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Introductory Chemistry - 1st Canadian Edition: Calculating Equilibrium Constant Values

Introductory Chemistry - 1st Canadian Edition
Calculating Equilibrium Constant Values
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table of contents
  1. Cover
  2. Title Page
  3. Copyright
  4. Table Of Contents
  5. Acknowledgments
  6. Dedication
  7. About BCcampus Open Education
  8. Chapter 1. What is Chemistry
    1. Some Basic Definitions
    2. Chemistry as a Science
  9. Chapter 2. Measurements
    1. Expressing Numbers
    2. Significant Figures
    3. Converting Units
    4. Other Units: Temperature and Density
    5. Expressing Units
    6. End-of-Chapter Material
  10. Chapter 3. Atoms, Molecules, and Ions
    1. Acids
    2. Ions and Ionic Compounds
    3. Masses of Atoms and Molecules
    4. Molecules and Chemical Nomenclature
    5. Atomic Theory
    6. End-of-Chapter Material
  11. Chapter 4. Chemical Reactions and Equations
    1. The Chemical Equation
    2. Types of Chemical Reactions: Single- and Double-Displacement Reactions
    3. Ionic Equations: A Closer Look
    4. Composition, Decomposition, and Combustion Reactions
    5. Oxidation-Reduction Reactions
    6. Neutralization Reactions
    7. End-of-Chapter Material
  12. Chapter 5. Stoichiometry and the Mole
    1. Stoichiometry
    2. The Mole
    3. Mole-Mass and Mass-Mass Calculations
    4. Limiting Reagents
    5. The Mole in Chemical Reactions
    6. Yields
    7. End-of-Chapter Material
  13. Chapter 6. Gases
    1. Pressure
    2. Gas Laws
    3. Other Gas Laws
    4. The Ideal Gas Law and Some Applications
    5. Gas Mixtures
    6. Kinetic Molecular Theory of Gases
    7. Molecular Effusion and Diffusion
    8. Real Gases
    9. End-of-Chapter Material
  14. Chapter 7. Energy and Chemistry
    1. Formation Reactions
    2. Energy
    3. Stoichiometry Calculations Using Enthalpy
    4. Enthalpy and Chemical Reactions
    5. Work and Heat
    6. Hess’s Law
    7. End-of-Chapter Material
  15. Chapter 8. Electronic Structure
    1. Light
    2. Quantum Numbers for Electrons
    3. Organization of Electrons in Atoms
    4. Electronic Structure and the Periodic Table
    5. Periodic Trends
    6. End-of-Chapter Material
  16. Chapter 9. Chemical Bonds
    1. Lewis Electron Dot Diagrams
    2. Electron Transfer: Ionic Bonds
    3. Covalent Bonds
    4. Other Aspects of Covalent Bonds
    5. Violations of the Octet Rule
    6. Molecular Shapes and Polarity
    7. Valence Bond Theory and Hybrid Orbitals
    8. Molecular Orbitals
    9. End-of-Chapter Material
  17. Chapter 10. Solids and Liquids
    1. Properties of Liquids
    2. Solids
    3. Phase Transitions: Melting, Boiling, and Subliming
    4. Intermolecular Forces
    5. End-of-Chapter Material
  18. Chapter 11. Solutions
    1. Colligative Properties of Solutions
    2. Concentrations as Conversion Factors
    3. Quantitative Units of Concentration
    4. Colligative Properties of Ionic Solutes
    5. Some Definitions
    6. Dilutions and Concentrations
    7. End-of-Chapter Material
  19. Chapter 12. Acids and Bases
    1. Acid-Base Titrations
    2. Strong and Weak Acids and Bases and Their Salts
    3. Brønsted-Lowry Acids and Bases
    4. Arrhenius Acids and Bases
    5. Autoionization of Water
    6. Buffers
    7. The pH Scale
    8. End-of-Chapter Material
  20. Chapter 13. Chemical Equilibrium
    1. Chemical Equilibrium
    2. The Equilibrium Constant
    3. Shifting Equilibria: Le Chatelier’s Principle
    4. Calculating Equilibrium Constant Values
    5. Some Special Types of Equilibria
    6. End-of-Chapter Material
  21. Chapter 14. Oxidation and Reduction
    1. Oxidation-Reduction Reactions
    2. Balancing Redox Reactions
    3. Applications of Redox Reactions: Voltaic Cells
    4. Electrolysis
    5. End-of-Chapter Material
  22. Chapter 15. Nuclear Chemistry
    1. Units of Radioactivity
    2. Uses of Radioactive Isotopes
    3. Half-Life
    4. Radioactivity
    5. Nuclear Energy
    6. End-of-Chapter Material
  23. Chapter 16. Organic Chemistry
    1. Hydrocarbons
    2. Branched Hydrocarbons
    3. Alkyl Halides and Alcohols
    4. Other Oxygen-Containing Functional Groups
    5. Other Functional Groups
    6. Polymers
    7. End-of-Chapter Material
  24. Chapter 17. Kinetics
    1. Factors that Affect the Rate of Reactions
    2. Reaction Rates
    3. Rate Laws
    4. Concentration–Time Relationships: Integrated Rate Laws
    5. Activation Energy and the Arrhenius Equation
    6. Reaction Mechanisms
    7. Catalysis
    8. End-of-Chapter Material
  25. Chapter 18. Chemical Thermodynamics
    1. Spontaneous Change
    2. Entropy and the Second Law of Thermodynamics
    3. Measuring Entropy and Entropy Changes
    4. Gibbs Free Energy
    5. Spontaneity: Free Energy and Temperature
    6. Free Energy under Nonstandard Conditions
    7. End-of-Chapter Material
  26. Appendix A: Periodic Table of the Elements
  27. Appendix B: Selected Acid Dissociation Constants at 25°C
  28. Appendix C: Solubility Constants for Compounds at 25°C
  29. Appendix D: Standard Thermodynamic Quantities for Chemical Substances at 25°C
  30. Appendix E: Standard Reduction Potentials by Value
  31. Glossary
  32. About the Authors
  33. Versioning History

Calculating Equilibrium Constant Values

Learning Objectives

  1. Calculate equilibrium concentrations from the values of the initial amounts and the Keq.

There are some circumstances in which, given some initial amounts and the Keq, you will have to determine the concentrations of all species when equilibrium is achieved. Such calculations are not difficult to do, especially if a consistent approach is applied. We will consider such an approach here.

Suppose we have this simple equilibrium. Its associated Keq is 4.0, and the initial concentration of each reactant is 1.0 M:

\begin{array}{rl} \begin{array}{ccccc} \ce{H2(g)}&+&\ce{Cl2(g)}&\rightleftharpoons &\ce{2HCl(g)} \\ 1.0\text{ M}&&1.0\text{ M}&& \end{array} &\hspace{2em} K_{\text{eq}}=4.0 \end{array}

Because we have concentrations for the reactants but not the products, we presume that the reaction will proceed in the forward direction to make products. But by how much will it proceed? We don’t know, so let us assign it a variable. Let us assume that x M H2 reacts as the reaction goes to equilibrium. This means that at equilibrium, we have (1.0 − x) M H2 left over.

According to the balanced chemical equation, H2 and Cl2 react in a 1:1 ratio. How do we know that? The coefficients of these two species in the balanced chemical equation are 1 (unwritten, of course). This means that if x M H2 reacts, x M Cl2 reacts as well. If we start with 1.0 M Cl2 at the beginning and we react x M, we have (1.0 − x) M Cl2 left at equilibrium.

How much HCl is made? We start with zero, but we also see that 2 mol of HCl are made for every mole of H2 (or Cl2) that reacts (from the coefficients in the balanced chemical equation), so if we lose x M H2, we gain 2x M HCl. So now we know the equilibrium concentrations of our species:

\begin{array}{rl} \begin{array}{ccccc} \ce{H2(g)}&+&\ce{Cl2(g)}&\rightleftharpoons &\ce{2HCl(g)} \\ (1.0-x)\text{M}&&(1.0-x)\text{M}&&2x\text{M} \end{array} &\hspace{2em} K_{\text{eq}}=4.0 \end{array}

We can substitute these concentrations into the Keq expression for this reaction and combine it with the known value of Keq:

K_{\text{eq}}=\dfrac{[\ce{HCl}]^2}{[\ce{H2}][\ce{Cl2}]}=\dfrac{(2x)^2}{(1-x)(1-x)}=4.0

This is an equation in one variable, so we should be able to solve for the unknown value. This expression may look formidable, but first we can simplify the denominator and write it as a perfect square as well:

\dfrac{(2x)^2}{(1-x)^2}=4.0

The fraction is a perfect square, as is the 4.0 on the right. So we can take the square root of both sides:

\dfrac{2x}{1-x}=2.0

Now we rearrange and solve (be sure you can follow each step):

\begin{array}{rcccc} 2x&=&2.0&-&2x \\ +2x&&&+&2x \\ \hline 4x&=&2.0&& \\ \\ \dfrac{4x}{4}&=&\dfrac{2.0}{4}&& \\ \\ x&=&0.50&& \end{array}

Now we have to remind ourselves what x is — the amount of H2 and Cl2 that reacted — and 2x is the equilibrium [HCl]. To determine the equilibrium concentrations, we need to go back and evaluate the expressions 1 − x and 2x to get the equilibrium concentrations of our species:

\begin{array}{rcccrcl} 1.0 - x &=& 1.0 - 0.50&=&0.50\text{ M}&=&[\ce{H2}]=[\ce{Cl2}] \\ 2x &=&2(0.50)&=&1.0\text{ M}&=&[\ce{HCl}] \end{array}

The units are assumed to be molarity. To check, we simply substitute these concentrations and verify that we get the numerical value of the Keq, in this case 4.0:

\dfrac{(1.0)^2}{(0.50)(0.50)}=4.0

We formalize this process by introducing the ICE chart, where ICE stands for initial, change, and equilibrium.

The initial values go in the first row of the chart. The change values, usually algebraic expressions because we do not yet know their exact numerical values, go in the next row. However, the change values must be in the proper stoichiometric ratio as indicated by the balanced chemical equation. Finally, the equilibrium expressions in the last row are a combination of the initial value and the change value for each species. The expressions in the equilibrium row are substituted into the Keq expression, which yields an algebraic equation that we try to solve.

The ICE chart for the above example would look like this:

H2(g)+Cl2(g)⇄2HCl(g)Keq = 4.0
I1.01.00
C−x−x+2x
E1.0 − x1.0 − x+2x

Substituting the last row into the expression for the Keq yields

K_{\text{eq}}=\dfrac{[\ce{HCl}]^2}{[\ce{H2}][\ce{Cl2}]}=\dfrac{(2x)^2}{(1-x)(1-x)}=4.0

which, of course, is the same expression we have already solved and yields the same answers for the equilibrium concentrations. The ICE chart is a more formalized way to do these types of problems. The + sign is included explicitly in the change row of the ICE chart to avoid any confusion.

Sometimes when an ICE chart is set up and the Keq expression is constructed, a more complex algebraic equation will result. One of the more common equations has an x2 term in it and is called a quadratic equation. There will be two values possible for the unknown x, and for a quadratic equation with the general formula ax2 + bx + c = 0 (where a, b, and c are the coefficients of the quadratic equation), the two possible values are as follows:

x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

One value of x is the + sign used in the numerator, and the other value of x is the − sign used in the numerator. In this case, one value of x typically makes no sense as an answer and can be discarded as physically impossible, leaving only one possible value and the resulting set of concentrations. Example 13.9 illustrates this.

Example 13.9

Set up an ICE chart and solve for the equilibrium concentrations in this chemical reaction.

\begin{array}{rl} \begin{array}{ccccc} \ce{COI2(g)}&\leftrightharpoons&\ce{CO(g)}&+&\ce{I2(g)} \\ 0.55\text{ M}&&0&&0 \end{array} &\hspace{2em} K_{\text{eq}}=0.00088 \end{array}

Solution
The ICE chart is set up like this. First, the initial values:

COI2(g)⇄CO(g)+I2(g)
I0.5500
C
E

Some of the COI2 will be lost, but how much? We don’t know, so we represent it by the variable x. So x M COI2 will be lost, and for each COI2 that is lost, x M CO and x M I2 will be produced. These expressions go into the change row:

COI2(g)⇄CO(g)+I2(g)
I0.5500
C−x+x+x
E

At equilibrium, the resulting concentrations will be a combination of the initial amount and the changes:

COI2(g)⇄CO(g)+I2(g)
I0.5500
C−x+x+x
E0.55 − x+x+x

The expressions in the equilibrium row go into the Keq expression:

K_{\text{eq}}=\dfrac{[\ce{CO}][\ce{I2}]}{\ce{COI2}}=0.00088=\dfrac{(x)(x)}{(0.55-x)}

We rearrange this into a quadratic equation that equals 0:

\begin{array}{rcl} (0.55-x)0.00088&=&\dfrac{(x)(x)}{(\cancel{0.55-x})}\times (\cancel{0.55-x}) \\ \\ 0.000484-0.00088x&=&x^2 \\ x^2+0.00088x-0.000484&=&0 \end{array}

Now we use the quadratic equation to solve for the two possible values of x:

x=\dfrac{-0.00088\pm \sqrt{(0.00088)^2-4(1)(-0.000484)}}{2(1)}

Evaluate for both signs in the numerator — first the + sign and then the − sign:

x=0.0216\text{ or }x=-0.0224

Because x is the final concentration of both CO and I2, it cannot be negative, so we discount the second numerical answer as impossible. Thus x = 0.0216.

Going back to determine the final concentrations using the expressions in the E row of our ICE chart, we have:

    \begin{align*} [\ce{COI2}]&=0.55-x=0.55-0.0216=0.53\text{ M} \\ {[\ce{CO}]}&=x=0.0216\text{ M} \\ {[\ce{I2}]}&=x=0.0216\text{ M} \end{align*}

You can verify that these numbers are correct by substituting them into the Keq expression and evaluating and comparing to the known Keq value.

Test Yourself
Set up an ICE chart and solve for the equilibrium concentrations in this chemical reaction.

\begin{array}{rl} \begin{array}{ccccc} \ce{N2H2(g)}&\leftrightharpoons&\ce{N2(g)}&+&\ce{H2(g)} \\ 0.075\text{ M}&&0&&0 \end{array} &\hspace{2em} K_{\text{eq}}=0.052 \end{array}

Answer
The completed ICE chart is as follows:

N2H2(g)⇄N2(g)+H2(g)
I0.07500
C−x+x+x
E0.075 − x+x+x

Solving for x gives the equilibrium concentrations as [N2H2] = 0.033 M; [N2] = 0.042 M; and [H2] = 0.042 M

Key Takeaways

  • An ICE chart is a convenient way to determine equilibrium concentrations from starting amounts.

Exercises

Questions

  1. Describe the three parts of an ICE chart.
  2. What is the relationship between the equilibrium row in an ICE chart and the other two rows?
  3. Set up (but do not solve) an ICE chart for this reaction, given the initial conditions.

    \begin{array}{ccc} \ce{3O2(g)}&\leftrightharpoons&\ce{2O3(g)} \\ 0.075\text{ M}&& \end{array}

  4. Set up (but do not solve) an ICE chart for this reaction, given the initial conditions.

    \begin{array}{ccccccc} \ce{CH4(g)}&+&\ce{2O2(g)}&\leftrightharpoons&\ce{CO2(g)}&+&\ce{2H2O(g)} \\ 0.750\text{ M}&&0.450\text{ M}&&&& \end{array}

  5. Given that pure solids and liquids do not appear in Keq expressions, set up the ICE chart for this reaction, given the initial conditions.

    \begin{array}{ccccccc} \ce{CH4(g)}&+&\ce{2O2(g)}&\leftrightharpoons&\ce{CO2(g)}&+&\ce{2H2O(\ell)} \\ 0.0060\text{ M}&&0.055\text{ M}&&&& \end{array}

  6. Given that pure solids and liquids do not appear in Keq expressions, set up the ICE chart for this reaction, given the initial conditions.

    \begin{array}{ccccccc} \ce{N2H4(\ell)}&+&\ce{O2(g)}&\leftrightharpoons&\ce{N2(g)}&+&\ce{2H2O(\ell)} \\ 2.33\text{ M}&&1.09\text{ M}&&&& \end{array}

  7. Determine the equilibrium concentrations for this chemical reaction with the given Keq.

    \begin{array}{rl} \begin{array}{ccc} \ce{HCN(g)}&\leftrightharpoons&\ce{HNC(g)} \\ 2.00\text{ M}&& \end{array} &\hspace{2em} K_{\text{eq}}=4.50 \end{array}

  8. Determine the equilibrium concentrations for this chemical reaction with the given Keq.

    \begin{array}{rl} \begin{array}{ccccc} \ce{IF3(g)}&+&\ce{F2(g)}&\leftrightharpoons&\ce{IF5(g)} \\ 1.0\text{ M}&&0.50\text{ M}&& \end{array} &\hspace{2em} K_{\text{eq}}=7.59 \end{array}

  9. Determine the equilibrium concentrations for this chemical reaction with the given Keq.

    \begin{array}{rl} \begin{array}{ccccc} \ce{N2O3(g)}&\leftrightharpoons&\ce{NO(g)}&+&\ce{NO2(g)} \\ 0.0663\text{ M}&&&& \end{array} &\hspace{2em} K_{\text{eq}}=2.50 \end{array}

  10. Determine the equilibrium concentrations for this chemical reaction with the given Keq.

    \begin{array}{rl} \begin{array}{ccccccc} \ce{CO(g)}&+&\ce{H2O(g)}&\leftrightharpoons&\ce{CO2(g)}&+&\ce{H2(g)} \\ 0.750\text{ M}&&0.750\text{ M}&&&& \end{array} &\hspace{2em} K_{\text{eq}}=16.0 \end{array}

  11. Determine the equilibrium concentrations for this chemical reaction with the given Keq.

    \begin{array}{rl} \begin{array}{ccccc} \ce{H2S(g)}&\leftrightharpoons&\ce{H2(g)}&+&\ce{S(s)} \\ 0.882\text{ M}&&&& \end{array} &\hspace{2em} K_{\text{eq}}=0.055 \end{array}

  12. Determine the equilibrium concentrations for this chemical reaction with the given Keq.

    \begin{array}{rl} \begin{array}{ccccccc} \ce{AgCl2(s)}&+&\ce{F2(g)}&\leftrightharpoons&\ce{2AgF(s)}&+&\ce{Cl2(g)} \\ &&1.99\text{ M}&&&& \end{array} &\hspace{2em} K_{\text{eq}}=1.2\times 10^2 \end{array}

Answers

  1. I = initial concentrations; C = change in concentrations; E = equilibrium concentrations
  1. 3O2⇄2O3
    I0.0750
    C−3x+2x
    E0.075 − 3x+2x
  1. CH4+2O2⇄CO2+2H2O
    I0.00600.05500
    C−x−2x+x—
    E0.0060 − x0.055 − 2x+x—
  1. [HCN] = 0.364 M; [HNC] = 1.64 M
  1. [N2O3] = 0.0017 M; [NO] = [NO2] = 0.0646 M
  1. [H2S] = 0.836 M; [H2] = 0.046 M

Annotate

Next Chapter
Some Special Types of Equilibria
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Chemistry

Copyright © 2014

                                by Jessie A. Key

            Introductory Chemistry - 1st Canadian Edition by Jessie A. Key is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.
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