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Chemistry: Atoms First 2e

13.4 Equilibrium Calculations

Chemistry: Atoms First 2e13.4 Equilibrium Calculations
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  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
    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 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 Electrochemistry
    1. Introduction
    2. 16.1 Review of Redox Chemistry
    3. 16.2 Galvanic Cells
    4. 16.3 Electrode and Cell Potentials
    5. 16.4 Potential, Free Energy, and Equilibrium
    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
By the end of this section, you will be able to:
  • Identify the changes in concentration or pressure that occur for chemical species in equilibrium systems
  • Calculate equilibrium concentrations or pressures and equilibrium constants, using various algebraic approaches
  • Explain how temperature affects the spontaneity of some proceses
  • Relate standard free energy changes to equilibrium constants

Having covered the essential concepts of chemical equilibria in the preceding sections of this chapter, this final section will demonstrate the more practical aspect of using these concepts and appropriate mathematical strategies to perform various equilibrium calculations. These types of computations are essential to many areas of science and technology—for example, in the formulation and dosing of pharmaceutical products. After a drug is ingested or injected, it is typically involved in several chemical equilibria that affect its ultimate concentration in the body system of interest. Knowledge of the quantitative aspects of these equilibria is required to compute a dosage amount that will solicit the desired therapeutic effect.

Many of the useful equilibrium calculations that will be demonstrated here require terms representing changes in reactant and product concentrations. These terms are derived from the stoichiometry of the reaction, as illustrated by decomposition of ammonia:

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

As shown earlier in this chapter, this equilibrium may be established within a sealed container that initially contains either NH3 only, or a mixture of any two of the three chemical species involved in the equilibrium. Regardless of its initial composition, a reaction mixture will show the same relationships between changes in the concentrations of the three species involved, as dictated by the reaction stoichiometry (see also the related content on expressing reaction rates in the chapter on kinetics). For example, if the nitrogen concentration increases by an amount x:

Δ[N2]=+x,Δ[N2]=+x,

the corresponding changes in the other species concentrations are

Δ[H2]=Δ[N2](3molH21molN2)=+3xΔ[NH3]=Δ[N2](2molNH31molN2)=2x,Δ[H2]=Δ[N2](3molH21molN2)=+3xΔ[NH3]=Δ[N2](2molNH31molN2)=2x,

where the negative sign indicates a decrease in concentration.

Example 13.5

Determining Relative Changes in Concentration Derive the missing terms representing concentration changes 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

Calculation of an Equilibrium Constant

The equilibrium constant for a reaction is calculated from the equilibrium concentrations (or pressures) of its reactants and products. If these concentrations are known, the calculation simply involves their substitution into the K expression, as was illustrated by Example 13.2. A slightly more challenging example is provided next, in which the reaction stoichiometry is used to derive equilibrium concentrations from the information provided. The basic strategy of this computation is helpful for many types of equilibrium computations and relies on the use of terms for the reactant and product concentrations initially present, for how they change as the reaction proceeds, and for what they are when the system reaches equilibrium. The acronym ICE is commonly used to refer to this mathematical approach, and the concentrations terms are usually gathered in a tabular format called an ICE table.

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 To calculate the equilibrium constants, equilibrium concentrations are needed for all the reactants and products:

KC=[I3][I2][I]KC=[I3][I2][I]

Provided are the initial concentrations of the reactants and the equilibrium concentration of the product. Use this information to derive terms for the equilibrium concentrations of the reactants, presenting all the information in an 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 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.

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

The ICE table may now be updated with numerical values for all its concentrations:

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 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 3.39 times 10 to the negative fourth power, 6.61 times 10 to the negative fourth power. The third column has the following: 0, positive 3.39 times 10 to the negative fourth power, 3.39 times 10 to the negative fourth power.

Finally, substitute the equilibrium concentrations into the K expression and solve:

Kc=[I3][I2][I]Kc=[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 a solute in this reaction.)

Answer:

Kc = 4

Calculation of a Missing Equilibrium Concentration

When the equilibrium constant and all but one equilibrium concentration are provided, the other equilibrium concentration(s) may be calculated. A computation of this sort is illustrated in the next example exercise.

Example 13.7

Calculation of a Missing Equilibrium ConcentrationNitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At 2000 °C, the value of the Kc 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 Substitute the provided quantities into the equilibrium constant expression and solve for [NO]:

Kc=[NO]2[N2][O2]Kc=[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.

To confirm this result, it may be used along with the provided equilibrium concentrations to calculate a value for K:

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

This result is consistent with the provided value for K within nominal uncertainty, differing by just 1 in the least significant digit’s place.

Check Your Learning The equilibrium constant Kc 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

Perhaps the most challenging type of equilibrium calculation can be one in which equilibrium concentrations are derived from initial concentrations and an equilibrium constant. For these calculations, a four-step approach is typically useful:

  1. Identify the direction in which the reaction will proceed to reach equilibrium.
  2. Develop an ICE table.
  3. Calculate the concentration changes and, subsequently, the equilibrium concentrations.
  4. Confirm the calculated equilibrium concentrations.

The last two example exercises of this chapter demonstrate the application of this strategy.

Example 13.8

Calculation of Equilibrium Concentrations Under certain conditions, the equilibrium constant Kc 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 in a mixture that initially contained only PCl5 at a concentration of 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 only the reactant is present initially Qc = 0 and the reaction will proceed to the right.

  2. Step 2.

    Develop an 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 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, x. The third column has the following: 0, positive x, 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)
    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 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

    The two roots of the quadratic are, therefore,

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

    and

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

    For this scenario, only the positive root is physically meaningful (concentrations are either zero or positive), and so 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.

    Confirm the calculated equilibrium concentrations.

    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).

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 for a mixture that is initially 0.15 M in CH3CO2H, 0.15 M in C2H5OH, 0.40 M in CH3CO2C2H5, and 0.40 M in H2O?

Answer:

[CH3CO2H] = 0.37 M, [C2H5OH] = 0.37 M, [CH3CO2C2H5] = 0.18 M, [H2O] = 0.18 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

Example 13.9

Calculation of Equilibrium Concentrations Using an Algebra-Simplifying Assumption 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, positive x, x. The third column has the following: 0, positive x, x.

Substitute the equilibrium concentration terms into the Kc expression

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

Rearrange to the quadratic form and solve for x

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.

Note in this case that the change in concentration is significantly less than the initial concentration (a consequence of the small K), and so the initial concentration experiences a negligible change:

ifx0.15M,then(0.15x)0.15ifx0.15M,then(0.15x)0.15

This approximation allows for a more expedient mathematical approach to the calculation that avoids the need to solve for the roots of a quadratic equation:

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

The value of x calculated is, indeed, much less than the initial concentration

8.6×10−60.158.6×10−60.15

and so the approximation was justified. If this simplified approach were to yield a value for x that did not justify the approximation, the calculation would need to be repeated without making the approximation.

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

Answer:

[OH]=[NH4+]=0.0021M;[OH]=[NH4+]=0.0021M; [NH3] = 0.25 M

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 13.8.

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 13.8 There are four possibilities regarding the signs of enthalpy and entropy changes.

Example 13.10

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 13.9. 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

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 13.9 These plots show the variation in ΔG with temperature for the four possible combinations of arithmetic sign for ΔH and ΔS.

Example 13.11

Equilibrium Temperature for a Phase TransitionAs 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°=1 mol×ΔHf°(H2O(g))1 mol×ΔHf°(H2O(l))=(1 mol)241.82 kJ/mol(1 mol)(−241.82 kJ/mol)=44.01 kJ ΔS°=1 mol×ΔS°(H2O(g))1 mol×ΔS°(H2O(l))=(1 mol)188.8 J/K·mol(1 mol)70.0 J/K·mol=118.8 J/K T=ΔH°ΔS°=44.01×103J118.8J/K=370.5K=97.3°CΔH°=1 mol×ΔHf°(H2O(g))1 mol×ΔHf°(H2O(l))=(1 mol)241.82 kJ/mol(1 mol)(−241.82 kJ/mol)=44.01 kJ ΔS°=1 mol×ΔS°(H2O(g))1 mol×ΔS°(H2O(l))=(1 mol)188.8 J/K·mol(1 mol)70.0 J/K·mol=118.8 J/K T=ΔH°ΔS°=44.01×103J118.8J/K=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 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 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 (pressures other than 1 bar; concentrations other than 1 M) is related to the standard free energy change, 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. This equation may be used to predict the spontaneity for a process under any given set of conditions as illustrated in Example 13.12.

Example 13.12

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° Composition of an Equilibrium Mixture
> 1 < 0 Products are more abundant
< 1 > 0 Reactants are more abundant
= 1 = 0 Reactants and products are comparably abundant
Table 13.1

Example 13.13

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°=[ΔGf°(Ag+(aq))+ΔGf°(Cl(aq))][ΔGf°(AgCl(s))]=[77.1 kJ/mol131.2 kJ/mol][109.8 kJ/mol]=55.7 kJ/molΔG°=[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 13.10). 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 13.10 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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