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Chemistry 2e

13.2 Equilibrium Constants

Chemistry 2e13.2 Equilibrium Constants
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Table of contents
  1. Preface
  2. 1 Essential Ideas
    1. Introduction
    2. 1.1 Chemistry in Context
    3. 1.2 Phases and Classification of Matter
    4. 1.3 Physical and Chemical Properties
    5. 1.4 Measurements
    6. 1.5 Measurement Uncertainty, Accuracy, and Precision
    7. 1.6 Mathematical Treatment of Measurement Results
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  3. 2 Atoms, Molecules, and Ions
    1. Introduction
    2. 2.1 Early Ideas in Atomic Theory
    3. 2.2 Evolution of Atomic Theory
    4. 2.3 Atomic Structure and Symbolism
    5. 2.4 Chemical Formulas
    6. 2.5 The Periodic Table
    7. 2.6 Ionic and Molecular Compounds
    8. 2.7 Chemical Nomenclature
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  4. 3 Composition of Substances and Solutions
    1. Introduction
    2. 3.1 Formula Mass and the Mole Concept
    3. 3.2 Determining Empirical and Molecular Formulas
    4. 3.3 Molarity
    5. 3.4 Other Units for Solution Concentrations
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  5. 4 Stoichiometry of Chemical Reactions
    1. Introduction
    2. 4.1 Writing and Balancing Chemical Equations
    3. 4.2 Classifying Chemical Reactions
    4. 4.3 Reaction Stoichiometry
    5. 4.4 Reaction Yields
    6. 4.5 Quantitative Chemical Analysis
    7. Key Terms
    8. Key Equations
    9. Summary
    10. Exercises
  6. 5 Thermochemistry
    1. Introduction
    2. 5.1 Energy Basics
    3. 5.2 Calorimetry
    4. 5.3 Enthalpy
    5. Key Terms
    6. Key Equations
    7. Summary
    8. Exercises
  7. 6 Electronic Structure and Periodic Properties of Elements
    1. Introduction
    2. 6.1 Electromagnetic Energy
    3. 6.2 The Bohr Model
    4. 6.3 Development of Quantum Theory
    5. 6.4 Electronic Structure of Atoms (Electron Configurations)
    6. 6.5 Periodic Variations in Element Properties
    7. Key Terms
    8. Key Equations
    9. Summary
    10. Exercises
  8. 7 Chemical Bonding and Molecular Geometry
    1. Introduction
    2. 7.1 Ionic Bonding
    3. 7.2 Covalent Bonding
    4. 7.3 Lewis Symbols and Structures
    5. 7.4 Formal Charges and Resonance
    6. 7.5 Strengths of Ionic and Covalent Bonds
    7. 7.6 Molecular Structure and Polarity
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  9. 8 Advanced Theories of Covalent Bonding
    1. Introduction
    2. 8.1 Valence Bond Theory
    3. 8.2 Hybrid Atomic Orbitals
    4. 8.3 Multiple Bonds
    5. 8.4 Molecular Orbital Theory
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  10. 9 Gases
    1. Introduction
    2. 9.1 Gas Pressure
    3. 9.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law
    4. 9.3 Stoichiometry of Gaseous Substances, Mixtures, and Reactions
    5. 9.4 Effusion and Diffusion of Gases
    6. 9.5 The Kinetic-Molecular Theory
    7. 9.6 Non-Ideal Gas Behavior
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  11. 10 Liquids and Solids
    1. Introduction
    2. 10.1 Intermolecular Forces
    3. 10.2 Properties of Liquids
    4. 10.3 Phase Transitions
    5. 10.4 Phase Diagrams
    6. 10.5 The Solid State of Matter
    7. 10.6 Lattice Structures in Crystalline Solids
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  12. 11 Solutions and Colloids
    1. Introduction
    2. 11.1 The Dissolution Process
    3. 11.2 Electrolytes
    4. 11.3 Solubility
    5. 11.4 Colligative Properties
    6. 11.5 Colloids
    7. Key Terms
    8. Key Equations
    9. Summary
    10. Exercises
  13. 12 Kinetics
    1. Introduction
    2. 12.1 Chemical Reaction Rates
    3. 12.2 Factors Affecting Reaction Rates
    4. 12.3 Rate Laws
    5. 12.4 Integrated Rate Laws
    6. 12.5 Collision Theory
    7. 12.6 Reaction Mechanisms
    8. 12.7 Catalysis
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  14. 13 Fundamental Equilibrium Concepts
    1. Introduction
    2. 13.1 Chemical Equilibria
    3. 13.2 Equilibrium Constants
    4. 13.3 Shifting Equilibria: Le Châtelier’s Principle
    5. 13.4 Equilibrium Calculations
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  15. 14 Acid-Base Equilibria
    1. Introduction
    2. 14.1 Brønsted-Lowry Acids and Bases
    3. 14.2 pH and pOH
    4. 14.3 Relative Strengths of Acids and Bases
    5. 14.4 Hydrolysis of Salts
    6. 14.5 Polyprotic Acids
    7. 14.6 Buffers
    8. 14.7 Acid-Base Titrations
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  16. 15 Equilibria of Other Reaction Classes
    1. Introduction
    2. 15.1 Precipitation and Dissolution
    3. 15.2 Lewis Acids and Bases
    4. 15.3 Coupled Equilibria
    5. Key Terms
    6. Key Equations
    7. Summary
    8. Exercises
  17. 16 Thermodynamics
    1. Introduction
    2. 16.1 Spontaneity
    3. 16.2 Entropy
    4. 16.3 The Second and Third Laws of Thermodynamics
    5. 16.4 Free Energy
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  18. 17 Electrochemistry
    1. Introduction
    2. 17.1 Review of Redox Chemistry
    3. 17.2 Galvanic Cells
    4. 17.3 Electrode and Cell Potentials
    5. 17.4 Potential, Free Energy, and Equilibrium
    6. 17.5 Batteries and Fuel Cells
    7. 17.6 Corrosion
    8. 17.7 Electrolysis
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  19. 18 Representative Metals, Metalloids, and Nonmetals
    1. Introduction
    2. 18.1 Periodicity
    3. 18.2 Occurrence and Preparation of the Representative Metals
    4. 18.3 Structure and General Properties of the Metalloids
    5. 18.4 Structure and General Properties of the Nonmetals
    6. 18.5 Occurrence, Preparation, and Compounds of Hydrogen
    7. 18.6 Occurrence, Preparation, and Properties of Carbonates
    8. 18.7 Occurrence, Preparation, and Properties of Nitrogen
    9. 18.8 Occurrence, Preparation, and Properties of Phosphorus
    10. 18.9 Occurrence, Preparation, and Compounds of Oxygen
    11. 18.10 Occurrence, Preparation, and Properties of Sulfur
    12. 18.11 Occurrence, Preparation, and Properties of Halogens
    13. 18.12 Occurrence, Preparation, and Properties of the Noble Gases
    14. Key Terms
    15. Summary
    16. Exercises
  20. 19 Transition Metals and Coordination Chemistry
    1. Introduction
    2. 19.1 Occurrence, Preparation, and Properties of Transition Metals and Their Compounds
    3. 19.2 Coordination Chemistry of Transition Metals
    4. 19.3 Spectroscopic and Magnetic Properties of Coordination Compounds
    5. Key Terms
    6. Summary
    7. Exercises
  21. 20 Organic Chemistry
    1. Introduction
    2. 20.1 Hydrocarbons
    3. 20.2 Alcohols and Ethers
    4. 20.3 Aldehydes, Ketones, Carboxylic Acids, and Esters
    5. 20.4 Amines and Amides
    6. Key Terms
    7. Summary
    8. Exercises
  22. 21 Nuclear Chemistry
    1. Introduction
    2. 21.1 Nuclear Structure and Stability
    3. 21.2 Nuclear Equations
    4. 21.3 Radioactive Decay
    5. 21.4 Transmutation and Nuclear Energy
    6. 21.5 Uses of Radioisotopes
    7. 21.6 Biological Effects of Radiation
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  23. A | The Periodic Table
  24. B | Essential Mathematics
  25. C | Units and Conversion Factors
  26. D | Fundamental Physical Constants
  27. E | Water Properties
  28. F | Composition of Commercial Acids and Bases
  29. G | Standard Thermodynamic Properties for Selected Substances
  30. H | Ionization Constants of Weak Acids
  31. I | Ionization Constants of Weak Bases
  32. J | Solubility Products
  33. K | Formation Constants for Complex Ions
  34. L | Standard Electrode (Half-Cell) Potentials
  35. M | Half-Lives for Several Radioactive Isotopes
  36. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
    18. Chapter 18
    19. Chapter 19
    20. Chapter 20
    21. Chapter 21
  37. Index

Learning Objectives

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

The status of a reversible reaction is conveniently assessed by evaluating its reaction quotient (Q). For a reversible reaction described by

mA+nB+⇌xC+yDmA+nB+⇌xC+yD

the reaction quotient is derived directly from the stoichiometry of the balanced equation as

Qc=[C]x[D]y[A]m[B]nQc=[C]x[D]y[A]m[B]n

where the subscript c denotes the use of molar concentrations in the expression. If the reactants and products are gaseous, a reaction quotient may be similarly derived using partial pressures:

Qp=PCxPDyPAmPBnQp=PCxPDyPAmPBn

Note that the reaction quotient equations above are a simplification of more rigorous expressions that use relative values for concentrations and pressures rather than absolute values. These relative concentration and pressure values are dimensionless (they have no units); consequently, so are the reaction quotients. For purposes of this introductory text, it will suffice to use the simplified equations and to disregard units when computing Q. In most cases, this will introduce only modest errors in calculations involving reaction quotients.

Example 13.1

Writing Reaction Quotient Expressions

Write the concentration-based reaction quotient expression for each of the following reactions:

(a) 3O2(g)⇌2O3(g)3O2(g)⇌2O3(g)

(b) N2(g)+3H2(g)⇌2NH3(g)N2(g)+3H2(g)⇌2NH3(g)

(c) 4NH3(g)+7O2(g)⇌4NO2(g)+6H2O(g)4NH3(g)+7O2(g)⇌4NO2(g)+6H2O(g)

Solution

(a) Qc=[O3]2[O2]3Qc=[O3]2[O2]3

(b) Qc=[NH3]2[N2][H2]3Qc=[NH3]2[N2][H2]3

(c) Qc=[NO2]4[H2O]6[NH3]4[O2]7Qc=[NO2]4[H2O]6[NH3]4[O2]7

Check Your Learning

Write the concentration-based reaction quotient expression for each of the following reactions:

(a) 2SO2(g)+O2(g)⇌2SO3(g)2SO2(g)+O2(g)⇌2SO3(g)

(b) C4H8(g)⇌2C2H4(g)C4H8(g)⇌2C2H4(g)

(c) 2C4H10(g)+13O2(g)⇌8CO2(g)+10H2O(g)2C4H10(g)+13O2(g)⇌8CO2(g)+10H2O(g)

Answer:

(a) Qc=[SO3]2[SO2]2[O2];Qc=[SO3]2[SO2]2[O2]; (b) Qc=[C2H4]2[C4H8];Qc=[C2H4]2[C4H8]; (c) Qc=[CO2]8[H2O]10[C4H10]2[O2]13Qc=[CO2]8[H2O]10[C4H10]2[O2]13

Four graphs are shown and labeled, “a,” “b,” “c,” and “d.” All four graphs have a vertical dotted line running through the middle labeled, “Equilibrium is reached.” The y-axis on graph a is labeled, “Concentration,” and the x-axis is labeled, “Time.” Three curves are plotted on graph a. The first is labeled, “[ S O subscript 2 ];” this line starts high on the y-axis, ends midway down the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, “[ O subscript 2 ];” this line mimics the first except that it starts and ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, “[ S O subscript 3 ].” The y-axis on graph b is labeled, “Concentration,” and the x-axis is labeled, “Time.” Three curves are plotted on graph b. The first is labeled, “[ S O subscript 2 ];” this line starts low on the y-axis, ends midway up the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, “[ O subscript 2 ];” this line mimics the first except that it ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, “[ S O subscript 3 ].” The y-axis on graph c is labeled, “Reaction Quotient,” and the x-axis is labeled, “Time.” A single curve is plotted on graph c. This curve begins at the bottom of the y-axis and rises steeply up near the top of the y-axis, then levels off into a horizontal line. The top point of this line is labeled, “k.” The y-axis on graph d is labeled, “Reaction Quotient,” and the x-axis is labeled, “Time.” A single curve is plotted on graph d. This curve begins near the edge of the top of the y-axis and falls steeply toward the x-axis, then levels off into a horizontal line. The bottom point of this line is labeled, “k.”
Figure 13.5 Changes in concentrations and Qc for a chemical equilibrium achieved beginning with (a) a mixture of reactants only and (b) products only.

The numerical value of Q varies as a reaction proceeds towards equilibrium; therefore, it can serve as a useful indicator of the reaction’s status. To illustrate this point, consider the oxidation of sulfur dioxide:

2SO2(g)+O2(g)⇌2SO3(g)2SO2(g)+O2(g)⇌2SO3(g)

Two different experimental scenarios are depicted in Figure 13.5, one in which this reaction is initiated with a mixture of reactants only, SO2 and O2, and another that begins with only product, SO3. For the reaction that begins with a mixture of reactants only, Q is initially equal to zero:

Qc=[SO3]2[SO2]2[O2]=02[SO2]2[O2]=0Qc=[SO3]2[SO2]2[O2]=02[SO2]2[O2]=0

As the reaction proceeds toward equilibrium in the forward direction, reactant concentrations decrease (as does the denominator of Qc), product concentration increases (as does the numerator of Qc), and the reaction quotient consequently increases. When equilibrium is achieved, the concentrations of reactants and product remain constant, as does the value of Qc.

If the reaction begins with only product present, the value of Qc is initially undefined (immeasurably large, or infinite):

Qc=[SO3]2[SO2]2[O2]=[SO3]20→∞Qc=[SO3]2[SO2]2[O2]=[SO3]20→∞

In this case, the reaction proceeds toward equilibrium in the reverse direction. The product concentration and the numerator of Qc decrease with time, the reactant concentrations and the denominator of Qc increase, and the reaction quotient consequently decreases until it becomes constant at equilibrium.

The constant value of Q exhibited by a system at equilibrium is called the equilibrium constant, K:

K≡Qat equilibriumK≡Qat equilibrium

Comparison of the data plots in Figure 13.5 shows that both experimental scenarios resulted in the same value for the equilibrium constant. This is a general observation for all equilibrium systems, known as the law of mass action: At a given temperature, the reaction quotient for a system at equilibrium is constant.

Example 13.2

Evaluating a Reaction Quotient

Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:
2NO2(g)⇌N2O4(g)2NO2(g)⇌N2O4(g)

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

As for all equilibrium calculations in this text, use the simplified equations for Q and K and disregard any concentration or pressure units, as noted previously in this section.

(a) Before any product is formed, [NO2]=0.10mol1.0L=0.10M,[NO2]=0.10mol1.0L=0.10M, and [N2O4] = 0 M. Thus,

Qc=[N2O4][NO2]2=00.102=0Qc=[N2O4][NO2]2=00.102=0

(b) At equilibrium, Kc=Qc=[N2O4][NO2]2=0.0420.0162=1.6×102.Kc=Qc=[N2O4][NO2]2=0.0420.0162=1.6×102. The equilibrium constant is 1.6 ×× 102.

Check Your Learning

For the reaction 2SO2(g)+O2(g)⇌2SO3(g),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

By its definition, the magnitude of an equilibrium constant explicitly reflects the composition of a reaction mixture at equilibrium, and it may be interpreted with regard to the extent of the forward reaction. A reaction exhibiting a large K will reach equilibrium when most of the reactant has been converted to product, whereas a small K indicates the reaction achieves equilibrium after very little reactant has been converted. It’s important to keep in mind that the magnitude of K does not indicate how rapidly or slowly equilibrium will be reached. Some equilibria are established so quickly as to be nearly instantaneous, and others so slowly that no perceptible change is observed over the course of days, years, or longer.

The equilibrium constant for a reaction can be used to predict the behavior of mixtures containing its reactants and/or products. As demonstrated by the sulfur dioxide oxidation process described above, a chemical reaction will proceed in whatever direction is necessary to achieve equilibrium. Comparing Q to K for an equilibrium system of interest allows prediction of what reaction (forward or reverse), if any, will occur.

To further illustrate this important point, consider the reversible reaction shown below:

CO(g)+H2O(g)⇌CO2(g)+H2(g)Kc=0.640T=800°CCO(g)+H2O(g)⇌CO2(g)+H2(g)Kc=0.640T=800°C

The bar charts in Figure 13.6 represent changes in reactant and product concentrations for three different reaction mixtures. The reaction quotients for mixtures 1 and 3 are initially lesser than the reaction’s equilibrium constant, so each of these mixtures will experience a net forward reaction to achieve equilibrium. The reaction quotient for mixture 2 is initially greater than the equilibrium constant, so this mixture will proceed in the reverse direction until equilibrium is established.

Two sets of bar graphs are shown. The left is labeled, “Before reaction,” and the right is labeled, “At equilibrium.” Both graphs have y-axes labeled, “Concentration ( M ),” and three bars on the x-axes labeled, “Mixture 1,” “Mixture 2,” and “Mixture 3.” The y-axis has a scale beginning at 0.00 and ending at 0.10, with measurement increments of 0.02. The bars on the graphs are color coded and a key is provided with a legend. Red is labeled, “C O;” blue is labeled, “H subscript 2 O;” green is labeled, “C O subscript 2,” and yellow is labeled, “H subscript 2.” The graph on the left shows the red bar for mixture one just above 0.02, labeled “0.0243,” and the blue bar near 0.05, labeled “0.0243.” For mixture two, the green bar is near 0.05, labeled “0.0468,” and the yellow bar is near 0.09, labeled “0.0468.” For mixture 3, the red bar is near 0.01, labeled “0.0330,” the blue bar is slightly above that, labeled “0.190,” with green and yellow topping it off at 0.02. Green is labeled “0.00175” and yellow is labeled “0.00160.” On the right graph, the bar for mixture one shows the red bar slightly above 0.01, labeled “0.0135,” the blue bar stacked on it rising slightly above 0.02, labeled “0.0135,” the green rising near 0.04, labeled “0.0108,” and the yellow bar reaching near 0.05, labeled “0.0108.” A label above this bar reads, “Q equals 0.640.” The bar for mixture two shows the red bar slightly above 0.02, labeled “0.0260,” the blue bar stacked on it rising near 0.05, labeled “0.0260,” the green rising near 0.07, labeled “0.0208,” and the yellow bar reaching near 0.10, labeled “0.0208.” A label above this bar reads “Q equals 0.640.” The bar for mixture three shows the red bar near 0.01, labeled “0.0231,” the blue bar stacked on it rising slightly above 0.01, labeled “0.00909,” the green rising near 0.02, labeled “0.0115,” and the yellow bar reaching 0.02, labeled “0.0117.” A label above this bar reads “Q equals 0.640”.
Figure 13.6 Compositions of three mixtures before (Qc ≠ Kc) and after (Qc = Kc) equilibrium is established for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g).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:
CO(g)+H2O(g)⇌CO2(g)+H2(g)CO(g)+H2O(g)⇌CO2(g)+H2(g)
Kc=0.64Kc=0.64

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

Reactants/Products Experiment 1 Experiment 2 Experiment 3
[CO]i 0.020 M 0.011 M 0.0094 M
[H2O]i 0.020 M 0.0011 M 0.0025 M
[CO2]i 0.0040 M 0.037 M 0.0015 M
[H2]i 0.0040 M 0.046 M 0.0076 M

Solution

Experiment 1:
Qc=[CO2][H2][CO][H2O]=(0.0040)(0.0040)(0.020)(0.020)=0.040.Qc=[CO2][H2][CO][H2O]=(0.0040)(0.0040)(0.020)(0.020)=0.040.

Qc < Kc (0.040 < 0.64)

The reaction will proceed in the forward direction.

Experiment 2:

Qc=[CO2][H2][CO][H2O]=(0.037)(0.046)(0.011)(0.0011)=1.4×102Qc=[CO2][H2][CO][H2O]=(0.037)(0.046)(0.011)(0.0011)=1.4×102

Qc > Kc (140 > 0.64)

The reaction will proceed in the reverse direction.

Experiment 3:

Qc=[CO2][H2][CO][H2O]=(0.0015)(0.0076)(0.0094)(0.0025)=0.48Qc=[CO2][H2][CO][H2O]=(0.0015)(0.0076)(0.0094)(0.0025)=0.48

Qc < Kc (0.48 < 0.64)

The reaction will proceed in the forward direction.

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:

2NO(g)+Cl2(g)⇌2NOCl(g)Kc=4.6×1042NO(g)+Cl2(g)⇌2NOCl(g)Kc=4.6×104

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

N2(g)+3H2(g)⇌2NH3(g)Kc=0.060N2(g)+3H2(g)⇌2NH3(g)Kc=0.060

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

2SO3(g)⇌2SO2(g)+O2(g)Kc=0.2302SO3(g)⇌2SO2(g)+O2(g)Kc=0.230

Answer:

(a) Qc = 6.45 ×× 103, forward. (b) Qc = 0.23, reverse. (c) Qc = 0, forward.

Homogeneous Equilibria

A homogeneous equilibrium is one in which all reactants and products (and any catalysts, if applicable) are present in the same phase. By this definition, homogeneous equilibria take place in solutions. These solutions are most commonly either liquid or gaseous phases, as shown by the examples below:

C2H2(aq)+2Br2(aq)⇌C2H2Br4(aq)Kc=[C2H2Br4][C2H2][Br2]2 I2(aq)+I−(aq)⇌I3−(aq)Kc=[I3−][I2][I−] HF(aq)+H2O(l)⇌H3O+(aq)+F−(aq)Kc=[H3O+][F−][HF] NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)Kc=[NH4+][OH−][NH3]C2H2(aq)+2Br2(aq)⇌C2H2Br4(aq)Kc=[C2H2Br4][C2H2][Br2]2 I2(aq)+I−(aq)⇌I3−(aq)Kc=[I3−][I2][I−] HF(aq)+H2O(l)⇌H3O+(aq)+F−(aq)Kc=[H3O+][F−][HF] NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)Kc=[NH4+][OH−][NH3]

These examples all involve aqueous solutions, those in which water functions as the solvent. In the last two examples, water also functions as a reactant, but its concentration is not included in the reaction quotient. The reason for this omission is related to the more rigorous form of the Q (or K) expression mentioned previously in this chapter, in which relative concentrations for liquids and solids are equal to 1 and needn’t be included. Consequently, reaction quotients include concentration or pressure terms only for gaseous and solute species.

The equilibria below all involve gas-phase solutions:

C2H6(g)⇌C2H4(g)+H2(g)Kc=[C2H4][H2][C2H6] 3O2(g)⇌2O3(g)Kc=[O3]2[O2]3 N2(g)+3H2(g)⇌2NH3(g)Kc=[NH3]2[N2][H2]3 C3H8(g)+5O2(g)⇌3CO2(g)+4H2O(g)Kc=[CO2]3[H2O]4[C3H8][O2]5C2H6(g)⇌C2H4(g)+H2(g)Kc=[C2H4][H2][C2H6] 3O2(g)⇌2O3(g)Kc=[O3]2[O2]3 N2(g)+3H2(g)⇌2NH3(g)Kc=[NH3]2[N2][H2]3 C3H8(g)+5O2(g)⇌3CO2(g)+4H2O(g)Kc=[CO2]3[H2O]4[C3H8][O2]5

For gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations (Kc) or partial pressures (Kp) of the reactants and products. A relation between these two K values may be simply derived from the ideal gas equation and the definition of molarity:

PV=nRTPV=nRT
P=(nV)RTP=(nV)RT
=MRT=MRT

where P is partial pressure, V is volume, n is molar amount, R is the gas constant, T is temperature, and M is molar concentration.

For the gas-phase reaction mA+nB⇌xC+yD:mA+nB⇌xC+yD:

KP=(PC)x(PD)y(PA)m(PB)nKP=(PC)x(PD)y(PA)m(PB)n
=([C]×RT)x([D]×RT)y([A]×RT)m([B]×RT)n=([C]×RT)x([D]×RT)y([A]×RT)m([B]×RT)n
=[C]x[D]y[A]m[B]n×(RT)x+y(RT)m+n=[C]x[D]y[A]m[B]n×(RT)x+y(RT)m+n
=Kc(RT)(x+y)−(m+n)=Kc(RT)(x+y)−(m+n)
=Kc(RT)Δn=Kc(RT)Δn

And so, the relationship between Kc and KP is

KP=Kc(RT)ΔnKP=Kc(RT)Δn

where Δn is the difference in the molar amounts of product and reactant gases, in this case:

Δn=(x+y)−(m+n)Δn=(x+y)−(m+n)

Example 13.4

Calculation of KP

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

(a) C2H6(g)⇌C2H4(g)+H2(g)C2H6(g)⇌C2H4(g)+H2(g)

(b) CO(g)+H2O(g)⇌CO2(g)+H2(g)CO(g)+H2O(g)⇌CO2(g)+H2(g)

(c) N2(g)+3H2(g)⇌2NH3(g)N2(g)+3H2(g)⇌2NH3(g)

(d) Kc is equal to 0.28 for the following reaction at 900 °C:

CS2(g)+4H2(g)⇌CH4(g)+2H2S(g)CS2(g)+4H2(g)⇌CH4(g)+2H2S(g)

What is KP at this temperature?

Solution

(a) Δn = (2) − (1) = 1
 KP = Kc (RT)Δn = Kc (RT)1 = Kc (RT)

(b) Δn = (2) − (2) = 0
 KP = Kc (RT)Δn = Kc (RT)0 = Kc

(c) Δn = (2) − (1 + 3) = −2
 KP = Kc (RT)Δn = Kc (RT)−2 = Kc(RT)2Kc(RT)2

(d) KP = Kc (RT)Δn = (0.28)[(0.0821)(1173)]−2 = 3.0 ×× 10−5

Check Your Learning

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

(a) 2SO2(g)+O2(g)⇌2SO3(g)2SO2(g)+O2(g)⇌2SO3(g)

(b) N2O4(g)⇌2NO2(g)N2O4(g)⇌2NO2(g)

(c) C3H8(g)+5O2(g)⇌3CO2(g)+4H2O(g)C3H8(g)+5O2(g)⇌3CO2(g)+4H2O(g)

(d) At 227 °C, the following reaction has Kc = 0.0952:

CH3OH(g)⇌CO(g)+2H2(g)CH3OH(g)⇌CO(g)+2H2(g)

What would be the value of KP at this temperature?

Answer:

(a) KP = Kc (RT)−1; (b) KP = Kc (RT); (c) KP = Kc (RT); (d) 160 or 1.6 ×× 102

Heterogeneous Equilibria

A heterogeneous equilibrium involves reactants and products in two or more different phases, as illustrated by the following examples:

PbCl2(s)⇌Pb2+(aq)+2Cl−(aq)Kc=[Pb2+][Cl−]2 CaO(s)+CO2(g)⇌CaCO3(s)Kc=1[CO2] C(s)+2S(g)⇌CS2(g)Kc=[CS2][S]2 Br2(l)⇌Br2(g)Kc=[Br2(g)]PbCl2(s)⇌Pb2+(aq)+2Cl−(aq)Kc=[Pb2+][Cl−]2 CaO(s)+CO2(g)⇌CaCO3(s)Kc=1[CO2] C(s)+2S(g)⇌CS2(g)Kc=[CS2][S]2 Br2(l)⇌Br2(g)Kc=[Br2(g)]

Again, note that concentration terms are only included for gaseous and solute species, as discussed previously.

Two of the above examples include terms for gaseous species only in their equilibrium constants, and so Kp expressions may also be written:

CaO(s)+CO2(g)⇌CaCO3(s)KP=1PCO2 C(s)+2S(g)⇌CS2(g)KP=PCS2(PS)2CaO(s)+CO2(g)⇌CaCO3(s)KP=1PCO2 C(s)+2S(g)⇌CS2(g)KP=PCS2(PS)2

Coupled Equilibria

The equilibrium systems discussed so far have all been relatively simple, involving just single reversible reactions. Many systems, however, involve two or more coupled equilibrium reactions, those which have in common one or more reactant or product species. Since the law of mass action allows for a straightforward derivation of equilibrium constant expressions from balanced chemical equations, the K value for a system involving coupled equilibria can be related to the K values of the individual reactions. Three basic manipulations are involved in this approach, as described below.

1. Changing the direction of a chemical equation essentially swaps the identities of “reactants” and “products,” and so the equilibrium constant for the reversed equation is simply the reciprocal of that for the forward equation.

A⇌BKc=[B][A]B⇌AKc'=[A][B]A⇌BKc=[B][A]B⇌AKc'=[A][B]
Kc'=1KcKc'=1Kc

2. Changing the stoichiometric coefficients in an equation by some factor x results in an exponential change in the equilibrium constant by that same factor:

A⇌BKc=[B][A]xA⇌xBKc'=[B]x[A]xA⇌BKc=[B][A]xA⇌xBKc'=[B]x[A]x
Kc'=KcxKc'=Kcx

3. Adding two or more equilibrium equations together yields an overall equation whose equilibrium constant is the mathematical product of the individual reaction’s K values:

A⇌BKc1=[B][A]B⇌CKc2=[C][B]A⇌BKc1=[B][A]B⇌CKc2=[C][B]

The net reaction for these coupled equilibria is obtained by summing the two equilibrium equations and canceling any redundancies:

A+B⇌B+CA+B⇌B+CA⇌CKc'=[C][A]A+B⇌B+CA+B⇌B+CA⇌CKc'=[C][A]

Comparing the equilibrium constant for the net reaction to those for the two coupled equilibrium reactions reveals the following relationship:

Kc1Kc2=[B][A]×[C][B]=[B][C][A][B]=[C][A]=Kc'Kc1Kc2=[B][A]×[C][B]=[B][C][A][B]=[C][A]=Kc'
Kc'=Kc1Kc2Kc'=Kc1Kc2

Example 13.5 demonstrates the use of this strategy in describing coupled equilibrium processes.

Example 13.5

Equilibrium Constants for Coupled Reactions

A mixture containing nitrogen, hydrogen, and iodine established the following equilibrium at 400 °C:
2NH3(g)+3I2(g)⇌N2(g)+6HI(g)2NH3(g)+3I2(g)⇌N2(g)+6HI(g)

Use the information below to calculate Kc for this reaction.

N2(g)+3H2(g)⇌2NH3(g)Kc1=0.50at400°CH2(g)+I2(g)⇌2HI(g)Kc2=50at400°CN2(g)+3H2(g)⇌2NH3(g)Kc1=0.50at400°CH2(g)+I2(g)⇌2HI(g)Kc2=50at400°C

Solution

The equilibrium equation of interest and its K value may be derived from the equations for the two coupled reactions as follows.

Reverse the first coupled reaction equation:

2NH3(g)⇌N2(g)+3H2(g)Kc1'=1Kc1=10.50=2.02NH3(g)⇌N2(g)+3H2(g)Kc1'=1Kc1=10.50=2.0

Multiply the second coupled reaction by 3:

3H2(g)+3I2(g)⇌6HI(g)Kc2'=Kc23=503=1.2×1053H2(g)+3I2(g)⇌6HI(g)Kc2'=Kc23=503=1.2×105

Finally, add the two revised equations:

2NH3(g)+3H2(g)+3I2(g)⇌N2(g)+3H2(g)+6HI(g)2NH3(g)+3I2(g)⇌N2(g)+6HI(g)Kc=Kc1' Kc2'=(2.0)(1.2×105)=2.5×1052NH3(g)+3H2(g)+3I2(g)⇌N2(g)+3H2(g)+6HI(g)2NH3(g)+3I2(g)⇌N2(g)+6HI(g)Kc=Kc1' Kc2'=(2.0)(1.2×105)=2.5×105

Check Your Learning

Use the provided information to calculate Kc for the following reaction at 550 °C:
H2(g)+CO2(g)⇌CO(g)+H2O(g)Kc=?CoO(s)+CO(g)⇌Co(s)+CO2(g)Kc1=490CoO(s)+H2(g)⇌Co(s)+H2O(g)Kc1=67H2(g)+CO2(g)⇌CO(g)+H2O(g)Kc=?CoO(s)+CO(g)⇌Co(s)+CO2(g)Kc1=490CoO(s)+H2(g)⇌Co(s)+H2O(g)Kc1=67

Answer:

Kc = 0.14

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