Chapter 14: Chemical Equilibrium 1
chapter 14
Chemical Equilibrium
Chapter Terms and Definitions
Numbers in parentheses after definitions give the text sections in which the terms are explained. Starred terms are italicized in the text. Where a term does not fall directly under a text section heading, additional information is given for you to locate it.
reversible* describes chemical reactions in which products formed can themselves react, giving back the original reactants (chapter introduction)
catalytic methanation* conversion of carbon monoxide and hydrogen to methane and water in the presence of a catalyst (chapter introduction)
steam re-forming* preparing carbon monoxide and hydrogen by reacting hydrocarbons with steam (chapter introduction)
dynamic equilibrium* state in which the reactants and products of a reversible reaction or process are being formed at the same rate, such that there is no apparent change in the system (14.1)
chemical equilibrium state reached by a reaction mixture when the rates of forward and reverse reactions have become equal so that net change no longer occurs (14.1)
equilibrium constant* quantity relating equilibrium compositions for a particular reaction at a given temperature (14.2)
equilibrium-constant expression arrangement of symbols showing multiplication of the concentrations of reaction products and division by the concentrations of reactants, the concentration of each raised to a power equal to its coefficient in the chemical equation (14.2)
equilibrium constant (Kc) value obtained for the equilibrium-constant expression when equilibrium concentrations are substituted (14.2)
equilibrium constant (Kp) equilibrium constant for a gaseous reaction expressed in terms of partial pressures (14.2)
law of mass action relation stating that the values of the equilibrium-constant expression Kc are constant for a particular reaction at a given temperature whatever equilibrium concentrations are substituted (14.2)
activities* dimensionless quantities defining the equilibrium constant; for an ideal mixture, the activity of a substance is the ratio of its concentration (or partial pressure if a gas) to a standard concentration of 1 M (or partial pressure of 1 atm) so that units cancel (14.2, marginal note)
homogeneous equilibrium equilibrium that involves reactants and products in a single phase (14.3)
heterogeneous equilibrium equilibrium that involves reactants and products in more than one phase (14.3)
oscillating reaction* a reaction that cycles back and forth over time (A Chemist Looks at: Slime Molds and Leopard’s Spots)
reaction quotient (Qc) expression identical to the equilibrium-constant expression but with concentrations not necessarily those at equilibrium (14.5)
quadratic formula* solutions to a quadratic equation of the form ax2 + bx + c = 0;
x = (14.6, marginal note)
Le Châtelier’s principle when a system in chemical equilibrium is disturbed by a change of temperature, pressure, or a concentration, the system shifts in equilibrium composition in a way that tends to counteract this change of variable (14.7)
contact process* industrial method of preparing sulfuric acid by the catalytic oxidation of sulfur dioxide (14.9)
catalyst* substance that speeds up the attainment of equilibrium, is not consumed by the reaction, and has no effect on the equilibrium composition of the reaction mixture (14.9)
acid rain* rain with increased acidity owing to the presence of sulfuric and nitric acids (14.9, marginal note)
Ostwald process* industrial method of preparing nitric acid by the catalytic oxidation of ammonia (14.9)
Chapter Diagnostic Test
1.Write equilibrium-constant expressions for the following equilibria in terms of Kc.
a.2HCl(g) +½O2(g) H2O(g) + Cl2(g)
b.2NO(g) + Br2(g) 2NOBr(g)
c.Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
d.HCN(aq) + H2O(l) H3O+(aq) + CN–(aq)
e.4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g)
f.I3–(aq) + H2O(l) HOI(aq) + 2I–(aq) + H+(aq)
2.The equilibrium concentrations for the decomposition of PCl5(g) at 433 K,
PCl5(g) PCl3(g) + Cl2(g)
are [PCl5] = 0.865 mol/L, [PCl3] = [Cl2] = 0.135 mol/L. Calculate Kc.
3.What effect would an increase in pressure have on the equilibrium of the system in Problem 2?
4.A system containing nitrogen, hydrogen, and ammonia is allowed to come to equilibrium. The total equilibrium pressure is 5 atm. The partial pressures are = 1 atm, = 2 atm, and = 2 atm. Calculate Kp for the reaction
N2(g) + 3H2(g) 2NH3(g)
5.Using the data from Problem 4, calculate Kp for
N2(g) + H2(g) NH3(g)
6.If 1.00 mol CO2 and 1.00 mol H2 are placed in a 1.00-L flask at 825 K and react according to
CO2(g) + H2(g) CO(g) + H2O(g)
analysis of the equilibrium mixture shows 0.27 mol CO present. Determine Kc at this temperature.
7.Consider the following equilibrium:
2H2S(g) 2H2(g) + S2(g)
For a 5.00-L vessel containing the following amounts of gases, determine whether the initial concentrations of these gases will remain fixed or change (and if they change, indicate which gases will show an increase in concentration):
0.0131 mol H2, 0.00650 mol S2, 0.0383 mol H2S. Kc equals 2.3 104.
8.Consider the following reaction:
4HCl(g) + O2(g) 2H2O(g) + 2Cl2(g) ΔH = +28 kcal
Describe what happens to the composition of the equilibrium mixture and to the equilibrium constant K with each of the following changes to the system at equilibrium.
a.Addition of oxygen gas
b.An increase in temperature
c.Reduction of the volume of the reaction container
d.Addition of a catalyst
e.Removal of HCl(g) from the reaction vessel
9.I2 vapor is a deep purple color. The dissociation of HI(g) into H2(g) and I2(g) in a closed vessel can be followed qualitatively by observing changes in the relative intensity of the purple color of I2 vapor. When H2 gas is added to the reaction at equilibrium, the vapor slowly takes on a less intense purple color. Explain this observation in terms of Le Châtelier’s principle.
10.Consider the reaction
2SiO(g) 2Si(l) + O2(g) Kc = 9.62 101
If 1.00 mol SiO is placed into a 1.00-L container, what are the equilibrium concentrations of SiO and O2?
11.Once the equilibrium in Problem 10 is reached, how would adding 10.0 g of Si affect that equilibrium?
12.Consider the reaction
CO2(g) + H2(g) CO(g) + H2O(g); Kc = 0.137
If 5.0 mol each of CO2 and H2 are placed in a 10.0-L flask, what are the equilibrium concentrations?
13.Consider the following equilibrium:
2HI(g) H2(g) + I2(g)
At equilibrium, a 2.00-L vessel contains 1.25 mol I2, 1.25 mol H2, and an unknown amount of HI. Kc for this equilibrium is 0.0183. Calculate the equilibrium concentration of HI.
Answers to Chapter Diagnostic Test
If you missed an answer, study the text section and problem-solving skill given in parentheses after the answer.
1.
a.
(Note that [H2O] is included because the reaction is in the gas phase and [H2O] is not a constant.)
b.
c.
d.
e.
f. (14.2 , 14.3, PS Sk. 2)
2.Kc = 0.0211 (14.2, PS Sk. 3)
3.An increase in pressure would cause a shift to the side of the reaction that has the smaller number of moles of gaseous materials, which would reduce the increased pressure. Therefore, the rate of the reverse reaction would increase, and we say that we would observe a shift to the left. (14.8, PS Sk. 7)
4.Kp = 0.5 (14.2, PS Sk. 2, 3)
5.Kp = 0.7 (14.2, PS Sk. 2, 3)
6.Kc = 0.14 (14.1, 14.2, PS Sk. 1, 3)
7.Qc = 1.52 104, which is < Kc. Therefore, initial concentrations will change, and the concentrations of H2 and S2 gases will increase as the equilibrium shifts to the right. (14.4, 14.5, PS Sk. 4)
8.
a.Shift right to consume the added oxygen; no change in K. (14.7, PS Sk. 7)
b.Favors endothermic reaction, so it shifts right; K increases. (14.8, PS Sk. 7)
c.This increases the pressure, so reaction shifts right, toward fewer moles of gas; no change in K. (14.8, PS Sk. 7)
d.No effect on the equilibrium composition; no change in K. (14.9, PS Sk. 7)
e.Shift left to replace the HCl lost; no change in K. (14.7, 14.8, PS Sk. 7)
9.The equilibrium may be expressed
2HI(g) H2(g) + I2(g)
When H2 is added to the reaction at equilibrium, the rate of the reverse reaction increases. We say that the equilibrium shifts to decrease the concentration of H2 and thus to form more HI. I2 vapor is consumed in this equilibrium shift to the left. A loss of I2 vapor results in a less intense purple color. (14.5, 14.7, PS Sk. 7)
10.[SiO] = 0.51 mol/L, [O2] = 0.247 mol/L (14.6, PS Sk. 6)
11.It would not. The Si formed is a pure liquid with fixed density and thus fixed concentration. Adding more of it will not change the Si concentration. This constant value is, in effect, included in the value of Kc for the reaction and is not part of the equilibrium-constant expression. (14.3, 14.7, PS Sk. 7)
12.[CO2] = [H2] = 0.37 mol/L
[CO] = [H2O] = 0.14 mol/L (14.6, PS Sk. 6)
13.[HI] = 4.62 mol/L (14.6, PS Sk. 5)
Summary of Chapter Topics
Chapter 14 is the first of four chapters devoted to the study of chemical equilibrium. Most of the problems in these chapters, although involving different species, are worked in essentially the same manner. The key to performing these calculations is to set up the table shown in Example 14.1 in the text and used in the solutions in this study guide. You may balk at using this table format because it takes extra time to set it up. Take this extra time. It is extremely worthwhile.
In preparing the table, there usually will be an unknown quantity, which we designatex. For each problem, be sure to define, and write down, what you are letting x be. This is another crucial step to ensure that you work the problem correctly.
Pay close attention to the coefficients in the reaction when working with unknowns. If two molecules of a product substance are formed when one molecule reacts, then be sure your amount of substance that reacts is x and the amount of product formed is 2x.
Make sure that you then have the correct form of the equilibrium-constant expression. Many students forget exponents. Always go back and check that they are there if necessary. And remember that the form of the expression is always “products over reactants.”
We will not follow our typical “Wanted, Given, etc.,” format too closely in solving the exercises in these chapters because the table gives an equally useful structure to problem solving. You should be quite familiar with the necessary steps of problem solving by now.
14.1 Chemical Equilibrium—A Dynamic Equilibrium
Learning Objectives
- Define dynamic equilibrium and chemical equilibrium.
- Apply stoichiometry to an equilibrium mixture. (Example 14.1)
Problem-Solving Skill
1.Applying stoichiometry to an equilibrium mixture. Given the starting amounts of reactants and the amount of one substance at equilibrium, find the equilibrium composition (Example 14.1).
Exercise 14.1
Synthesis gas (a mixture of CO and H2) is increased in concentration of hydrogen by passing it with steam over a catalyst. This is the so-called water–gas shift reaction. Some of the CO is converted to CO2, which can be removed:
CO(g) + H2O(g) CO2(g) + H2(g)
Suppose that you start with a gaseous mixture containing 1.00 mol CO and 1.00 mol H2O. When equilibrium is reached at 1000C, the mixture contains 0.43 mol H2. What is the molar composition of the equilibrium mixture?
Solution: Set up the table under the equation; let x = moles CO that react.
Amounts (mol): CO(g) + H2O(g) CO2(g) + H2(g)
Starting / 1.00 / 1.00 / 0 / 0Change / x / x / +x / +x
Equilibrium / 1.00 x / 1.00 x / x / 0.43
For every mole of CO that reacts, 1 mol of H2 is produced. Therefore, x = 0.43 mol.
Equilibrium amount of CO = 1.00 – 0.43 = 0.57 mol CO
Equilibrium amount of H2O = 1.00 – 0.43 = 0.57 mol H2O
Equilibrium amount of CO2 = x = 0.43 mol
Equilibrium amount of H2 = 0.43 mol (as given)
14.2 The Equilibrium Constant
Learning Objectives
- Define equilibrium-constant expression and equilibrium constant.
- State the law of mass action.
- Write equilibrium-constant expressions. (Example 14.2)
- Describe the kinetics argument for the approach to chemical equilibrium.
- Obtain an equilibrium constant from reaction composition. (Example 14.3)
- Describe the equilibrium constant Kp; indicate how Kpand Kcare related.
- Obtain Kcfor a reaction that can be written as a sum of other reactions of known Kcvalues.
Problem-Solving Skills
2.Writing equilibrium-constant expressions. Given the chemical equation, write the equilibrium-constant expression (Example 14.2). (See also Section 14.3)
3.Obtaining an equilibrium constant from reaction composition. Given the equilibrium composition, find Kc (Example 14.3).
The equilibrium constant for a given reaction and equation is constant for that equation as long as the temperature remains unchanged. No matter what the starting mixture or how the equilibrium system is perturbed, the value of this constant will be the same as long as the temperature is not changed. This concept is called the law of mass action.
Exercise 14.2
a.Write the equilibrium-constant expression Kc for the equation
2NO2(g) + 7H2(g) 2NH3(g) + 4H2O(g)
b.Write the equilibrium-constant expression Kc when this reaction is written
NO2(g) + H2(g) NH3(g) + 2H2O(g)
Known:The equilibrium-constant expression is equal to product concentrations over reactant concentrations, each to the coefficient power.
Solution: For (a):
Kc =
For (b):
Kc =
Exercise 14.3
When 1.00 mol each of carbon monoxide and water reach equilibrium at 1000C in a 10.0-L vessel, the equilibrium mixture contains 0.57molCO, 0.57 mol H2O, 0.43 mol CO2, and 0.43 mol H2. Write the chemical equation for the equilibrium. What is the value of Kc?
Solution: The equation is
CO(g) + H2O(g) CO2(g) + H2(g)
The equilibrium-constant expression is
Kc =
Each concentration must be calculated at equilibrium and then placed in this expression:
[CO] = [H2O] = = 0.057 M
[CO2] = [H2] = = 0.043 M
Kc = = 0.57
In this case, the units cancel. It is conventional, however, that even when they do not cancel, we do not write units for an equilibrium constant. The reason for this is discussed in text Section 18.6.
Exercise 14.4
Hydrogen sulfide, a colorless gas with a foul odor, dissociates on heating:
2H2S(g) 2H2(g) + S2(g)
When 0.100 mol H2S was put into a 10.0-L vessel and heated to 1132C, it gave an equilibrium mixture containing 0.0285 mol H2. What is the value of Kc at this temperature?
Solution: First, change amounts to concentrations:
Starting concentration of H2S = = 0.0100 M
Equilibrium concentration of H2 = = 0.00285 M
Second, set up a table under the equation, letting x = mol/L of H2S that react.
Concentration (M) 2H2S(g) 2H2(g) + S2(g)
Starting / 0.0100 / 0 / 0Change / 2x / +2x / +x
Equilibrium / 0.0100 2x / 0.00285 = 2x / x
Third, calculate equilibrium concentrations from the bottom line in the table:
[H2S] = (0.0100 – 0.00285) M = 0.00715 M
[H2] = 0.00285 M (as given)
[S2] = = 0.00143 M
Finally, write the equilibrium-constant expression from the equation and substitute in the calculated molarities:
Kc = = = = 2.3 104
Exercise 14.5
Phosphorus pentachloride dissociates on heating:
PCl5(g) PCl3(g) + Cl2(g)
If Kc equals 3.26 102 at 191C, what is Kp at this temperature?
Known:Kp = Kc(RT)n, where n is the sum of gaseous product coefficients minus gaseous reactant coefficients, and R = 0.0821(L ∙ atm)/(K ∙ mol).
Solution:
n = 2 – 1 = 1
T = 191 + 273 = 464 K
Kp = Kc(RT) = 3.26 1020.0821 464 = 1.24
Note that we do not use units because Kc had no units. However, R must be expressed as (L ∙ atm)/(K ∙ mol) and T in kelvins.
14.3 Heterogeneous Equilibria; Solvents in Homogeneous Equilibria
Learning Objectives
- Define homogeneous equilibrium and heterogeneous equilibrium.
- Write Kc for a reaction with pure solids or liquids. (Example 14.4)
Problem-Solving Skill
2.Writing equilibrium-constant expressions. Given the chemical equation, write the equilibrium-constant expression (Example 14.4).*
In working problems with gas reactions that may include heterogeneous equilibria, pay careful attention to the states of the substances in the equations. Only the gases are included in an equilibrium-constant expression.
Exercise 14.6
The Mond process for purifying nickel involves the formation of nickel tetracarbonyl, Ni(CO)4, a volatile liquid, from nickel metal and carbon monoxide. Carbon monoxide is passed over impure nickel to form nickel carbonyl vapor, which, when heated, decomposes and deposits pure nickel.
Ni(s) + 4CO(g) Ni(CO)4(g)
Write the expression for Kc for this reaction.
Solution: Kc =
*Note that Problem-Solving Skill 2 is used again in this section. Here, as well as in later chapters, problem-solving skills are repeated as needed.
14.4 Qualitatively Interpreting the Equilibrium Constant
Learning Objective
- Give a qualitative interpretation of the equilibrium constant based on its value.
Exercise 14.7
The equilibrium constant Kc for the reaction
2NO(g) + O2(g) 2NO2(g)
equals 4.0 1013 at 25C. Does the equilibrium mixture contain predominantly reactants or products? If [NO] = [O2] = 2.0 10–6M at equilibrium, what is the equilibrium concentration of NO2?
Solution: Since Kc is large, the equilibrium mixture contains mostly products. Determine [NO2] at equilibrium by substituting values in the equilibrium-constant expression and solving for [NO2]:
Kc = = = 4.0 1013
[NO2] = 1.8 10–2M
14.5 Predicting the Direction of Reaction
Learning Objective
- Use the reaction quotient Q.
- Describe the direction of reaction after comparing Q with Kc.
- Use the reaction quotient. (Example 14.5)
Problem-Solving Skill
4.Using the reaction quotient. Given the concentrations of substances in a reaction mixture, predict the direction of reaction (Example 14.5).
Exercise 14.8
A 10.0-L vessel contains 0.0015 mol CO2 and 0.10 mol CO. If a small amount of carbon is added to this vessel and the temperature raised to 1000C, will more CO form? The reaction is
CO2(g) + C(s) 2CO(g)
The value of Kc for this reaction is 1.17 at 1000C. Assume that the volume of gas in the vessel is 10.0 L.
Wanted: Will more CO form?
Given:10.0-L vessel, 0.0015 mol CO2, 0.10 mol CO; Kc at 1000C = 1.17
Known:Calculate the value of the reaction quotient Qc and compare with Kc.
Qc =
Solution: Calculate concentrations:
[CO2]I = = 0.00015 M CO2
[CO] I = = 0.010 M CO
Qc = = 0.667
Qc is less than Kc. Thus, yes, the rate of the forward reaction will increase to produce more CO.
14.6 Calculating Equilibrium Concentrations
Learning Objectives
- Obtain one equilibrium concentration given the others. (Example 14.6)
- Solve an equilibrium problem (involving a linear equation in x). (Example 14.7)
- Solve an equilibrium problem (involving a quadratic equation in x). (Example 14.8)
Problem-Solving Skills
5.Obtaining one equilibrium concentration given the others. Given Kc and all concentrations of substances but one in an equilibrium mixture, calculate the concentration of this one substance (Example 14.6).
6.Solving equilibrium problems. Given the starting composition and Kc of a reaction mixture, calculate the equilibrium composition (Examples 14.7 and 14.8).
Exercise 14.9
Phosphorus pentachloride gives an equilibrium mixture of PCl5, PCl3, and Cl2 when heated.
PCl5(g) PCl3(g) + Cl2(g)
A 1.00-L vessel contains an unknown amount of PCl5 and 0.020 mol each of PCl3 and Cl2 at equilibrium at 250C. How many moles of PCl5 are in the vessel if Kc for this reaction is 0.0415 at 250C?
Wanted:moles PCl5
Given:V = 1.00 L; 0.020 mol PCl3, 0.020 mol Cl2; Kc = 0.0415
Known:Kc =
Solution: Solve the preceding for [PCl5] and substitute in known values:
[PCl5] = =
= 0.0096
Moles PCl5 = 0.0096
Note that in substituting values into the equilibrium-constant expression in Exercise 14.9, we did not write [0.020/1.00]2. We wrote the number twice. The reason for doing this, when you are in a hurry, such as when taking a test, is that you will not forget to square the value. If you write it twice, you will avoid making that error.
Exercise 14.10
What is the equilibrium composition of a reaction mixture if you start with 0.500 mol each of H2 and I2 in a 1.0-L vessel? The reaction is
H2(g) + I2(g) 2HI(g)
Kc = 49.7 at 458C
Solution: First, set up a table of concentrations. Since it is a 1-L vessel, the concentrations are the mole amounts to two significant figures.
Concentrations (M): H2(g) + I2(g) 2HI(g)
Starting / 0.500 / 0.500 / 0Change / x / x / 2x
Equilibrium / 0.500 x / 0.500 x / 2x
Second, substitute values into the equilibrium-constant expression:
Kc = = = = 49.7
Third, take the square root of both sides and solve for x:
= ±7.050 (only the positive value is possible)
2x = 3.52 – 7.050x
9.050x = 3.52
x = = 0.389 M
Thus equilibrium concentrations are
[H2] = [I2] = 0.500 0.389 = 0.11 M
[HI] = 2(0.389) = 0.78 M
When calculating equilibrium concentrations given only starting amounts, be sure to look for ways to simplify your work so that you do not have to deal with x2. First, check to see whether the expression with the x is a perfect square. If it appears not to be, check to see that you’ve written the equilibrium-constant expression with the proper exponents. If you still do not have a perfect square, then see if you can eliminate the x that is subtracted from a beginning concentration. You can ignore this change when it is very small compared with the other number. This often will be the case when the equilibrium constant divided by the initial concentration is 103 or less. If you cannot simplify the work, then you will have to use the quadratic formula to solve for x. It is very useful but takes a long time to solve, and when you use it, you must take great care to avoid errors.
Exercise 14.11
Phosphorus pentachloride, PCl5, decomposes when heated.
PCl5(g) PCl3(g) + Cl2(g)
If the initial concentration of PCl5 is 1.00 mol/L, what is the equilibrium composition of the gaseous mixture at 160C? The equilibrium constant Kc at 160C is 0.0211.
Solution: Set up the table of concentrations.
Concentrations (M): PCl5(g) PCl3(g) + Cl2(g)