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Chemistry: Atoms First

13.4 Equilibrium Calculations

Chemistry: Atoms First13.4 Equilibrium Calculations
  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. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  4. 3 Electronic Structure and Periodic Properties of Elements
    1. Introduction
    2. 3.1 Electromagnetic Energy
    3. 3.2 The Bohr Model
    4. 3.3 Development of Quantum Theory
    5. 3.4 Electronic Structure of Atoms (Electron Configurations)
    6. 3.5 Periodic Variations in Element Properties
    7. 3.6 The Periodic Table
    8. 3.7 Molecular and Ionic Compounds
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  5. 4 Chemical Bonding and Molecular Geometry
    1. Introduction
    2. 4.1 Ionic Bonding
    3. 4.2 Covalent Bonding
    4. 4.3 Chemical Nomenclature
    5. 4.4 Lewis Symbols and Structures
    6. 4.5 Formal Charges and Resonance
    7. 4.6 Molecular Structure and Polarity
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  6. 5 Advanced Theories of Bonding
    1. Introduction
    2. 5.1 Valence Bond Theory
    3. 5.2 Hybrid Atomic Orbitals
    4. 5.3 Multiple Bonds
    5. 5.4 Molecular Orbital Theory
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  7. 6 Composition of Substances and Solutions
    1. Introduction
    2. 6.1 Formula Mass and the Mole Concept
    3. 6.2 Determining Empirical and Molecular Formulas
    4. 6.3 Molarity
    5. 6.4 Other Units for Solution Concentrations
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  8. 7 Stoichiometry of Chemical Reactions
    1. Introduction
    2. 7.1 Writing and Balancing Chemical Equations
    3. 7.2 Classifying Chemical Reactions
    4. 7.3 Reaction Stoichiometry
    5. 7.4 Reaction Yields
    6. 7.5 Quantitative Chemical Analysis
    7. Key Terms
    8. Key Equations
    9. Summary
    10. Exercises
  9. 8 Gases
    1. Introduction
    2. 8.1 Gas Pressure
    3. 8.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law
    4. 8.3 Stoichiometry of Gaseous Substances, Mixtures, and Reactions
    5. 8.4 Effusion and Diffusion of Gases
    6. 8.5 The Kinetic-Molecular Theory
    7. 8.6 Non-Ideal Gas Behavior
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  10. 9 Thermochemistry
    1. Introduction
    2. 9.1 Energy Basics
    3. 9.2 Calorimetry
    4. 9.3 Enthalpy
    5. 9.4 Strengths of Ionic and Covalent Bonds
    6. Key Terms
    7. Key Equations
    8. Summary
    9. 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 Thermodynamics
    1. Introduction
    2. 12.1 Spontaneity
    3. 12.2 Entropy
    4. 12.3 The Second and Third Laws of Thermodynamics
    5. 12.4 Free Energy
    6. Key Terms
    7. Key Equations
    8. Summary
    9. 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 Salt Solutions
    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 Multiple Equilibria
    5. Key Terms
    6. Key Equations
    7. Summary
    8. Exercises
  17. 16 Electrochemistry
    1. Introduction
    2. 16.1 Balancing Oxidation-Reduction Reactions
    3. 16.2 Galvanic Cells
    4. 16.3 Standard Reduction Potentials
    5. 16.4 The Nernst Equation
    6. 16.5 Batteries and Fuel Cells
    7. 16.6 Corrosion
    8. 16.7 Electrolysis
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  18. 17 Kinetics
    1. Introduction
    2. 17.1 Chemical Reaction Rates
    3. 17.2 Factors Affecting Reaction Rates
    4. 17.3 Rate Laws
    5. 17.4 Integrated Rate Laws
    6. 17.5 Collision Theory
    7. 17.6 Reaction Mechanisms
    8. 17.7 Catalysis
    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 Nuclear Chemistry
    1. Introduction
    2. 20.1 Nuclear Structure and Stability
    3. 20.2 Nuclear Equations
    4. 20.3 Radioactive Decay
    5. 20.4 Transmutation and Nuclear Energy
    6. 20.5 Uses of Radioisotopes
    7. 20.6 Biological Effects of Radiation
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  22. 21 Organic Chemistry
    1. Introduction
    2. 21.1 Hydrocarbons
    3. 21.2 Alcohols and Ethers
    4. 21.3 Aldehydes, Ketones, Carboxylic Acids, and Esters
    5. 21.4 Amines and Amides
    6. Key Terms
    7. Summary
    8. 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:
  • Write equations representing changes in concentration and pressure for chemical species in equilibrium systems
  • Use algebra to perform various types of equilibrium calculations
  • Explain how temperature affects the spontaneity of some proceses
  • Relate standard free energy changes to equilibrium constants

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:

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

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.

Δ[H2]=3×Δ[N2]Δ[H2]=3×Δ[N2]
=3×(0.11M)=0.33M=3×(0.11M)=0.33M

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.

Δ[NH3]=−2×Δ[N2]=−2×(0.11M)=−0.22MΔ[NH3]=−2×Δ[N2]=−2×(0.11M)=−0.22M

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

2NH3(g)N2(g)+3H2(g)Δ[NH3]=−2×Δ[N2]Δ[N2]=0.11MΔ[H2]=3×Δ[N2]2NH3(g)N2(g)+3H2(g)Δ[NH3]=−2×Δ[N2]Δ[N2]=0.11MΔ[H2]=3×Δ[N2]

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.

Δ[N2]=xΔ[N2]=x

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

Δ[H2]=3×Δ[N2]=3xΔ[H2]=3×Δ[N2]=3x
Δ[NH3]=−2×Δ[N2]=−2xΔ[NH3]=−2×Δ[N2]=−2x

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

2NH3(g)N2(g)+3H2(g)−2xx3x2NH3(g)N2(g)+3H2(g)−2xx3x

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

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

(a) C2H2(g)+2Br2(g)C2H2Br4(g)x__________C2H2(g)+2Br2(g)C2H2Br4(g)x__________

(b) I2(aq)+I(aq)I3(aq)__________xI2(aq)+I(aq)I3(aq)__________x

(c) C3H8(g)+5O2(g)3CO2(g)+4H2O(g)x_______________C3H8(g)+5O2(g)3CO2(g)+4H2O(g)x_______________

Solution

(a) C2H2(g)+2Br2(g)C2H2Br4(g)x2xxC2H2(g)+2Br2(g)C2H2Br4(g)x2xx

(b) I2(aq)+I(aq)I3(aq)xxxI2(aq)+I(aq)I3(aq)xxx

(c) C3H8(g)+5O2(g)3CO2(g)+4H2O(g)x5x−3x−4xC3H8(g)+5O2(g)3CO2(g)+4H2O(g)x5x−3x−4x

Check Your Learning

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

(a) 2SO2(g)+O2(g)2SO3(g)_____x_____2SO2(g)+O2(g)2SO3(g)_____x_____

(b) C4H8(g)2C2H4(g)_____−2xC4H8(g)2C2H4(g)_____−2x

(c) 4NH3(g)+7H2O(g)4NO2(g)+6H2O(g)____________________4NH3(g)+7H2O(g)4NO2(g)+6H2O(g)____________________

Answer:

(a) 2x, x, −2x; (b) x, −2x; (c) 4x, 7x, −4x, −6x or −4x, −7x, 4x, 6x

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

In order to calculate an equilibrium constant, enough information must be available to determine the equilibrium concentrations of all reactants and products. Armed with the concentrations, we can solve the equation for Kc, as it will be the only unknown.

Example 13.2 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. The initial concentrations of the reactants and products are provided in the first row of the ICE table (these essentially time-zero concentrations that assume no reaction has taken place). The next row of data contains the changes in concentrations that result when the reaction proceeds toward equilibrium (don’t forget to account for the reaction stoichiometry). 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.
I2(aq)+I(aq)I3(aq)I2(aq)+I(aq)I3(aq)

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. This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header, “I subscript 2 plus sign I superscript negative sign equilibrium arrow I subscript 3 superscript negative sign.” Under the second column is a subgroup of three rows and three columns. The first column has the following: 1.000 times 10 to the negative third power, negative x, [ I subscript 2 ] subscript i minus x. The second column has the following: 1.000 times 10 to the negative third power, negative x, [ I superscript negative sign ] subscript i minus x. The third column has the following: 0, positive x, [ I superscript negative sign ] subscript i plus x.

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

1.000×10−3x=6.61×10−41.000×10−3x=6.61×10−4
x=1.000×10−36.61×10−4x=1.000×10−36.61×10−4
=3.39×10−4M=3.39×10−4M

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

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header, “I subscript 2 plus sign I superscript negative sign equilibrium arrow I subscript 3 superscript negative sign.” Under the second column is a subgroup of three rows and three columns. The first column has the following: 1.000 times 10 to the negative third power, negative x equals negative 3.39 times 10 to the negative fourth power, 6.61 times 10 to the negative fourth power. The second column has the following: 1.000 times 10 to the negative third power, negative x, 6.61 times 10 to the negative fourth power. The third column has the following: 0, positive x, 3.39 times 10 to the negative fourth power.

We now calculate the value of the equilibrium constant.

Kc=Qc=[I3][I2][I]Kc=Qc=[I3][I2][I]
=3.39×10−4M(6.61×10−4M)(6.61×10−4M)=776=3.39×10−4M(6.61×10−4M)(6.61×10−4M)=776

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.
C2H5OH+CH3CO2HCH3CO2C2H5+H2OC2H5OH+CH3CO2HCH3CO2C2H5+H2O

When 1 mol each of C2H5OH and CH3CO2H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when 1313 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

If we know the equilibrium constant for a reaction and know the concentrations at equilibrium of all reactants and products except one, we can calculate the missing concentration.

Example 13.7

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),N2(g)+O2(g)2NO(g), is 4.1 ×× 10−4. Calculate the equilibrium concentration of NO(g) in air at 1 atm pressure and 2000 °C. The equilibrium concentrations of N2 and O2 at this pressure and temperature are 0.036 M and 0.0089 M, respectively.

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.
Kc=Qc=[NO]2[N2][O2]Kc=Qc=[NO]2[N2][O2]
[NO]2=Kc[N2][O2][NO]2=Kc[N2][O2]
[NO]=Kc[N2][O2][NO]=Kc[N2][O2]
=(4.1×10−4)(0.036)(0.0089)=(4.1×10−4)(0.036)(0.0089)
=1.31×10−7=1.31×10−7
=3.6×10−4=3.6×10−4

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.

Qc=[NO]2[N2][O2]Qc=[NO]2[N2][O2]
=(3.6×10−4)2(0.036)(0.0089)=(3.6×10−4)2(0.036)(0.0089)
Qc=4.0×10−4=KcQc=4.0×10−4=Kc

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 Equilibrium Concentrations from Initial Concentrations

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.
    1. Write a balanced chemical equation for the reaction.
    2. 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.
    1. 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.
    2. 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.
    1. Substitute the equilibrium concentrations into the expression for the equilibrium constant, solve for x, and check any assumptions used to find x.
    2. Calculate the equilibrium concentrations.
  4. Check the arithmetic.
    1. 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.
  1. Step 1.

    Determine the direction the reaction proceeds.

    The balanced equation for the decomposition of PCl5 is

    PCl5(g)PCl3(g)+Cl2(g)PCl5(g)PCl3(g)+Cl2(g)

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

  2. 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.

    PCl5(g)PCl3(g)+Cl2(g)xxxPCl5(g)PCl3(g)+Cl2(g)xxx

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

    This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header, “P C l subscript 5 equilibrium arrow P C l subscript 3 plus C l subscript 2.” Under the second column is a subgroup of three rows and three columns. The first column has the following: 1.00, negative x, 1.00 minus x. The second column has the following: 0, positive x, 0 plus x equals x. The third column has the following: 0, positive x, 0 plus x equals x.
  3. Step 3.

    Solve for the change and the equilibrium concentrations.

    Substituting the equilibrium concentrations into the equilibrium constant equation gives

    Kc=[PCl3][Cl2][PCl5]=0.0211Kc=[PCl3][Cl2][PCl5]=0.0211
    =(x)(x)(1.00x)=(x)(x)(1.00x)

    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.

    0.0211=(x)(x)(1.00x)0.0211=(x)(x)(1.00x)
    0.0211(1.00x)=x20.0211(1.00x)=x2
    x2+0.0211x0.0211=0x2+0.0211x0.0211=0

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

    x=b±b24ac2ax=b±b24ac2a

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

    x=0.0211±(0.0211)24(1)(−0.0211)2(1)x=0.0211±(0.0211)24(1)(−0.0211)2(1)
    =0.0211±(4.45×10−4)+(8.44×10−2)2=0.0211±(4.45×10−4)+(8.44×10−2)2
    =0.0211±0.2912=0.0211±0.2912

    Hence

    x=0.0211+0.2912=0.135x=0.0211+0.2912=0.135

    or

    x=0.02110.2912=−0.156x=0.02110.2912=−0.156

    Quadratic equations often have two different solutions, one that is physically possible and one that is physically impossible (an extraneous root). In this case, the second solution (−0.156) is physically impossible because we know the change must be a positive number (otherwise we would end up with negative values for concentrations of the products). Thus, x = 0.135 M.

    The equilibrium concentrations are

    [PCl5]=1.000.135=0.87M[PCl5]=1.000.135=0.87M
    [PCl3]=x=0.135M[PCl3]=x=0.135M
    [Cl2]=x=0.135M[Cl2]=x=0.135M
  4. Step 4.

    Check the arithmetic.

    Substitution into the expression for Kc (to check the calculation) gives

    Kc=[PCl3][Cl2][PCl5]=(0.135)(0.135)0.87=0.021Kc=[PCl3][Cl2][PCl5]=(0.135)(0.135)0.87=0.021

    The equilibrium constant calculated from the equilibrium concentrations is equal to the value of Kc given in the problem (when rounded to the proper number of significant figures). Thus, the calculated equilibrium concentrations check.

Check Your Learning

Acetic acid, CH3CO2H, reacts with ethanol, C2H5OH, to form water and ethyl acetate, CH3CO2C2H5.
CH3CO2H+C2H5OHCH3CO2C2H5+H2OCH3CO2H+C2H5OHCH3CO2C2H5+H2O

The equilibrium constant for this reaction with dioxane as a solvent is 4.0. What are the equilibrium concentrations when a mixture that is 0.15 M in CH3CO2H, 0.15 M in C2H5OH, 0.40 M in CH3CO2C2H5, and 0.40 M in H2O are mixed in enough dioxane to make 1.0 L of solution?

Answer:

[CH3CO2H] = 0.36 M, [C2H5OH] = 0.36 M, [CH3CO2C2H5] = 0.17 M, [H2O] = 0.17 M

Check Your Learning

A 1.00-L flask is filled with 1.00 moles of H2 and 2.00 moles of I2. The value of the equilibrium constant for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of H2, I2, and HI in moles/L?
H2(g)+I2(g)2HI(g)H2(g)+I2(g)2HI(g)

Answer:

[H2] = 0.06 M, [I2] = 1.06 M, [HI] = 1.88 M

Sometimes it is possible to use chemical insight to find solutions to equilibrium problems without actually solving a quadratic (or more complicated) equation. First, however, it is useful to verify that equilibrium can be obtained starting from two extremes: all (or mostly) reactants and all (or mostly) products (similar to what was shown in Figure 13.7).

Consider the ionization of 0.150 M HA, a weak acid.

HA(aq)H+(aq)+A(aq)Kc=6.80×10−4HA(aq)H+(aq)+A(aq)Kc=6.80×10−4

The most obvious way to determine the equilibrium concentrations would be to start with only reactants. This could be called the “all reactant” starting point. Using x for the amount of acid ionized at equilibrium, this is the ICE table and solution.

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial pressure ( M ), Change ( M ), Equilibrium ( M ). The second column has the header, “H A ( a q ) equilibrium arrow H superscript plus sign ( a q ) plus A subscript negative sign ( a q ).” Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.150, negative x, 0.150 minus x. The second column has the following: 0, x, x. The third column has the following: 0, x, x.

Setting up and solving the quadratic equation gives

Kc=[H+][A][HA]=(x)(x)(0.150x)=6.80×10−4Kc=[H+][A][HA]=(x)(x)(0.150x)=6.80×10−4
x2+6.80×10−4x−1.02×10−4=0x2+6.80×10−4x−1.02×10−4=0
x=6.80×10−4±(6.80×10−4)2(4)(1)(−1.02×10−4)(2)(1)x=6.80×10−4±(6.80×10−4)2(4)(1)(−1.02×10−4)(2)(1)
x=0.00977Mor−0.0104Mx=0.00977Mor−0.0104M

Using the positive (physical) root, the equilibrium concentrations are

[HA]=0.150x=0.140M[HA]=0.150x=0.140M
[H+]=[A]=x=0.00977M[H+]=[A]=x=0.00977M

A less obvious way to solve the problem would be to assume all the HA ionizes first, then the system comes to equilibrium. This could be called the “all product” starting point. Assuming all of the HA ionizes gives

[HA]=0.1500.150=0M[HA]=0.1500.150=0M
[H+]=0+0.150=0.150M[H+]=0+0.150=0.150M
[A]=0+0.150=0.150M[A]=0+0.150=0.150M

Using these as initial concentrations and “y” to represent the concentration of HA at equilibrium, this is the ICE table for this starting point.

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial pressure ( M ), Change ( M ), Equilibrium ( M ). The second column has the header, “H A ( a q ) equilibrium arrow H superscript plus sign ( a q ) plus A subscript negative sign ( a q ).” Under the second column is a subgroup of three columns and three rows. The first column has the following: 0, positive y, y. The second column has the following: 0.150, negative y, 0.150 minus y. The third column has the following: 0.150, negative y, and 0.150 minus y.

Setting up and solving the quadratic equation gives

Kc=[H+][A][HA]=(0.150y)(0.150y)(y)=6.80×10−4Kc=[H+][A][HA]=(0.150y)(0.150y)(y)=6.80×10−4
6.80×10−4y=0.02250.300y+y26.80×10−4y=0.02250.300y+y2

Retain a few extra significant figures to minimize rounding problems.

y20.30068y+0.022500=0y20.30068y+0.022500=0
y=0.30068±(0.30068)2(4)(1)(0.022500)(2)(1)y=0.30068±(0.30068)2(4)(1)(0.022500)(2)(1)
y=0.30068±0.0202102y=0.30068±0.0202102

Rounding each solution to three significant figures gives

y=0.160Mory=0.140My=0.160Mory=0.140M

Using the physically significant root (0.140 M) gives the equilibrium concentrations as

[HA]=y=0.140M[HA]=y=0.140M
[H+]=0.150y=0.010M[H+]=0.150y=0.010M
[A]=0.150y=0.010M[A]=0.150y=0.010M

Thus, the two approaches give the same results (to three decimal places), and show that both starting points lead to the same equilibrium conditions. The “all reactant” starting point resulted in a relatively small change (x) because the system was close to equilibrium, while the “all product” starting point had a relatively large change (y) that was nearly the size of the initial concentrations. It can be said that a system that starts “close” to equilibrium will require only a ”small” change in conditions (x) to reach equilibrium.

Recall that a small Kc means that very little of the reactants form products and a large Kc means that most of the reactants form products. If the system can be arranged so it starts “close” to equilibrium, then if the change (x) is small compared to any initial concentrations, it can be neglected. Small is usually defined as resulting in an error of less than 5%. The following two examples demonstrate this.

Example 13.9

Approximate Solution Starting Close to Equilibrium

What are the concentrations at equilibrium of a 0.15 M solution of HCN?
HCN(aq)H+(aq)+CN(aq)Kc=4.9×10−10HCN(aq)H+(aq)+CN(aq)Kc=4.9×10−10

Solution

Using “x” to represent the concentration of each product at equilibrium gives this ICE table. This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial pressure ( M ), Change ( M ), Equilibrium ( M ). The second column has the header, “H C N ( a q ) equilibrium arrow H superscript plus sign ( a q ) plus C N subscript negative sign ( a q ).” Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.15, negative x, 0.15 minus x. The second column has the following: 0, x, x. The third column has the following: 0, x, x.

The exact solution may be obtained using the quadratic formula with

Kc=(x)(x)0.15xKc=(x)(x)0.15x

solving

x2+4.9×10−107.35×10−11=0x2+4.9×10−107.35×10−11=0
x=8.56×10−6M(3 sig. figs.)=8.6×10−6M(2 sig. figs.)x=8.56×10−6M(3 sig. figs.)=8.6×10−6M(2 sig. figs.)

Thus [H+] = [CN] = x = 8.6 ×× 10–6 M and [HCN] = 0.15 – x = 0.15 M.

In this case, chemical intuition can provide a simpler solution. From the equilibrium constant and the initial conditions, x must be small compared to 0.15 M. More formally, if x0.15,x0.15, then 0.15 – x ≈ 0.15. If this assumption is true, then it simplifies obtaining x

Kc=(x)(x)0.15xx20.15Kc=(x)(x)0.15xx20.15
4.9×10−10=x20.154.9×10−10=x20.15
x2=(0.15)(4.9×10−10)=7.4×10−11x2=(0.15)(4.9×10−10)=7.4×10−11
x=7.4×10−11=8.6×10−6Mx=7.4×10−11=8.6×10−6M

In this example, solving the exact (quadratic) equation and using approximations gave the same result to two significant figures. While most of the time the approximation is a bit different from the exact solution, as long as the error is less than 5%, the approximate solution is considered valid. In this problem, the 5% applies to IF (0.15 – x) ≈ 0.15 M, so if

x0.15×100%=8.6×10−60.15×100%=0.006%x0.15×100%=8.6×10−60.15×100%=0.006%

is less than 5%, as it is in this case, the assumption is valid. The approximate solution is thus a valid solution.

Check Your Learning

What are the equilibrium concentrations in a 0.25 M NH3 solution?
NH3(aq)+H2O(l)NH4+(aq)+OH(aq)Kc=1.8×10−5NH3(aq)+H2O(l)NH4+(aq)+OH(aq)Kc=1.8×10−5

Assume that x is much less than 0.25 M and calculate the error in your assumption.

Answer:

[OH]=[NH4+]=0.0021M;[OH]=[NH4+]=0.0021M; [NH3] = 0.25 M, error = 0.84%

The second example requires that the original information be processed a bit, but it still can be solved using a small x approximation.

Example 13.10

Approximate Solution After Shifting Starting Concentration

Copper(II) ions form a complex ion in the presence of ammonia
Cu2+(aq)+4NH3(aq)Cu(NH3)42+(aq)Kc=5.0×1013=[Cu(NH3)42+][Cu2+(aq)][NH3]4Cu2+(aq)+4NH3(aq)Cu(NH3)42+(aq)Kc=5.0×1013=[Cu(NH3)42+][Cu2+(aq)][NH3]4

If 0.010 mol Cu2+ is added to 1.00 L of a solution that is 1.00 M NH3 what are the concentrations when the system comes to equilibrium?

Solution

The initial concentration of copper(II) is 0.010 M. The equilibrium constant is very large so it would be better to start with as much product as possible because “all products” is much closer to equilibrium than “all reactants.” Note that Cu2+ is the limiting reactant; if all 0.010 M of it reacts to form product the concentrations would be
[Cu2+]=0.0100.010=0M[Cu2+]=0.0100.010=0M
[Cu(NH3)42+]=0.010M[Cu(NH3)42+]=0.010M
[NH3]=1.004×0.010=0.96M[NH3]=1.004×0.010=0.96M

Using these “shifted” values as initial concentrations with x as the free copper(II) ion concentration at equilibrium gives this ICE table.

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial pressure ( M ), Change ( M ), Equilibrium ( M ). The second column has the header, “C u superscript 2 positive sign ( a q ) plus 4 N H subscript 3 ( a q ) equilibrium arrow C u ( N H subscript 3 ) subscript 4 superscript 2 positive sign ( a q ).” Under the second column is a subgroup of three columns and three rows. The first column has the following: 0, positive x, x. The second column has the following: 0.96, positive 4 x, 0.96 plus 4 x. The third column has the following: 0.010, negative x, 0.010 minus x.

Since we are starting close to equilibrium, x should be small so that

0.96+4x0.96M0.96+4x0.96M
0.010x0.010M0.010x0.010M
Kc=(0.010x)x(0.964x)4(0.010)x(0.96)4=5.0×1013Kc=(0.010x)x(0.964x)4(0.010)x(0.96)4=5.0×1013
x=(0.010)Kc(0.96)4=2.4×10−16Mx=(0.010)Kc(0.96)4=2.4×10−16M

Select the smallest concentration for the 5% rule.

2.4×10−160.010×100%=2×10−12%2.4×10−160.010×100%=2×10−12%

This is much less than 5%, so the assumptions are valid. The concentrations at equilibrium are

[Cu2+]=x=2.4×10−16M[Cu2+]=x=2.4×10−16M
[NH3]=0.964x=0.96M[NH3]=0.964x=0.96M
[Cu(NH3)42+]=0.010x=0.010M[Cu(NH3)42+]=0.010x=0.010M

By starting with the maximum amount of product, this system was near equilibrium and the change (x) was very small. With only a small change required to get to equilibrium, the equation for x was greatly simplified and gave a valid result well within the 5% error maximum.

Check Your Learning

What are the equilibrium concentrations when 0.25 mol Ni2+ is added to 1.00 L of 2.00 M NH3 solution?
Ni2+(aq)+6NH3(aq)Ni(NH3)62+(aq)Kc=5.5×108Ni2+(aq)+6NH3(aq)Ni(NH3)62+(aq)Kc=5.5×108

With such a large equilibrium constant, first form as much product as possible, then assume that only a small amount (x) of the product shifts left. Calculate the error in your assumption.

Answer:

[Ni(NH3)62+]=0.25M,[Ni(NH3)62+]=0.25M, [NH3] = 0.50 M, [Ni2+] = 2.9 ×× 10–8 M, error = 1.2 ×× 10–5%

Temperature Dependence of Spontaneity

As was previously demonstrated in the section on entropy in an earlier chapter, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. Likewise, some chemical reactions can also exhibit temperature dependent spontaneities. To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:

ΔG=ΔHTΔSΔG=ΔHTΔS

The spontaneity of a process, as reflected in the arithmetic sign of its free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since T is the absolute (kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes:

  1. Both ΔH and ΔS are positive. This condition describes an endothermic process that involves an increase in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is greater than ΔH. If the TΔS term is less than ΔH, the free energy change will be positive. Such a process is spontaneous at high temperatures and nonspontaneous at low temperatures.
  2. Both ΔH and ΔS are negative. This condition describes an exothermic process that involves a decrease in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is less than ΔH. If the TΔS term’s magnitude is greater than ΔH, the free energy change will be positive. Such a process is spontaneous at low temperatures and nonspontaneous at high temperatures.
  3. ΔH is positive and ΔS is negative. This condition describes an endothermic process that involves a decrease in system entropy. In this case, ΔG will be positive regardless of the temperature. Such a process is nonspontaneous at all temperatures.
  4. ΔH is negative and ΔS is positive. This condition describes an exothermic process that involves an increase in system entropy. In this case, ΔG will be negative regardless of the temperature. Such a process is spontaneous at all temperatures.

These four scenarios are summarized in Figure 13.9.

A table with three columns and four rows is shown. The first column has the phrase, “Delta S greater than zero ( increase in entropy ),” in the third row and the phrase, “Delta S less than zero ( decrease in entropy),” in the fourth row. The second and third columns have the phrase, “Summary of the Four Scenarios for Enthalpy and Entropy Changes,” written above them. The second column has, “delta H greater than zero ( endothermic ),” in the second row, “delta G less than zero at high temperature, delta G greater than zero at low temperature, Process is spontaneous at high temperature,” in the third row, and “delta G greater than zero at any temperature, Process is nonspontaneous at any temperature,” in the fourth row. The third column has, “delta H less than zero ( exothermic ),” in the second row, “delta G less than zero at any temperature, Process is spontaneous at any temperature,” in the third row, and “delta G less than zero at low temperature, delta G greater than zero at high temperature, Process is spontaneous at low temperature.”
Figure 13.9 There are four possibilities regarding the signs of enthalpy and entropy changes.

Example 13.11

Predicting the Temperature Dependence of Spontaneity

The incomplete combustion of carbon is described by the following equation:
2C(s)+O2(g)2CO(g)2C(s)+O2(g)2CO(g)

How does the spontaneity of this process depend upon temperature?

Solution

Combustion processes are exothermic (ΔH < 0). This particular reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (net gain of one mole of gas, ΔS > 0). The reaction is therefore spontaneous (ΔG < 0) at all temperatures.

Check Your Learning

Popular chemical hand warmers generate heat by the air-oxidation of iron:
4Fe(s)+3O2(g)2Fe2O3(s)4Fe(s)+3O2(g)2Fe2O3(s)

How does the spontaneity of this process depend upon temperature?

Answer:

ΔH and ΔS are negative; the reaction is spontaneous at low temperatures.

When considering the conclusions drawn regarding the temperature dependence of spontaneity, it is important to keep in mind what the terms “high” and “low” mean. Since these terms are adjectives, the temperatures in question are deemed high or low relative to some reference temperature. A process that is nonspontaneous at one temperature but spontaneous at another will necessarily undergo a change in “spontaneity” (as reflected by its ΔG) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which ΔG is plotted on the y axis versus T on the x axis:

ΔG=ΔHTΔSΔG=ΔHTΔS
y=b+mxy=b+mx

Such a plot is shown in Figure 13.10. A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependent spontaneity as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative ΔG) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the x-intercept of the line, that is, the value of T for which ΔG is zero:

ΔG=0=ΔHTΔSΔG=0=ΔHTΔS
T=ΔHΔST=ΔHΔS

And so, saying a process is spontaneous at “high” or “low” temperatures means the temperature is above or below, respectively, that temperature at which ΔG for the process is zero. As noted earlier, the condition of ΔG = 0 describes a system at equilibrium.

A graph is shown where the y-axis is labeled, “Free energy,” and the x-axis is labeled, “Increasing temperature ( K ).” The value of zero is written midway up the y-axis with the label, “delta G greater than 0,” written above this line and, “delta G less than 0,” written below it. The bottom half of the graph is labeled on the right as, “Spontaneous,” and the top half is labeled on the right as, “Nonspontaneous.” A green line labeled, “delta H less than 0, delta S greater than 0,” extends from a quarter of the way up the y-axis to the bottom right of the graph. A yellow line labeled, “delta H less than 0, delta S less than 0,” extends from a quarter of the way up the y-axis to the middle right of the graph. A second yellow line labeled, “delta H greater than 0, delta S greater than 0,” extends from three quarters of the way up the y-axis to the middle right of the graph. A red line labeled, “delta H greater than 0, delta S less than 0,” extends from three quarters of the way up the y-axis to the top right of the graph.
Figure 13.10 These plots show the variation in ΔG with temperature for the four possible combinations of arithmetic sign for ΔH and ΔS.

Example 13.12

Equilibrium Temperature for a Phase Transition

As defined in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its liquid and gaseous phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Use the information in Appendix G to estimate the boiling point of water.

Solution

The process of interest is the following phase change:
H2O(l)H2O(g)H2O(l)H2O(g)

When this process is at equilibrium, ΔG = 0, so the following is true:

0=ΔH°TΔS°orT=ΔH°ΔS°0=ΔH°TΔS°orT=ΔH°ΔS°

Using the standard thermodynamic data from Appendix G,

ΔH°=ΔHf°(H2O(g))ΔHf°(H2O(l))=241.82 kJ/mol(285.83 kJ/mol)=44.01 kJ/molΔH°=ΔHf°(H2O(g))ΔHf°(H2O(l))=241.82 kJ/mol(285.83 kJ/mol)=44.01 kJ/mol
ΔS°=ΔS298°(H2O(g))ΔS298°(H2O(l))=188.8 J/K·mol70.0 J/K·mol=118.8 J/K·molΔS°=ΔS298°(H2O(g))ΔS298°(H2O(l))=188.8 J/K·mol70.0 J/K·mol=118.8 J/K·mol
T=ΔH°ΔS°=44.01×103J/mol118.8J/K·mol=370.5K=97.3°CT=ΔH°ΔS°=44.01×103J/mol118.8J/K·mol=370.5K=97.3°C

The accepted value for water’s normal boiling point is 373.2 K (100.0 °C), and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K (Appendix G). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point.

Check Your Learning

Use the information in Appendix G to estimate the boiling point of CS2.

Answer:

313 K (accepted value 319 K)

Free Energy and Equilibrium

The free energy change for a process may be viewed as a measure of its driving force. A negative value for ΔG represents a finite driving force for the process in the forward direction, while a positive value represents a driving force for the process in the reverse direction. When ΔG is zero, the forward and reverse driving forces are equal, and so the process occurs in both directions at the same rate (the system is at equilibrium).

In the section on equilibrium, the reaction quotient, Q, was introduced as a convenient measure of the status of an equilibrium system. Recall that Q is the numerical value of the mass action expression for the system, and that you may use its value to identify the direction in which a reaction will proceed in order to achieve equilibrium. When Q is lesser than the equilibrium constant, K, the reaction will proceed in the forward direction until equilibrium is reached and Q = K. Conversely, if Q > K, the process will proceed in the reverse direction until equilibrium is achieved.

The free energy change for a process taking place with reactants and products present under nonstandard conditions, ΔG, is related to the standard free energy change, ΔG°, according to this equation:

ΔG=ΔG°+RTlnQΔG=ΔG°+RTlnQ

R is the gas constant (8.314 J/K mol), T is the kelvin or absolute temperature, and Q is the reaction quotient. We may use this equation to predict the spontaneity for a process under any given set of conditions as illustrated in Example 13.13.

Example 13.13

Calculating ΔG under Nonstandard Conditions

What is the free energy change for the process shown here under the specified conditions?

T = 25 °C, PN2=0.870 atm,PN2=0.870 atm, PH2=0.250 atm,PH2=0.250 atm, and PNH3=12.9 atmPNH3=12.9 atm

2NH3(g)3H2(g)+N2(g)ΔG°=33.0 kJ/mol2NH3(g)3H2(g)+N2(g)ΔG°=33.0 kJ/mol

Solution

The equation relating free energy change to standard free energy change and reaction quotient may be used directly:
ΔG=ΔG°+RTlnQ=33.0kJmol+(8.314Jmol K×298 K×ln(0.2503)×0.87012.92)=9680Jmolor 9.68 kJ/molΔG=ΔG°+RTlnQ=33.0kJmol+(8.314Jmol K×298 K×ln(0.2503)×0.87012.92)=9680Jmolor 9.68 kJ/mol

Since the computed value for ΔG is positive, the reaction is nonspontaneous under these conditions.

Check Your Learning

Calculate the free energy change for this same reaction at 875 °C in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these conditions?

Answer:

ΔG = −47 kJ/mol; yes

For a system at equilibrium, Q = K and ΔG = 0, and the previous equation may be written as

0=ΔG°+RTlnK(at equilibrium)0=ΔG°+RTlnK(at equilibrium)
ΔG°=RTlnKorK=eΔG°RTΔG°=RTlnKorK=eΔG°RT

This form of the equation provides a useful link between these two essential thermodynamic properties, and it can be used to derive equilibrium constants from standard free energy changes and vice versa. The relations between standard free energy changes and equilibrium constants are summarized in Table 13.1.

Relations between Standard Free Energy Changes and Equilibrium Constants
K ΔG° Comments
> 1 < 0 Products are more abundant at equilibrium.
< 1 > 0 Reactants are more abundant at equilibrium.
= 1 = 0 Reactants and products are equally abundant at equilibrium.
Table 13.1

Example 13.14

Calculating an Equilibrium Constant using Standard Free Energy Change

Given that the standard free energies of formation of Ag+(aq), Cl(aq), and AgCl(s) are 77.1 kJ/mol, −131.2 kJ/mol, and −109.8 kJ/mol, respectively, calculate the solubility product, Ksp, for AgCl.

Solution

The reaction of interest is the following:
AgCl(s)Ag+(aq)+Cl(aq)Ksp=[Ag+][Cl]AgCl(s)Ag+(aq)+Cl(aq)Ksp=[Ag+][Cl]

The standard free energy change for this reaction is first computed using standard free energies of formation for its reactants and products:

ΔG°=ΔG298°=[ΔGf°(Ag+(aq))+ΔGf°(Cl(aq))][ΔGf°(AgCl(s))]=[77.1 kJ/mol131.2 kJ/mol][109.8 kJ/mol]=55.7 kJ/molΔG°=ΔG298°=[ΔGf°(Ag+(aq))+ΔGf°(Cl(aq))][ΔGf°(AgCl(s))]=[77.1 kJ/mol131.2 kJ/mol][109.8 kJ/mol]=55.7 kJ/mol

The equilibrium constant for the reaction may then be derived from its standard free energy change:

Ksp=eΔG°RT=exp(ΔG°RT)=exp(55.7×103J/mol8.314J/mol·K×298.15K)=exp(22.470)=e22.470=1.74×10−10Ksp=eΔG°RT=exp(ΔG°RT)=exp(55.7×103J/mol8.314J/mol·K×298.15K)=exp(22.470)=e22.470=1.74×10−10

This result is in reasonable agreement with the value provided in Appendix J.

Check Your Learning

Use the thermodynamic data provided in Appendix G to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 °C.
2NO2(g)N2O4(g)2NO2(g)N2O4(g)

Answer:

K = 6.9

To further illustrate the relation between these two essential thermodynamic concepts, consider the observation that reactions spontaneously proceed in a direction that ultimately establishes equilibrium. As may be shown by plotting the free energy change versus the extent of the reaction (for example, as reflected in the value of Q), equilibrium is established when the system’s free energy is minimized (Figure 13.11). If a system consists of reactants and products in nonequilibrium amounts (QK), the reaction will proceed spontaneously in the direction necessary to establish equilibrium.

Three graphs, labeled, “a,” “b,” and “c” are shown where the y-axis is labeled, “Gibbs free energy ( G ),” and, “G superscript degree sign ( reactants ),” while the x-axis is labeled, “Reaction progress,” and “Reactants,” on the left and, “Products,” on the right. In graph a, a line begins at the upper left side and goes steadily down to a point about halfway up the y-axis and two thirds of the way on the x-axis, then rises again to a point labeled, “G superscript degree sign ( products ),” that is slightly higher than halfway up the y-axis. The distance between the beginning and ending points of the graph is labeled as, “delta G less than 0,” while the lowest point on the graph is labeled, “Q equals K greater than 1.” In graph b, a line begins at the middle left side and goes steadily down to a point about two fifths up the y-axis and one third of the way on the x-axis, then rises again to a point labeled, “G superscript degree sign ( products ),” that is near the top of the y-axis. The distance between the beginning and ending points of the graph is labeled as, “delta G greater than 0,” while the lowest point on the graph is labeled, “Q equals K less than 1.” In graph c, a line begins at the upper left side and goes steadily down to a point near the bottom of the y-axis and half way on the x-axis, then rises again to a point labeled, “G superscript degree sign ( products ),” that is equal to the starting point on the y-axis which is labeled, “G superscript degree sign ( reactants ).” The lowest point on the graph is labeled, “Q equals K equals 1.” At the top of the graph is the label, “Delta G superscript degree sign equals 0.”
Figure 13.11 These plots show the free energy versus reaction progress for systems whose standard free changes are (a) negative, (b) positive, and (c) zero. Nonequilibrium systems will proceed spontaneously in whatever direction is necessary to minimize free energy and establish equilibrium.
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