward and reverse reactions are equal?

Click here to see Answer

Question 13.5

13.5

If the concentrations of products and reactants are equal, is the system at equilibrium?

Click here to see Answer


13.2 Equilibrium Constants

By the end of this section, you will be able to: 

  • Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions 
  • Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures 
  • Relate the magnitude of an equilibrium constant to properties of the chemical system 

Now that we have a symbol (⇌) to designate reversible reactions, we will need a way to express mathematically how the amounts of reactants and products affect the equilibrium of the system. A general equation for a reversible reaction may be written as follows:

We can write the reaction quotient (Q) for this equation. When evaluated using concentrations, it is called Qc. We use brackets to indicate molar concentrations of reactants and products.

The reaction quotient is equal to the molar concentrations of the products of the chemical equation (multiplied together) over the reactants (also multiplied together), with each concentration raised to the power of the coefficient of that substance in the balanced chemical equation. For example, the reaction quotient for the reversible reaction 2NO2(g)⇌N2O4(g) is given by this expression:

Example 13.1

Writing Reaction Quotient Expressions 

Write the expression for the reaction quotient for each of the following reactions: 

Solution

Check Your Learning

Write the expression for the reaction quotient for each of the following reactions:

 

Answer:  


The numeric value of Qc for a given reaction varies; it depends on the concentrations of products and reactants present at the time when Qc is determined. When pure reactants are mixed, Qc is initially zero because there are no products present at that point. As the reaction proceeds, the value of Qc increases as the concentrations of the products increase and the concentrations of the reactants simultaneously decrease (Figure 13.6). When the reaction reaches equilibrium, the value of the reaction quotient no longer changes because the concentrations no longer change.

Figure 13.6 (a) The change in the concentrations of reactants and products is depicted as the 2SO2(g)+O2(g)⇌2SO3(g) reaction approaches equilibrium. (b) The change in concentrations of reactants and products is depicted as the reaction 2SO3(g)⇌2SO2(g)+O2(g) approaches equilibrium. (c) The graph shows the change in the value of the reaction quotient as the reaction approaches equilibrium.

When a mixture of reactants and products of a reaction reaches equilibrium at a given temperature, its reaction quotient always has the same value. This value is called the equilibrium constant (K) of the reaction at that temperature. As for the reaction quotient, when evaluated in terms of concentrations, it is noted as Kc.

That a reaction quotient always assumes the same value at equilibrium can be expressed as:

This equation is a mathematical statement of the law of mass action: When a reaction has attained equilibrium at a given temperature, the reaction quotient for the reaction always has the same value.

Example 13.2

Evaluating a Reaction Quotient

Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:

When 0.10 mol NO2 is added to a 1.0-L flask at 25 °C, the concentration changes so that at equilibrium, [NO2] = 0.016 M and [N2O4] = 0.042 M.

(a) What is the value of the reaction quotient before any reaction occurs?

(b) What is the value of the equilibrium constant for the reaction?

Solution

(a) Before any product is formed,

(b) At equilibrium, the value of the equilibrium constant is equal to the value of the reaction quotient. At equilibrium, 

The equilibrium constant is 1.6 × 102.


Note that dimensional analysis would suggest the unit for this Kc value should be M−1. However, it is common practice to omit units for Kc values computed as described here, since it is the magnitude of an equilibrium constant that relays useful information. As will be discussed later in this module, the rigorous approach to computing equilibrium constants uses dimensionless quantities derived from concentrations instead of actual concentrations, and so Kc values are truly unitless.

Check Your Learning

For the reaction 2SO2(g)+O2(g)⇌2SO3(g), the concentrations at equilibrium are [SO2] = 0.90 M, [O2] = 0.35 M, and [SO3] = 1.1 M. What is the value of the equilibrium constant, Kc?

Answer: Kc =4.3

The magnitude of an equilibrium constant is a measure of the yield of a reaction when it reaches equilibrium. A large value for Kc indicates that equilibrium is attained only after the reactants have been largely converted into products. A small value of Kc—much less than 1—indicates that equilibrium is attained when only a small proportion of the reactants have been converted into products.

Once a value of Kc is known for a reaction, it can be used to predict directional shifts when compared to the value of Qc. A system that is not at equilibrium will proceed in the direction that establishes equilibrium. The data in Figure 13.7 illustrate this. When heated to a consistent temperature, 800 °C, different starting mixtures of CO, H2O, CO2, and H2 react to reach compositions adhering to the same equilibrium (the value of Qc changes until it equals the value of Kc). This value is 0.640, the equilibrium constant for the reaction under these conditions.

It is important to recognize that an equilibrium can be established starting either from reactants or from products, or from a mixture of both. For example, equilibrium was established from Mixture 2 in Figure 13.7 when the products of the reaction were heated in a closed container. In fact, one technique used to determine whether a reaction is truly at equilibrium is to approach equilibrium starting with reactants in one experiment and starting with products in another. If the same value of the reaction quotient is observed when the concentrations stop changing in both experiments, then we may be certain that the system has reached equilibrium.

Figure 13.7 Concentrations of three mixtures are shown before and after reaching equilibrium at 800 °C for the socalled water gas shift reaction: CO(g)+H2O(g)⇌CO2(g)+H2(g).

Example 13.3

Predicting the Direction of Reaction

Given here are the starting concentrations of reactants and products for three experiments involving this reaction:

Determine in which direction the reaction proceeds as it goes to equilibrium in each of the three experiments shown.

Solution

Experiment 1:

Qc < Kc (0.039 < 0.64)

The reaction will shift to the right.

Experiment 2:

Qc > Kc (140 > 0.64)

The reaction will shift to the left.

Experiment 3:

Qc < Kc (0.48 < 0.64)

The reaction will shift to the right.

Check Your Learning

Calculate the reaction quotient and determine the direction in which each of the following reactions will proceed to reach equilibrium.

(a) A 1.00-L flask containing 0.0500 mol of NO(g), 0.0155 mol of Cl2(g), and 0.500 mol of NOCl:

(b) A 5.0-L flask containing 17 g of NH3, 14 g of N2, and 12 g of H2:

(c) A 2.00-L flask containing 230 g of SO3(g):


Answer:

In Example 13.2, it was mentioned that the common practice is to omit units when evaluating reaction quotients and equilibrium constants. It should be pointed out that using concentrations in these computations is a convenient but simplified approach that sometimes leads to results that seemingly conflict with the law of mass action. For example, equilibria involving aqueous ions often exhibit equilibrium constants that vary quite significantly (are not constant) at high solution concentrations. This may be avoided by computing Kc values using the activities of the reactants and products in the equilibrium system instead of their concentrations. The activity of a substance is a measure of its effective concentration under specified conditions. While a detailed discussion of this important quantity is beyond the scope of an introductory text, it is necessary to be aware of a few important aspects:

  • Activities are dimensionless (unitless) quantities and are in essence “adjusted” concentrations.
  • For relatively dilute solutions, a substance's activity and its molar concentration are roughly equal.
  • Activities for pure condensed phases (solids and liquids) are equal to 1.

As a consequence of this last consideration, Qc and Kc expressions do not contain terms for solids or liquids (being numerically equal to 1, these terms have no effect on the expression's value). Several examples of equilibria yielding such expressions will be encountered in this section.

Homogeneous Equilibria

A homogeneous equilibrium is one in which all of the reactants and products are present in a single solution (by definition, a homogeneous mixture). In this chapter, we will concentrate on the two most common types of homogeneous equilibria: those occurring in liquid-phase solutions and those involving exclusively gaseous species. Reactions between solutes in liquid solutions belong to one type of homogeneous equilibria. The chemical species involved can be molecules, ions, or a mixture of both. Several examples are provided here.

In each of these examples, the equilibrium system is an aqueous solution, as denoted by the aq annotations on the solute formulas. Since H2O(l) is the solvent for these solutions, its concentration does not appear as a term in the Kc expression, as discussed earlier, even though it may also appear as a reactant or product in the chemical equation.

Reactions in which all reactants and products are gases represent a second class of homogeneous equilibria. We use molar concentrations in the following examples, but we will see shortly that partial pressures of the gases may be used as well.

Note that the concentration of H2O(g) has been included in the last example because water is not the solvent in this gas-phase reaction and its concentration (and activity) changes.

Whenever gases are involved in a reaction, the partial pressure of each gas can be used instead of its concentration in the equation for the reaction quotient because the partial pressure of a gas is directly proportional to its concentration at constant temperature. This relationship can be derived from the ideal gas equation, where M is the molar concentration of gas, n/v.

Thus, at constant temperature, the pressure of a gas is directly proportional to its concentration.

Using the partial pressures of the gases, we can write the reaction quotient for the system C2H6(g) ⇌ C2H4(g) + H2(g) by following the same guidelines for deriving concentration-based expressions:

In this equation we use QP to indicate a reaction quotient written with partial pressures: PC2H6 is the partial pressure of C2H6; PH2, the partial pressure of H2; and PC2H6, the partial pressure of C2H4. At equilibrium:

The subscript P in the symbol KP designates an equilibrium constant derived using partial pressures instead of concentrations. The equilibrium constant, KP, is still a constant, but its numeric value may differ from the equilibrium constant found for the same reaction by using concentrations.

Conversion between a value for Kc, an equilibrium constant expressed in terms of concentrations, and a value for KP, an equilibrium constant expressed in terms of pressures, is straightforward (a K or Q without a subscript could be either concentration or pressure).

The equation relating Kc and KP is derived as follows. For the gas-phase reaction 


The relationship between Kc and KP is

In this equation, Δn is the difference between the sum of the coefficients of the gaseous products and the sum of the coefficients of the gaseous reactants in the reaction (the change in moles of gas between the reactants and the products). For the gas-phase reaction 


Example 13.4

Calculation of KP

Write the equations for the conversion of Kc to KP for each of the following reactions:



What is KP at this temperature?

Solution




Check Your Learning

Write the equations for the conversion of Kc to KP for each of the following reactions, which occur in the gas phase:



What would be the value of KP at this temperature?

Answer:

Heterogeneous Equilibria

A heterogeneous equilibrium is a system in which reactants and products are found in two or more phases. The phases may be any combination of solid, liquid, or gas phases, and solutions. When dealing with these equilibria, remember that solids and pure liquids do not appear in equilibrium constant expressions (the activities of pure solids, pure liquids, and solvents are 1).

Some heterogeneous equilibria involve chemical changes; for example:

Other heterogeneous equilibria involve phase changes, for example, the evaporation of liquid bromine, as shown in the following equation:

We can write equations for reaction quotients of heterogeneous equilibria that involve gases, using partial pressures instead of concentrations. Two examples are:


Exercises

Question 13.6

13.6

Explain why there may be an infinite number of values for the reaction quotient of a reaction at a given temperature but there can be only one value for the equilibrium constant at that temperature.

Click here to see Answer

Question 13.7

13.7

Explain why an equilibrium between Br2(l)Br_{2}(l) and Br2(g) would not be established if the container were not a closed vessel shown in Figure 13.5.

Click here to see Answer

Question 13.8

13.8

If you observe the following reaction at equilibrium, is it possible to tell whether the reaction started with pure NO2NO_{2} or with pure N2O4N_{2}O_{4}? 2NO2(g)N2O4(g)2NO_{2}\left ( g \right )\rightleftharpoons N_{2}O_{4}\left ( g \right )

Click here to see Answer

Question 13.9

13.9

Among the solubility rules previously discussed is the statement: All chlorides are soluble except Hg2Cl2Hg_{2}Cl_{2}, AgCl, PBCl2PBCl_{2}, and CuCl.

(a) Write the expression for the equilibrium constant for the reaction represented by the equation AgCl(s)Ag+(aq)+Cl(aq)AgCl(s)\rightleftharpoons Ag^{+}\left ( aq \right )+Cl^{-}\left ( aq \right ). Is KcK_{c} > 1, < 1, or ≈ 1? Explain your answer.

(b) Write the expression for the equilibrium constant for the reaction represented by the equation Pb2+(aq)+2Cl(aq)PbCl2(s)Pb^{2+}\left ( aq \right )+2Cl^{-}\left ( aq \right )\rightleftharpoons PbCl_{2}\left ( s \right ). Is KcK_{c} > 1, < 1, or ≈ 1? Explain your answer.

Click here to see Answer

Question 13.10

13.10

Among the solubility rules previously discussed is the statement: Carbonates, phosphates, borates, and arsenates—except those of the ammonium ion and the alkali metals—are insoluble.

(a) Write the expression for the equilibrium constant for the reaction represented by the equation CaCO3(s)Ca2+(aq)+CO3(aq)CaCO_{3}\left ( s \right )\rightleftharpoons Ca^{2+}\left ( aq \right )+CO^{3-}\left ( aq \right ). Is KcK_{c} > 1, < 1, or ≈ 1? Explain your answer.

(b) Write the expression for the equilibrium constant for the reaction represented by the equation 3Ba2+(aq)+2PO43(aq)Ba3(PO4)2(s)3Ba^{2+}\left ( aq \right )+2PO_{4}^{3-}\left ( aq \right )\rightleftharpoons Ba_{3}\left ( PO_{4} \right )_{2}\left ( s \right ). Is KcK_{c} > 1, < 1, or ≈ 1? Explain your answer.

Click here to see Answer

Question 13.11

13.11

Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: 3C2H2(g)C6H6(g)3C_{2}H_{2}\left ( g \right )\rightarrow C_{6}H_{6}\left ( g \right ).Which value ofKc would make this reaction most useful commercially? KcK_{c} ≈ 0.01,Kc ≈ 1, or KcK_{c} ≈ 10. Explain your answer.

Click here to see Answer

Question 13.12

13.12

Show that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation KI(aq)+I2(aq)KI3(aq)KI\left ( aq \right )+I_{2}\left ( aq \right )\rightleftharpoons KI_{3}\left ( aq \right ) give the same expression for the reaction quotient. KI3KI_{3} is composed of the ions K+K^{+} and I3I_{3}^{-}.

Click here to see Answer

Question 13.13

13.13

For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be 100%. Is KcK_{c} > 1, < 1, or ≈ 1 for a titration reaction?

Click here to see Answer

Question 13.14

13.14

For a precipitation reaction to be useful in a gravimetric analysis, the product of the reaction must be insoluble. Is KcK_{c} > 1, < 1, or ≈ 1 for a useful precipitation reaction?

Click here to see Answer

Question 13.15

13.15

Write the mathematical expression for the reaction quotient, QcQ_{c}, for each of the following reactions:

(a)CH4(g)+Cl2(g)CH3Cl(g)+HCl(g)\left ( a \right ) CH_{4}\left ( g \right )+Cl_{2}\left ( g \right )\rightleftharpoons CH_{3}Cl\left ( g \right )+HCl\left ( g \right )

(b)N2(g)+O2(g)2NO(g)\left ( b \right )N_{2}\left ( g \right )+O_{2}\left ( g \right )\rightleftharpoons 2NO\left ( g \right )

(c)2SO2(g)+O2(g)2SO3(g)\left ( c \right ) 2SO_{2}\left ( g \right )+O_{2}\left ( g \right )\rightleftharpoons 2SO_{3}\left ( g \right )

(d)BaSO3(s)BaO(s)+SO2(g)\left ( d \right ) BaSO_{3}\left ( s \right )\rightleftharpoons BaO\left ( s \right )+SO_{2}\left ( g \right )

(e)P4(g)+5O2(g)P4O10(s)\left ( e \right )P_{4}\left ( g \right )+5O_{2}\left ( g \right )\rightleftharpoons P_{4}O_{10}\left ( s \right )

(f)Br2(g)2Br(g)\left ( f \right ) Br_{2}\left ( g \right )\rightleftharpoons 2B_{r}\left ( g \right )

(g)CH4(g)+2O2(g)CO2(g)+2H2O(l)\left ( g \right ) CH_{4}\left ( g \right )+2O_{2}\left ( g \right )\rightleftharpoons CO_{2}\left ( g \right )+2H_{2}O\left ( l \right )

(h)CuSO45H2O(s)CuSO4(s)+5H2O(g)\left ( h \right ) CuSO_{4}\cdot 5H_{2}O\left ( s \right )\rightleftharpoons CuSO_{4}\left ( s \right )+5H_{2}O\left ( g \right )

Click here to see Answer

Question 13.16

13.16

Write the mathematical expression for the reaction quotient, QcQ_{c}, for each of the following reactions:

(a)N2(g)+3H2(g)2NH3(g)\left ( a \right ) N_{2}\left ( g \right )+3H_{2}\left ( g \right )\rightleftharpoons 2NH_{3}\left ( g \right )

(b)4NH3(g)+5O2(g)4NO(g)+6H2O(g)\left ( b \right ) 4NH_{3}\left ( g \right )+5O_{2}\left ( g \right )\rightleftharpoons 4NO\left ( g \right )+6H_{2}O\left ( g \right )

(c)N2O4(g)2NO2(g)\left ( c \right ) N_{2}O_{4}\left ( g \right )\rightleftharpoons 2NO_{2}\left ( g \right )

(d)CO2(g)+H2(g)CO(g)+H2O(g)\left ( d \right ) CO_{2}\left ( g \right )+H_{2}\left ( g \right )\rightleftharpoons CO\left ( g \right )+H_{2}O\left ( g \right )

(e)NH4Cl(s)NH3(g)+HCl(g)\left ( e \right ) NH_{4}Cl\left ( s \right )\rightleftharpoons NH_{3}\left ( g \right )+HCl\left ( g \right )

(f)2Pb(NO3)2(s)2PbO(s)+4NO2(g)+O2(g)\left ( f \right ) 2Pb\left ( NO_{3} \right )_{2}\left ( s \right )\rightleftharpoons 2PbO\left ( s \right )+ 4NO_{2}\left ( g \right )+O_{2}\left ( g \right )

(g)2H2(g)+O2(g)2H2O(l)\left ( g \right ) 2H_{2}\left ( g \right )+O_{2}\left ( g \right )\rightleftharpoons 2H_{2}O\left ( l \right )

(h)S8(g)8S(g)\left ( h \right ) S_{8}\left ( g \right )\rightleftharpoons 8S\left ( g \right )

Click here to see Answer

Question 13.17

13.17

The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.

(a)2NH3(g)N2(g)+3H2(g)\left ( a \right ) 2NH_{3}\left ( g \right )\rightleftharpoons N_{2}\left ( g \right )+3H_{2}\left ( g \right ) KcK_{c} =17; [NH3]\left [ NH_{3} \right ] = 0.20 M, \left [math][ N_{2} \right ][/math] = 1.00 M, [H2]\left [ H_{2} \right ] = 1.00 M

(b)2NH3(g)N2(g)+3H2(g)\left ( b \right ) 2NH_{3}\left ( g \right )\rightleftharpoons N_{2}\left ( g \right )+3H_{2}\left ( g \right ) KP=6.8×104K_{P}=6.8\times 10^{4}; initial pressures: NH3=3.0atm,N2=2.0atm,H2=1.0atmNH_{3}= 3.0 atm, N_{2}= 2.0 atm, H_{2}= 1.0 atm

(c)2SO3(g)2SO2(g)+O2(g)\left ( c \right ) 2SO_{3}\left ( g \right )\rightleftharpoons 2SO_{2}\left ( g \right )+O_{2}\left ( g \right ) Kc=0.230;[SO3]=0.00M,[SO2]=1.00M,[O2]=1.00MK_{c}=0.230; \left [ SO_{3} \right ]= 0.00 M, \left [ SO_{2} \right ] = 1.00 M, \left [ O_{2} \right ] = 1.00 M

(d)2SO3(g)2SO2(g)+O2(g)\left ( d \right ) 2SO_{3}\left ( g \right )\rightleftharpoons 2SO_{2}\left ( g \right )+O_{2}\left ( g \right ) KP=16.5K_{P} =16.5; initial pressures: SO3=1.00atm,SO2=1.00atm,O2=1.00atmSO_{3} = 1.00 atm, SO_{2} = 1.00 atm, O_{2} = 1.00 atm

(e)2NO(g)+Cl2(g)2NOCl(g)\left ( e \right ) 2NO\left ( g \right ) +Cl_{2}\left ( g \right )\rightleftharpoons 2NOCl\left ( g \right ) Kc=4.6×104;[NO]=1.00M,[Cl2]=1.00M,[NOCl]=0MK_{c} =4.6\times 10^{4}; \left [ NO \right ] = 1.00M, \left [ Cl_{2} \right ] = 1.00 M, \left [ NOCl \right ] = 0 M

(f)N2(g)+O2(g)2NO(g)\left ( f \right ) N_{2}\left ( g \right ) +O_{2}\left ( g \right )\rightleftharpoons 2NO\left ( g \right ) KP=0.050K_{P} =0.050;initial pressures:NO = 10.0 atm, N2=O2=5atmN_{2} = O_{2} = 5 atm

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Question 13.18

13.18

The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.

(a)2NH3(g)N2(g)+3H2(g)\left ( a \right ) 2NH_{3}\left ( g \right )\rightleftharpoons N_{2}\left ( g \right )+3H_{2}\left ( g \right ) KcK_{c} =17; [NH3]\left [ NH_{3} \right ] = 0.50 M, \left [math][ N_{2} \right ][/math] = 0.15 M, [H2]\left [ H_{2} \right ] = 0.12 M

(b)2NH3(g)N2(g)+3H2(g)\left ( b \right ) 2NH_{3}\left ( g \right )\rightleftharpoons N_{2}\left ( g \right )+3H_{2}\left ( g \right ) KP=6.8×104K_{P}=6.8\times 10^{4}; initial pressures: NH3=2.00atm,N2=10.0atm,H2=10.00atmNH_{3}= 2.00 atm, N_{2}= 10.0 atm, H_{2}= 10.00 atm

(c)2SO3(g)2SO2(g)+O2(g)\left ( c \right ) 2SO_{3}\left ( g \right )\rightleftharpoons 2SO_{2}\left ( g \right )+O_{2}\left ( g \right ) Kc=0.230;[SO3]=2.00M,[SO2]=2.00M,[O2]=2.00MK_{c}=0.230; \left [ SO_{3} \right ]= 2.00 M, \left [ SO_{2} \right ] = 2.00 M, \left [ O_{2} \right ] = 2.00 M

(d)2SO3(g)2SO2(g)+O2(g)\left ( d \right ) 2SO_{3}\left ( g \right )\rightleftharpoons 2SO_{2}\left ( g \right )+O_{2}\left ( g \right ) KP=6.5K_{P} =6.5; initial pressures: SO2=1.00atm,O2=1.130atm,SO3=0atmSO_{2} = 1.00 atm, O_{2} = 1.130 atm, SO_{3} = 0 atm

(e)2NO(g)+Cl2(g)2NOCl(g)\left ( e \right ) 2NO\left ( g \right ) +Cl_{2}\left ( g \right )\rightleftharpoons 2NOCl\left ( g \right ) KP=2.5×103;initialpressures:[math][NO]K_{P} =2.5\times 10^{3}; initial pressures: [math]\left [ NO \right ] = 1.00 atm, \left [ Cl_{2} \right ] = 1.00 atm, \left [ NOCl \right ] = 0 atm[/math]

(f)N2(g)+O2(g)2NO(g)\left ( f \right ) N_{2}\left ( g \right ) +O_{2}\left ( g \right )\rightleftharpoons 2NO\left ( g \right ) Kc=0.050K_{c} =0.050;[math]\left [ N_{2} \right ]=1.00 M,\left [ O_{2} \right ]=0.200 M, \left [ NO \right ]=1.00 M

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Question 13.19

13.19

The following reaction has KP=4.50×105at720KKP= 4.50\times 10^{-5} at 720 K

N2(g)+3H2(g)2NH3(g)N_{2}\left ( g \right )+3H_{2}\left ( g \right )\rightleftharpoons 2NH_{3}\left ( g \right )

If a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium?P(NH3P(NH_{3}) = 93 atm,P(N2P(N_{2}) = 48 atm, and P(H2P(H_{2}) = 52

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Question 13.20

13.20

Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium? SO2Cl2(g)SO2(g)+Cl2(g)SO_{2}Cl_{2}\left ( g \right )\rightleftharpoons SO_{2}\left ( g \right )+Cl_{2}\left ( g \right )

[SO2Cl2]=0.12M,[Cl2]=0.16Mand[SO2]=0.050M.Kc\left [ SO_{2}Cl_{2} \right ] = 0.12M, \left [ Cl_{2} \right ] = 0.16M and \left [ SO_{2} \right ] = 0.050 M.K_{c} for the reaction is 0.078.

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Question 13.21

13.21

Which of the systems described in Exercise 13.15 give homogeneous equilibria? Which give heterogeneous equilibria?

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Question 13.22

13.22

Which of the systems described in Exercise 13.16 give homogeneous equilibria? Which give heterogeneous equilibria?

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Question 13.23

13.23

For which of the reactions in Exercise 13.15 does KcK_{c} (calculated using concentrations) equal KPK_{P} (calculated using pressures)?

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Question 13.24

13.24

For which of the reactions in Exercise 13.16 does KcK_{c} (calculated using concentrations) equal KPK_{P} (calculated using pressures)?

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Question 13.25

13.25

Convert the values of KcK_{c} to values of KPK_{P} or the values of KPK_{P} to values of KcK_{c}.

(a)N2(g)+3H2(g)2NH3(g)\left ( a \right ) N_{2}\left ( g \right ) + 3H_{2}\left ( g \right )\rightleftharpoons 2NH_{3}\left ( g \right ) Kc=0.50K_{c} = 0.50 at 400°C

(b)H2+I22HI\left ( b \right ) H_{2} +I_{2}\rightleftharpoons 2HI Kc=50.2K_{c} = 50.2 at 448°C

(c)Na2SO410H2O(s)Na2SO4(s)+10H2O(g)\left ( c \right ) Na_{2} SO_{4} \cdot 10H_{2} O\left ( s \right )\rightleftharpoons Na_{2}SO_{4}\left ( s \right ) + 10H_{2} O\left ( g \right ) KP=4.08××1025K_{P}=4.08×\times 10^{-25} at 25°C

(d)H2O(l)H2O(g)\left ( d \right ) H_{2} O\left ( l \right )\rightleftharpoons H_{2}O\left ( g \right ) KP=0.122K_{P}=0.122 at 50°C

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Question 13.26

13.26

Convert the values of KcK_{c} to values of KPK_{P} or the values of KPK_{P} to values of KcK_{c}

(a)Cl2(g)+Br2(g)2BrCl(g)\left ( a \right ) Cl_{2}\left ( g \right ) + Br_{2}\left ( g \right )\rightleftharpoons 2BrCl\left ( g \right ). Kc=4.7×102K_{c} = 4.7\times 10 ^{-2} at 25°C

(b)2SO2(g)+O2(g)2SO3(g)\left ( b \right ) 2SO_{2}\left ( g \right ) +O_{2}\left ( g \right )\rightleftharpoons 2SO_{3}\left ( g \right ) KP=48.2K_{P} = 48.2 at 500°C

(c)CaCl26H2O(s)CaCl2(s)+6H2O(g)\left ( c\right ) CaCl_{2}\cdot 6H_{2}O\left ( s \right )\rightleftharpoons CaCl_{2}\left ( s \right ) + 6H_{2}O\left ( g \right ) KP=5.09×1044K_{P} = 5.09\times 10 ^{-44} at 25°C

(d)H2O(l)H2O(g)\left ( d \right ) H_{2}O\left ( l \right )\rightleftharpoons H_{2}O\left ( g \right ) KP=0.196K_{P} = 0.196 at 60°C

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Question 13.27

13.27

What is the value of the equilibrium constant expression for the change H2O(l)H2O(g)H_{2}O\left ( l \right )\rightleftharpoons H_{2}O\left ( g \right ) at 30 °C?

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Question 13.28

13.28

Write the expression of the reaction quotient for the ionization of HOCN in water.

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Question 13.29

13.29

Write the reaction quotient expression for the ionization of NH3NH_{3} in water.

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Question 13.30

13.30

What is the approximate value of the equilibrium constant KPK_{P} for the change C2H5OC2H5(l)C2H5OC2H5(g)C_{2}H_{5}OC_{2}H_{5}\left ( l \right )\rightleftharpoons C_{2}H_{5}OC_{2}H_{5}\left ( g \right ) at 25 °C. (Vapor pressure was described in the previous chapter on liquids and solids; refer back to this chapter to find the relevant information needed to solve this problem.)

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13.3 Shifting Equilibria: Le Châtelier’s Principle

By the end of this section, you will be able to: 

  • Describe the ways in which an equilibrium system can be stressed 
  • Predict the response of a stressed equilibrium using Le Châtelier’s principle 

As we saw in the previous section, reactions proceed in both directions (reactants go to products and products go to reactants). We can tell a reaction is at equilibrium if the reaction quotient (Q) is equal to the equilibrium constant (K). We next address what happens when a system at equilibrium is disturbed so that Q is no longer equal to K. If a system at equilibrium is subjected to a perturbance or stress (such as a change in concentration) the position of equilibrium changes. Since this stress affects the concentrations of the reactants and the products, the value of Q will no longer equal the value of K. To re-establish equilibrium, the system will either shift toward the products (if Q < K) or the reactants (if Q > K) until Q returns to the same value as K.

This process is described by Le Châtelier's principle: When a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance. As described in the previous paragraph, the disturbance causes a change in Q; the reaction will shift to re-establish Q = K.

Predicting the Direction of a Reversible Reaction

Le Châtelier's principle can be used to predict changes in equilibrium concentrations when a system that is at equilibrium is subjected to a stress. However, if we have a mixture of reactants and products that have not yet reached equilibrium, the changes necessary to reach equilibrium may not be so obvious. In such a case, we can compare the values of Q and K for the system to predict the changes.

Effect of Change in Concentration on Equilibrium

A chemical system at equilibrium can be temporarily shifted out of equilibrium by adding or removing one or more of the reactants or products. The concentrations of both reactants and products then undergo additional changes to return the system to equilibrium.

The stress on the system in Figure 13.8 is the reduction of the equilibrium concentration of SCN (lowering the concentration of one of the reactants would cause Q to be larger than K). As a consequence, Le Châtelier's principle leads us to predict that the concentration of Fe(SCN)2+ should decrease, increasing the concentration of SCN part way back to its original concentration, and increasing the concentration of Fe3+ above its initial equilibrium concentration.

Figure 13.8 (a) The test tube contains 0.1 M Fe3+. (b) Thiocyanate ion has been added to solution in (a), forming the red Fe(SCN)2+ ion. Fe3+(aq)+SCN−(aq)⇌Fe(SCN)2+(aq). (c) Silver nitrate has been added to the solution in (b), precipitating some of the SCN− as the white solid AgSCN. Ag+(aq)+SCN−(aq)⇌AgSCN(s). The decrease in the SCN− concentration shifts the first equilibrium in the solution to the left, decreasing the concentration (and lightening color) of the Fe(SCN)2+. (credit: modification of work by Mark Ott)

The effect of a change in concentration on a system at equilibrium is illustrated further by the equilibrium of this chemical reaction:

The numeric values for this example have been determined experimentally. A mixture of gases at 400 °C with [H2] = [I2] = 0.221 M and [HI] = 1.563 M is at equilibrium; for this mixture, Qc = Kc = 50.0. If H2 is introduced into the system so quickly that its concentration doubles before it begins to react (new [H2] = 0.442 M), the reaction will shift so that a new equilibrium is reached, at which [H2] = 0.374 M, [I2] = 0.153 M, and [HI] = 1.692 M. This gives:

We have stressed this system by introducing additional H2. The stress is relieved when the reaction shifts to the right, using up some (but not all) of the excess H2, reducing the amount of uncombined I2, and forming additional HI.

Effect of Change in Pressure on Equilibrium

Sometimes we can change the position of equilibrium by changing the pressure of a system. However, changes in pressure have a measurable effect only in systems in which gases are involved, and then only when the chemical reaction produces a change in the total number of gas molecules in the system. An easy way to recognize such a system is to look for different numbers of moles of gas on the reactant and product sides of the equilibrium. While evaluating pressure (as well as related factors like volume), it is important to remember that equilibrium constants are defined with regard to concentration (for Kc) or partial pressure (for KP). Some changes to total pressure, like adding an inert gas that is not part of the equilibrium, will change the total pressure but not the partial pressures of the gases in the equilibrium constant expression. Thus, addition of a gas not involved in the equilibrium will not perturb the equilibrium.

Link to Learning

Check out this video to see a dramatic visual demonstration of how equilibrium changes with pressure changes.

As we increase the pressure of a gaseous system at equilibrium, either by decreasing the volume of the system or by adding more of one of the components of the equilibrium mixture, we introduce a stress by increasing the partial pressures of one or more of the components. In accordance with Le Châtelier's principle, a shift in the equilibrium that reduces the total number of molecules per unit of volume will be favored because this relieves the stress. The reverse reaction would be favored by a decrease in pressure.

Consider what happens when we increase the pressure on a system in which NO, O2, and NO2 are at equilibrium:

The formation of additional amounts of NO2 decreases the total number of molecules in the system because each time two molecules of NO2 form, a total of three molecules of NO and O2 are consumed. This reduces the total pressure exerted by the system and reduces, but does not completely relieve, the stress of the increased pressure. On the other hand, a decrease in the pressure on the system favors decomposition of NO2 into NO and O2, which tends to restore the pressure.

Now consider this reaction:

Because there is no change in the total number of molecules in the system during reaction, a change in pressure does not favor either formation or decomposition of gaseous nitrogen monoxide.

Effect of Change in Temperature on Equilibrium

Changing concentration or pressure perturbs an equilibrium because the reaction quotient is shifted away from the equilibrium value. Changing the temperature of a system at equilibrium has a different effect: A change in temperature actually changes the value of the equilibrium constant. However, we can qualitatively predict the effect of the temperature change by treating it as a stress on the system and applying Le Châtelier's principle.

When hydrogen reacts with gaseous iodine, heat is evolved.

Because this reaction is exothermic, we can write it with heat as a product.

Increasing the temperature of the reaction increases the internal energy of the system. Thus, increasing the temperature has the effect of increasing the amount of one of the products of this reaction. The reaction shifts to the left to relieve the stress, and there is an increase in the concentration of H2 and I2 and a reduction in the concentration of HI. Lowering the temperature of this system reduces the amount of energy present, favors the production of heat, and favors the formation of hydrogen iodide.

When we change the temperature of a system at equilibrium, the equilibrium constant for the reaction changes. Lowering the temperature in the HI system increases the equilibrium constant: At the new equilibrium the concentration of HI has increased and the concentrations of H2 and I2 decreased. Raising the temperature decreases the value of the equilibrium constant, from 67.5 at 357 °C to 50.0 at 400 °C.

Temperature affects the equilibrium between NO2 and N2O4 in this reaction

The positive ΔH value tells us that the reaction is endothermic and could be written

At higher temperatures, the gas mixture has a deep brown color, indicative of a significant amount of brown NO2 molecules. If, however, we put a stress on the system by cooling the mixture (withdrawing energy), the equilibrium shifts to the left to supply some of the energy lost by cooling. The concentration of colorless N2O4 increases, and the concentration of brown NO2 decreases, causing the brown color to fade.

Catalysts Do Not Affect Equilibrium

As we learned during our study of kinetics, a catalyst can speed up the rate of a reaction. Though this increase in reaction rate may cause a system to reach equilibrium more quickly (by speeding up the forward and reverse reactions), a catalyst has no effect on the value of an equilibrium constant nor on equilibrium concentrations.

The interplay of changes in concentration or pressure, temperature, and the lack of an influence of a catalyst on a chemical equilibrium is illustrated in the industrial synthesis of ammonia from nitrogen and hydrogen according to the equation

A large quantity of ammonia is manufactured by this reaction. Each year, ammonia is among the top 10 chemicals, by mass, manufactured in the world. About 2 billion pounds are manufactured in the United States each year.

Ammonia plays a vital role in our global economy. It is used in the production of fertilizers and is, itself, an important fertilizer for the growth of corn, cotton, and other crops. Large quantities of ammonia are converted to nitric acid, which plays an important role in the production of fertilizers, explosives, plastics, dyes, and fibers, and is also used in the steel industry.

Portrait of a Chemist

Fritz Haber 

In the early 20th century, German chemist Fritz Haber (Figure 13.9) developed a practical process for converting diatomic nitrogen, which cannot be used by plants as a nutrient, to ammonia, a form of nitrogen that is easiest for plants to absorb.

The availability of nitrogen is a strong limiting factor to the growth of plants. Despite accounting for 78% of air, diatomic nitrogen (N2) is nutritionally unavailable due the tremendous stability of the nitrogen-nitrogen triple bond. For plants to use atmospheric nitrogen, the nitrogen must be converted to a more bioavailable form (this conversion is called nitrogen fixation).

Haber was born in Breslau, Prussia (presently Wroclaw, Poland) in December 1868. He went on to study chemistry and, while at the University of Karlsruhe, he developed what would later be known as the Haber process: the catalytic formation of ammonia from hydrogen and atmospheric nitrogen under high temperatures and pressures. For this work, Haber was awarded the 1918 Nobel Prize in Chemistry for synthesis of ammonia from its elements. The Haber process was a boon to agriculture, as it allowed the production of fertilizers to no longer be dependent on mined feed stocks such as sodium nitrate. Currently, the annual production of synthetic nitrogen fertilizers exceeds 100 million tons and synthetic fertilizer production has increased the number of humans that arable land can support from 1.9 persons per hectare in 1908 to 4.3 in 2008.

Figure 13.9 The work of Nobel Prize recipient Fritz Haber revolutionized agricultural practices in the early 20th century. His work also affected wartime strategies, adding chemical weapons to the artillery.

In addition to his work in ammonia production, Haber is also remembered by history as one of the fathers of chemical warfare. During World War I, he played a major role in the development of poisonous gases used for trench warfare. Regarding his role in these developments, Haber said, “During peace time a scientist belongs to the World, but during war time he belongs to his country.”1 Haber defended the use of gas warfare against accusations that it was inhumane, saying that death was death, by whatever means it was inflicted. He stands as an example of the ethical dilemmas that face scientists in times of war and the double-edged nature of the sword of science.

Like Haber, the products made from ammonia can be multifaceted. In addition to their value for agriculture, nitrogen compounds can also be used to achieve destructive ends. Ammonium nitrate has also been used in explosives, including improvised explosive devices. Ammonium nitrate was one of the components of the bomb used in the attack on the Alfred P. Murrah Federal Building in downtown Oklahoma City on April 19, 1995.

It has long been known that nitrogen and hydrogen react to form ammonia. However, it became possible to manufacture ammonia in useful quantities by the reaction of nitrogen and hydrogen only in the early 20th century after the factors that influence its equilibrium were understood.

To be practical, an industrial process must give a large yield of product relatively quickly. One way to increase the yield of ammonia is to increase the pressure on the system in which N2, H2, and NH3 are at equilibrium or are coming to equilibrium.

The formation of additional amounts of ammonia reduces the total pressure exerted by the system and somewhat reduces the stress of the increased pressure.

Although increasing the pressure of a mixture of N2, H2, and NH3 will increase the yield of ammonia, at low temperatures, the rate of formation of ammonia is slow. At room temperature, for example, the reaction is so slow that if we prepared a mixture of N2 and H2, no detectable amount of ammonia would form during our lifetime. The formation of ammonia from hydrogen and nitrogen is an exothermic process:

Thus, increasing the temperature to increase the rate lowers the yield. If we lower the temperature to shift the equilibrium to favor the formation of more ammonia, equilibrium is reached more slowly because of the large decrease of reaction rate with decreasing temperature.

Part of the rate of formation lost by operating at lower temperatures can be recovered by using a catalyst. The net effect of the catalyst on the reaction is to cause equilibrium to be reached more rapidly.

In the commercial production of ammonia, conditions of about 500 °C, 150–900 atm, and the presence of a catalyst are used to give the best compromise among rate, yield, and the cost of the equipment necessary to produce and contain high-pressure gases at high temperatures (Figure 13.10).

Figure 13.10 Commercial production of ammonia requires heavy equipment to handle the high temperatures and pressures required. This schematic outlines the design of an ammonia plant.


Exercises

Question 13.31

13.31

The following equation represents a reversible decomposition: CaCO3(s)CaO(s)+CO2(g)CaCO_{3}\left ( s \right )\rightleftharpoons CaO\left ( s \right ) + CO_{2}\left ( g \right )

Under what conditions will decomposition in a closed container proceed to completion so that no CaCO3CaCO_{3} remains?

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Question 13.32

13.32

Explain how to recognize the conditions under which changes in pressure would affect systems at equilibrium.

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Question 13.33

13.33

What property of a reaction can we use to predict the effect of a change in temperature on the value of an equilibrium constant?

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Question 13.34

13.34

What would happen to the color of the solution in part (b) of Figure 13.8 if a small amount of NaOH were added and Fe(OH)3Fe(OH)_{3} precipitated? Explain your answer.

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Question 13.35

13.35

The following reaction occurs when a burner on a gas stove is lit: CH4(g)+2O2(g)CO2(g)+2H2O(g)CH_{4}\left ( g \right ) + 2O_{2}\left ( g \right )\rightleftharpoons CO_{2}\left ( g \right ) + 2H_{2}O\left ( g \right )

Is an equilibrium among CH4CH_{4}, O2O_{2}, CO2CO_{2}, and H2OH_{2}O established under these conditions? Explain your answer.

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Question 13.36

13.36

A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide, SO3SO_{3}, from sulfur dioxide, SO2SO_{2}, and oxygen, O2O_{2}, shown here. At high temperatures, the rate of formation of SO3SO_{3} is higher, but the equilibrium amount (concentration or partial pressure) of SO3SO_{3} is lower than it would be at lower temperatures.

2SO2(g)+O2(g)2SO3(g)2SO_{2}\left ( g \right ) +O_{2}\left ( g \right )\rightarrow 2SO_{3}\left ( g \right )

(a) Does the equilibrium constant for the reaction increase, decrease, or remain about the same as the temperature increases?

(b) Is the reaction endothermic or exothermic?

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Question 13.37

13.37

Suggest four ways in which the concentration of hydrazine, N2H4 N_{2}H_{4}, could be increased in an equilibrium described by the following equation:

N2(g)+2H2(g)N2H4(g)N_{2}\left ( g \right )+2H_{2}\left ( g \right )\rightleftharpoons N_{2}H_{4}\left ( g \right ) ΔH =95 kJ

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Question 13.38

13.38

Suggest four ways in which the concentration of PH3PH_{3} could be increased in an equilibrium described by the following equation:

P4(g)+6H2(g)4PH3(g)P_{4}\left ( g \right )+6H_{2}\left ( g \right )\rightleftharpoons 4PH_{3}\left ( g \right ) ΔH =110.5 kJ

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Question 13.39

13.39

How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?

(a)2NH3(g)N2(g)+3H2(g)\left ( a \right ) 2NH_{3}\left ( g \right )\rightleftharpoons N_{2}\left ( g \right )+3H_{2}\left ( g \right ) ΔH =92 kJ

(b)N2(g)+O2(g)2NO(g)\left ( b \right ) N_{2}\left ( g \right )+O_{2}\left ( g \right )\rightleftharpoons 2NO\left ( g \right ) ΔH =181 kJ

(c)2O3(g)3O2(g)\left ( c \right ) 2O_{3}\left ( g \right )\rightleftharpoons 3O_{2}\left ( g \right ) ΔH =−285 kJ

(d)CaOs+CO2(g)CaCO3(s)\left ( d \right ) CaO_{s} +CO_{2}\left ( g \right )\rightleftharpoons CaCO_{3}\left ( s \right ) ΔH =−176 kJ

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Question 13.40

13.40

How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?

(a)2H2O(g)2H2(g)+O2(g)\left ( a \right ) 2H_{2}O\left ( g \right )\rightleftharpoons 2H_{2}\left ( g \right )+O_{2}\left ( g \right ) ΔH =484 kJ

(b)N2(g)+3H2(g)2NH3(g)\left ( b \right ) N_{2}\left ( g \right )+3H_{2}\left ( g \right )\rightleftharpoons 2NH_{3}\left ( g \right ) ΔH =−92.2 kJ

(c)2Br(g)Br2(g)\left ( c \right ) 2Br\left ( g \right )\rightleftharpoons Br_{2}\left ( g \right ) ΔH =−224 kJ

(d)H2(g)+I2(s)2HI(g)\left ( d \right ) H_{2}\left ( g \right ) +I_{2}\left ( s \right )\rightleftharpoons 2HI\left ( g \right ) ΔH =53 kJ

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Question 13.41

13.41

Water gas is a 1:1 mixture of carbon monoxide and hydrogen gas and is called water gas because it is formed from steam and hot carbon in the following reaction: H2O(g)+C(s)H2(g)+CO(g)H_{2}O\left ( g \right )+C\left ( s \right )\rightleftharpoons H_{2}\left ( g \right ) +CO\left ( g \right ). Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and hydrogen at high temperature and pressure in the presence of a suitable catalyst.

(a) Write the expression for the equilibrium constant (Kc)(K_{c}) for the reversible reaction

2H2(g)+CO(g)CH3OH(g)2H_{2}\left ( g \right )+CO\left ( g \right )\rightleftharpoons CH_{3}OH\left ( g \right ) ΔH =−90.2 kJ

(b) What will happen to the concentrations of H2H_{2}, CO, and CH3OHCH_{3}OH at equilibrium if more H2H_{2} is added?

(c) What will happen to the concentrations of H2H_{2}, CO, and CH3OHCH_{3}OH at equilibrium if CO is removed?

(d) What will happen to the concentrations of H2H_{2}, CO, and CH3OHCH_{3}OH at equilibrium if CH3OHCH_{3}OH is added?

(e) What will happen to the concentrations of H2H_{2}, CO, and CH3OHCH_{3}OH at equilibrium if the temperature of the system is increased? (f) What will happen to the concentrations of H2H_{2}, CO, and CH3OHCH_{3}OH at equilibrium if more catalyst is added?

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Question 13.42

13.42

Nitrogen and oxygen react at high temperatures. (a) Write the expression for the equilibrium constant (Kc)(K_{c}) for the reversible reaction N2(g)+O2(g)2NO(g)ΔH=181kJN_{2}\left ( g \right )+O_{2}\left ( g \right )\rightleftharpoons 2NO\left ( g \right ) ΔH =181kJ

(b) What will happen to the concentrations of N2N_{2}, O2O_{2}, and NO at equilibrium if more O2O_{2} is added?

(c) What will happen to the concentrations of N2N_{2}, O2O_{2}, and NO at equilibrium if N2N_{2}is removed?

(d) What will happen to the concentrations of N2N_{2}, O2O_{2}, and NO at equilibrium if NO is added?

(e) What will happen to the concentrations of N2N_{2}, O2O_{2}, and NO at equilibrium if the pressure on the system is increased by reducing the volume of the reaction vessel?

(f) What will happen to the concentrations of N2N_{2}, O2O_{2}, and NO at equilibrium if the temperature of the system is increased?

(g) What will happen to the concentrations of N2N_{2}, O2O_{2}, and NO at equilibrium if a catalyst is added?

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Question 13.43

13.43

Water gas, a mixture of H2H_{2} and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon.

(a) Write the expression for the equilibrium constant for the reversible reaction

C(s)+H2O(g)CO(g)+H2(g)C\left ( s \right ) +H_{2} O\left ( g \right )\rightleftharpoons CO\left ( g \right )+H_{2}\left ( g \right )

(b) What will happen to the concentration of each reactant and product at equilibrium if more C is added?

(c) What will happen to the concentration of each reactant and product at equilibrium if H2OH_{2}O is removed?

(d) What will happen to the concentration of each reactant and product at equilibrium if CO is added?

(e) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?

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Question 13.44

13.44

Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas. (a) Write the expression for the equilibrium constant (Kc)\left (K_{c}\right ) for the reversible reaction

Fe2O3(s)+3H2(g)2Fe(s)+3H2O(g)Fe2O_{3}\left ( s \right )+3H_{2}\left ( g \right )\rightleftharpoons 2Fe\left ( s \right )+3H_{2}O\left ( g \right ) ΔH =98.7 kJ

(b) What will happen to the concentration of each reactant and product at equilibrium if more Fe is added?

(c) What will happen to the concentration of each reactant and product at equilibrium if H2OH_{2}O is removed?

(d) What will happen to the concentration of each reactant and product at equilibrium if H2H_{2} is added?

(e) What will happen to the concentration of each reactant and product at equilibrium if the pressure on the system is increased by reducing the volume of the reaction vessel?

(f) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?

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Question 13.45

13.45

Ammonia is a weak base that reacts with water according to this equation: [math]NH_{3}\left ( aq \right )+H_{2}O\left ( l \right )\rightleftharpoons NH_{4}\left ^{+}( aq \right )+OH^{-}\left ( aq \right )[/math]

Will any of the following increase the percent of ammonia that is converted to the ammonium ion in water?

(a) Addition of NaOH (b) Addition of HCl (c) Addition of NH4ClNH_{4}Cl

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Question 13.46

13.46

Acetic acid is a weak acid that reacts with water according to this equation:

[math]CH_{3}CO_{2}H\left ( aq \right )+H_{2}O\left ( aq \right )\rightleftharpoons H_{3}O^{+}(aq) +CH_{3}CO_{2}\left ^{-}( aq \right )[/math]

Will any of the following increase the percent of acetic acid that reacts and produces CH3CO2ionCH_{3}CO_{2}^{-}ion?

(a) Addition of HCl (b) Addition of NaOH (c) Addition of NaCH3CO2NaCH_{3}CO_{2}

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Question 13.47

13.47

Suggest two ways in which the equilibrium concentration of Ag+Ag^{+} can be reduced in a solution of Na+Na^{+}+, ClCl^{-}, Ag+Ag^{+} and NO3NO_{3}^{-}, in contact with solid AgCl.

[math]Na^{+}\left ( aq \right )+Cl^{-}\left ( aq \right )+Ag^{+}\left ( aq \right )+NO_{3}\left ^{-}( aq \right )\rightleftharpoons AgCl\left ( s \right )+Na\left ^{+} ( aq \right )+NO_{3}\left ^{-}( aq \right )[/math]

ΔH =−65.9 kJ

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Question 13.48

13.48

How can the pressure of water vapor be increased in the following equilibrium? H{2}O\left ( l \right )\rightleftharpoons H{}O\left ( g \right ) ΔH =41 kJ

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Question 13.49

13.49

Additional solid silver sulfate, a slightly soluble solid, is added to a solution of silver ion and sulfate ion at equilibrium with solid silver sulfate.

[math]2Ag^{+}\left ( aq \right )+SO_{4}\left ^{2-}( aq \right )\rightleftharpoons Ag_{2}SO_{4}\left ( s \right )[/math]

Which of the following will occur?

(a) Ag+Ag^{+} or SO42SO_{4}^{ 2-} concentrations will not change.

(b) The added silver sulfate will dissolve.

(c) Additional silver sulfate will form and precipitate from solution as Ag+Ag^{+} ions and SO42SO_{4}^{ 2-} ions combine.

(d) The Ag+Ag^{+} ion concentration will increase and the SO42SO_{4}^{ 2-} ion concentration will decrease.

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Question 13.50

13.50

The amino acid alanine has two isomers, α-alanine and β-alanine. When equal masses of these two compounds are dissolved in equal amounts of a solvent, the solution of α-alanine freezes at the lowest temperature. Which form, α-alanine or β-alanine, has the larger equilibrium constant for ionization [math]\left ( HX\rightleftharpoons H^{+}+X^{-})[/math] ?

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13.4 Equilibrium Calculations

By the end of this section, you will be able to: 

  • Write equations representing changes in concentration and pressure for chemical species in equilibrium systems 
  • Use algebra to perform various types of equilibrium calculations 

We know that at equilibrium, the value of the reaction quotient of any reaction is equal to its equilibrium constant. Thus, we can use the mathematical expression for Q to determine a number of quantities associated with a reaction at equilibrium or approaching equilibrium. While we have learned to identify in which direction a reaction will shift to reach equilibrium, we want to extend that understanding to quantitative calculations. We do so by evaluating the ways that the concentrations of products and reactants change as a reaction approaches equilibrium, keeping in mind the stoichiometric ratios of the reaction. This algebraic approach to equilibrium calculations will be explored in this section.

Changes in concentrations or pressures of reactants and products occur as a reaction system approaches equilibrium. In this section we will see that we can relate these changes to each other using the coefficients in the balanced chemical equation describing the system. We use the decomposition of ammonia as an example.

On heating, ammonia reversibly decomposes into nitrogen and hydrogen according to this equation:

If a sample of ammonia decomposes in a closed system and the concentration of N2 increases by 0.11 M, the change in the N2 concentration, Δ[N2], the final concentration minus the initial concentration, is 0.11 M. The change is positive because the concentration of N2 increases.

The change in the H2 concentration, Δ[H2], is also positive—the concentration of H2 increases as ammonia decomposes. The chemical equation tells us that the change in the concentration of H2 is three times the change in the concentration of N2 because for each mole of N2 produced, 3 moles of H2 are produced.

The change in concentration of NH3, Δ[NH3], is twice that of Δ[N2]; the equation indicates that 2 moles of NH3 must decompose for each mole of N2 formed. However, the change in the NH3 concentration is negative because the concentration of ammonia decreases as it decomposes.

We can relate these relationships directly to the coefficients in the equation

Note that all the changes on one side of the arrows are of the same sign and that all the changes on the other side of the arrows are of the opposite sign.

If we did not know the magnitude of the change in the concentration of N2, we could represent it by the symbol x.

The changes in the other concentrations would then be represented as:

The coefficients in the Δ terms are identical to those in the balanced equation for the reaction.

The simplest way for us to find the coefficients for the concentration changes in any reaction is to use the coefficients in the balanced chemical equation. The sign of the coefficient is positive when the concentration increases; it is negative when the concentration decreases.

Example 13.5

Determining Relative Changes in Concentration 





Solution 




Check Your Learning

Complete the changes in concentrations for each of the following reactions:




Answer:

Calculations Involving Equilibrium Concentrations

Because the value of the reaction quotient of any reaction at equilibrium is equal to its equilibrium constant, we can use the mathematical expression for Qc (i.e., the law of mass action) to determine a number of quantities associated with a reaction at equilibrium. It may help if we keep in mind that Qc = Kc (at equilibrium) in all of these situations and that there are only three basic types of calculations:

1.   Calculation of an equilibrium constant. If concentrations of reactants and products at equilibrium are known, the value of the equilibrium constant for the reaction can be calculated.

2.    Calculation of missing equilibrium concentrations. If the value of the equilibrium constant and all of the equilibrium concentrations, except one, are known, the remaining concentration can be calculated.

3.    Calculation of equilibrium concentrations from initial concentrations. If the value of the equilibrium constant and a set of concentrations of reactants and products that are not at equilibrium are known, the concentrations at equilibrium can be calculated.

A similar list could be generated using QP, KP, and partial pressure. We will look at solving each of these cases in sequence.

Calculation of an Equilibrium Constant

Since the law of mass action is the only equation we have to describe the relationship between Kc and the concentrations of reactants and products, any problem that requires us to solve for Kc must provide enough information to determine the reactant and product concentrations at equilibrium. Armed with the concentrations, we can solve the equation for Kc, as it will be the only unknown.

Example 13.6 showed us how to determine the equilibrium constant of a reaction if we know the concentrations of reactants and products at equilibrium. The following example shows how to use the stoichiometry of the reaction and a combination of initial concentrations and equilibrium concentrations to determine an equilibrium constant. This technique, commonly called an ICE chart—for Initial, Change, and Equilibrium–will be helpful in solving many equilibrium problems. A chart is generated beginning with the equilibrium reaction in question. Underneath the reaction the initial concentrations of the reactants and products are listed—these conditions are usually provided in the problem and we consider no shift toward equilibrium to have happened. The next row of data is the change that occurs as the system shifts toward equilibrium—do not forget to consider the reaction stoichiometry as described in a previous section of this chapter. The last row contains the concentrations once equilibrium has been reached.

Example 13.6

Calculation of an Equilibrium Constant

Iodine molecules react reversibly with iodide ions to produce triiodide ions.

If a solution with the concentrations of I2 and I both equal to 1.000 × 10−3 M before reaction gives an equilibrium concentration of I2 of 6.61 × 10−4 M, what is the equilibrium constant for the reaction?

Solution

We will begin this problem by calculating the changes in concentration as the system goes to equilibrium. Then we determine the equilibrium concentrations and, finally, the equilibrium constant. First, we set up a table with the initial concentrations, the changes in concentrations, and the equilibrium concentrations using −x as the change in concentration of I2.

Since the equilibrium concentration of I2 is given, we can solve for x. At equilibrium the concentration of I2 is 6.61 × 10−4 M so that

Now we can fill in the table with the concentrations at equilibrium.

We now calculate the value of the equilibrium constant.

Check Your Learning

Ethanol and acetic acid react and form water and ethyl acetate, the solvent responsible for the odor of some nail polish removers.

When 1 mol each of C2H5OH and CH3CO2H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when 1/3 mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is not a solvent in this reaction.)

Answer: Kc = 4

Calculation of a Missing Equilibrium Concentration 

Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At 2000 °C, the value of the equilibrium constant for the reaction, N2(g)+O2(g)⇌2NO(g), is 4.1 × 10−4. Find the concentration of NO(g) in an equilibrium mixture with air at 1 atm pressure at this temperature. In air, [N2] = 0.036 mol/L and [O2] 0.0089 mol/L.

Solution

We are given all of the equilibrium concentrations except that of NO. Thus, we can solve for the missing equilibrium concentration by rearranging the equation for the equilibrium constant.

Thus [NO] is 3.6 × 10−4 mol/L at equilibrium under these conditions.

We can check our answer by substituting all equilibrium concentrations into the expression for the reaction quotient to see whether it is equal to the equilibrium constant.

The answer checks; our calculated value gives the equilibrium constant within the error associated with the significant figures in the problem.

Check Your Learning

The equilibrium constant for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is 6.00 × 10−2. Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26 M and 2.09 M, respectively.

Answer: 1.53 mol/L

Calculation of Changes in Concentration 

If we know the equilibrium constant for a reaction and a set of concentrations of reactants and products that are not at equilibrium, we can calculate the changes in concentrations as the system comes to equilibrium, as well as the new concentrations at equilibrium. The typical procedure can be summarized in four steps.

1.    Determine the direction the reaction proceeds to come to equilibrium.

       a.  Write a balanced chemical equation for the reaction.

       b.  If the direction in which the reaction must proceed to reach equilibrium is not obvious,            calculate Qc from the initial concentrations and compare to Kc to determine the direction of        change.

2.   Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.

      a.   Define the changes in the initial concentrations that are needed for the reaction to reach         equilibrium. Generally, we represent the smallest change with the symbol x and express the         other changes in terms of the smallest change.

      b.   Define missing equilibrium concentrations in terms of the initial concentrations and the         changes in concentration determined in (a).

3.    Solve for the change and the equilibrium concentrations.

       a.   Substitute the equilibrium concentrations into the expression for the equilibrium                        constant, solve for x, and check any assumptions used to find x.

       b.   Calculate the equilibrium concentrations.

4.    Check the arithmetic.

        a.   Check the calculated equilibrium concentrations by substituting them into the                             equilibrium expression and determining whether they give the equilibrium constant.

        Sometimes a particular step may differ from problem to problem—it may be more complex         in some problems and less complex in others. However, every calculation of equilibrium               concentrations from a set of initial concentrations will involve these steps.

        In solving equilibrium problems that involve changes in concentration, sometimes it is                   convenient to set up an ICE table, as described in the previous section.

Example 13.8

Calculation of Concentration Changes as a Reaction Goes to Equilibrium 

Under certain conditions, the equilibrium constant for the decomposition of PCl5(g) into PCl3(g) and Cl2(g) is 0.0211. What are the equilibrium concentrations of PCl5, PCl3, and Cl2 if the initial concentration of PCl5 was 1.00 M?

Solution

Use the stepwise process described earlier.

Step 1.  Determine the direction the reaction proceeds.

                The balanced equation for the decomposition of PCl5 is

Because we have no products initially, Qc = 0 and the reaction will proceed to the right.

Step 2.  Determine the relative changes needed to reach equilibrium, then write the                                                equilibrium concentrations in terms of these changes.

Let us represent the increase in concentration of PCl3 by the symbol x. The other changes may be written in terms of x by considering the coefficients in the chemical equation.

The changes in concentration and the expressions for the equilibrium concentrations are:

Step 3. Solve for the change and the equilibrium concentrations. 

Substituting the equilibrium concentrations into the equilibrium constant equation gives

This equation contains only one variable, x, the change in concentration. We can write the equation as a quadratic equation and solve for x using the quadratic formula.

Appendix B shows us an equation of the form ax2 +bx+c= 0 can be rearranged to solve for x:

In this case, a = 1, b = 0.0211, and c = −0.0211. Substituting the appropriate values for a, b, and c yields: 

Hence