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Introductory Chemistry - 1st Canadian Edition: Hess’s Law

Introductory Chemistry - 1st Canadian Edition
Hess’s Law
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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

Hess’s Law

Learning Objectives

  1. Learn how to combine chemical equations and their enthalpy changes.

Now that we understand that chemical reactions occur with a simultaneous change in energy, we can apply the concept more broadly. To start, remember that some chemical reactions are rather difficult to perform. For example, consider the combustion of carbon to make carbon monoxide:

\ce{2C(s)}+\ce{O2(g)}\rightarrow \ce{2CO(g)}\quad \Delta H=?

In reality, this is extremely difficult to do; given the opportunity, carbon will react to make another compound, carbon dioxide:

\ce{2C(s)}+\ce{O2(g)}\rightarrow \ce{2CO2(g)}\quad \Delta H=-393.5\text{ kJ}

Is there a way around this? Yes. It comes from the understanding that chemical equations can be treated like algebraic equations, with the arrow acting like the equals sign. Like algebraic equations, chemical equations can be combined, and if the same substance appears on both sides of the arrow, it can be cancelled out (much like a spectator ion in ionic equations). For example, consider these two reactions:

\begin{array}{rrl} \ce{2C(s)}+\ce{2O2(g)}&\rightarrow&\ce{2CO2(g)} \\ \\ \ce{2CO2(g)}&\rightarrow&\ce{2CO(g)}+\ce{O2(g)} \end{array}

If we added these two equations by combining all the reactants together and all the products together, we would get:

\ce{2C(s)}+\ce{2O2(g)}+\ce{2CO2(g)}\rightarrow \ce{2CO2(g)}+\ce{2CO(g)}+\ce{O2(g)}

We note that 2CO2(g) appears on both sides of the arrow, so they cancel:

\ce{2C(s)}+\ce{2O2(g)}+\cancel{\ce{2CO2(g)}}\rightarrow \cancel{\ce{2CO2(g)}}+\ce{2CO(g)}+\ce{O2(g)}

We also note that there are 2 mol of O2 on the reactant side, and 1 mol of O2 on the product side. We can cancel 1 mol of O2 from both sides:

\ce{2C(s)}+\ce{\cancel{2}O2(g)}\rightarrow \ce{2CO(g)}+\cancel{\ce{O2(g)}}

What do we have left?

\ce{2C(s)}+\ce{O2(g)}\rightarrow \ce{2CO(g)}

This is the reaction we are looking for! So by algebraically combining chemical equations, we can generate new chemical equations that may not be feasible to perform.

What about the enthalpy changes? Hess’s law states that when chemical equations are combined algebraically, their enthalpies can be combined in exactly the same way. Two corollaries immediately present themselves:

  1. If a chemical reaction is reversed, the sign on ΔH is changed.
  2. If a multiple of a chemical reaction is taken, the same multiple of the ΔH is taken as well.

What are the equations being combined? The first chemical equation is the combustion of C, which produces CO2:

\ce{2C(s)}+\ce{2O2(g)}\rightarrow \ce{2CO2(g)}

This reaction is two times the reaction to make CO2 from C(s) and O2(g), whose enthalpy change is known:

\ce{C(s)}+\ce{O2(g)}\rightarrow \ce{CO2(g)}\quad \Delta H=-393.5\text{ kJ}

According to the first corollary, the first reaction has an energy change of two times −393.5 kJ, or −787.0 kJ:

\ce{2C(s)}+\ce{2O2(g)}\rightarrow \ce{2CO2(g)}\quad \Delta H=-787.0\text{ kJ}

The second reaction in the combination is related to the combustion of CO(g):

\ce{2CO(g)}+\ce{O2(g)}\rightarrow \ce{2CO2(g)}\quad \Delta H=-566.0\text{ kJ}

The second reaction in our combination is the reverse of the combustion of CO. When we reverse the reaction, we change the sign on the ΔH:

\ce{2CO2(g)}\rightarrow \ce{2CO(g)}+\ce{O2(g)}\quad \Delta H=+566.0\text{ kJ}

Now that we have identified the enthalpy changes of the two component chemical equations, we can combine the ΔH values and add them:

\begin{array}{ll} \begin{array}{rrl} \ce{2C(s)}+\ce{\cancel{2}O2(g)}&\rightarrow&\cancel{\ce{2CO2(g)}} \\ \cancel{\ce{2CO2(g)}}&\rightarrow&\ce{2CO(g)}+\cancel{\ce{O2(g)}} \\ \midrule \ce{2C(s)}+\ce{O2(g)}&\rightarrow&\ce{2CO(g)} \end{array} & \begin{array}{rrl} \Delta H&=&-787.0\text{ kJ} \\ \Delta H&=&+566.0\text{ kJ} \\ \Delta H&=&-787.0+566.0\text{ kJ}=-221.0\text{ kJ} \end{array} \end{array}

Hess’s law is very powerful. It allows us to combine equations to generate new chemical reactions whose enthalpy changes can be calculated, rather than directly measured.

Example 7.14

Problem

Determine the enthalpy change of:

\ce{C2H4}+\ce{3O2}\rightarrow \ce{2CO2}+\ce{2H2O}\quad \Delta H=?

from these reactions:

\begin{array}{rrll} \ce{C2H2}+\ce{H2}&\rightarrow &\ce{C2H4}&\Delta H = -174.5\text{ kJ} \\ \\ \ce{2C2H2}+\ce{5O2}&\rightarrow&\ce{4CO2}+\ce{2H2O}&\Delta H =-1,692.2\text{ kJ} \\ \\ \ce{2CO2}+\ce{H2}&\rightarrow&\ce{2O2}+\ce{C2H2}&\Delta H =-167.5\text{ kJ} \end{array}

Solution

We will start by writing chemical reactions that put the correct number of moles of the correct substance on the proper side. For example, our desired reaction has C2H4 as a reactant, and only one reaction from our data has C2H4. However, it has C2H4 as a product. To make it a reactant, we need to reverse the reaction, changing the sign on the ΔH:

\ce{C2H4}\rightarrow \ce{C2H2}+\ce{H2}\quad \Delta H=+174.5\text{ kJ}

We need CO2 and H2O as products. The second reaction has them on the proper side, so let us include one of these reactions (with the hope that the coefficients will work out when all our reactions are added):

\ce{2C2H2}+\ce{5O2}\rightarrow \ce{4CO2}+\ce{2H2O}\quad \Delta H=-1,692.2\text{ kJ}

We note that we now have 4 mol of CO2 as products; we need to get rid of 2 mol of CO2. The last reaction has 2CO2 as a reactant. Let us use it as written:

\ce{2CO2}+\ce{H2}\rightarrow \ce{2O2}+\ce{C2H2}\quad \Delta H=-167.5\text{ kJ}

We combine these three reactions, modified as stated:

\begin{array}{ll} \begin{array}{rrl} \\ \\ \ce{C2H4}&\rightarrow&\ce{C2H2}+\ce{H2} \\ \ce{2C2H2}+\ce{5O2}&\rightarrow&\ce{4CO2}+\ce{2H2O} \\ \ce{2CO2}+\ce{H2}&\rightarrow&\ce{2O2}+\ce{C2H2} \\ \midrule \ce{C2H4}+\ce{2C2H2}+\ce{5O2}+\ce{2CO2}+\ce{H2}&\rightarrow&\ce{C2H2}+\ce{H2}+\ce{4CO2}+\ce{2H2O}+\ce{2O2}+\ce{C2H2} \end{array} & \begin{array}{l} \Delta H=+174.5\text{ kJ} \\ \Delta H=-1,692.2\text{ kJ} \\ \Delta H=-167.5\text{ kJ} \end{array} \end{array}

What cancels? 2C2H2, H2, 2O2, and 2CO2. What is left is:

\ce{C2H4}+\ce{3O2}\rightarrow \ce{2CO2}+\ce{2H2O}

Which is the reaction we are looking for. The ΔH of this reaction is the sum of the three ΔH values:

\Delta H =+174.5-1,692.2-167.5=-1,685.2\text{ kJ}

Test Yourself

Given the following thermochemical equations:

\begin{array}{rrll} \ce{Pb}+\ce{Cl2}&\rightarrow&\ce{PbCl2}&\Delta H=-223\text{ kJ} \\ \ce{PbCl2}+\ce{Cl2}&\rightarrow&\ce{PbCl4}&\Delta H=-87\text{ kJ} \end{array}

Determine ΔH for:

\ce{2PbCl2}\rightarrow \ce{Pb}+\ce{PbCl4}

Answer

+136 kJ

Key Takeaways

  • Hess’s law allows us to combine reactions algebraically and then combine their enthalpy changes the same way.

Exercises

Questions

  1. Define Hess’s law.
  2. What does Hess’s law require us to do to the ΔH of a thermochemical equation if we reverse the equation?
  3. If the ΔH for C2H4 + H2 → C2H6 is −65.6 kJ, what is the ΔH for the reaction C2H6 → C2H4 + H2?
  4. If the ΔH for 2Na + Cl2 → 2NaCl is −772 kJ, what is the ΔH for the reaction 2NaCl → 2Na + Cl2?
  5. If the ΔH for C2H4 + H2 → C2H6 is −65.6 kJ, what is the ΔH for the reaction 2C2H4 + 2H2 → 2C2H6?
  6. If the ΔH for 2C2H6 + 7O2 → 4CO2 + 6H2O is −2,650 kJ, what is the ΔH for the reaction 6C2H6 + 21O2 → 12CO2 + 18H2O?
  7. The ΔH for C2H4 + H2O → C2H5OH is −44 kJ. What is the ΔH for the reaction 2C2H5OH → 2C2H4 + 2H2O?
  8. The ΔH for N2 + O2 → 2NO is 181 kJ. What is the ΔH for the reaction \ce{NO} \rightarrow \ce{\dfrac{1}{2}N2} + \ce{\dfrac{1}{2}O2}?
  9. Determine the ΔH for the reaction Cu + Cl2 → CuCl2 given these data:

        \begin{align*} \ce{2Cu}+\ce{Cl2}&\rightarrow \ce{2CuCl} & \Delta H&=-274\text{ kJ} \\ \ce{2CuCl}+\ce{Cl2}&\rightarrow \ce{2CuCl2} & \Delta H&=-166\text{ kJ} \end{align*}

  10. Determine ΔH for the reaction 2CH4 → 2H2 + C2H4 given these data:

        \begin{align*} \ce{CH4}+\ce{2O2}&\rightarrow\ce{CO2}+\ce{2H2O} & \Delta H &=-891\text{ kJ} \\ \ce{C2H4}+\ce{3O2}&\rightarrow \ce{2CO2}+\ce{2H2O} & \Delta H&=-1,411\text{ kJ} \\ \ce{2H2}+\ce{O2}&\rightarrow \ce{2H2O} & \Delta H&=-571\text{ kJ} \end{align*}

  11. Determine ΔH for the reaction Fe2(SO4)3 → Fe2O3 + 3SO3 given these data:

        \begin{align*} \ce{4Fe}+\ce{3O2}&\rightarrow \ce{2Fe2O3} & \Delta H&=-1,650\text{ kJ} \\ \ce{2S}+\ce{3O2}&\rightarrow \ce{2SO3} & \Delta H&=-792\text{ kJ} \\ \ce{2Fe}+\ce{3S}+\ce{6O2}&\rightarrow \ce{Fe2(SO4)3} & \Delta H&=-2,583\text{ kJ} \end{align*}

  12. Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data:

        \begin{align*} \ce{2Ca}+\ce{2C}+\ce{3O2}&\rightarrow \ce{2CaCO3} & \Delta H&=-2,414\text{ kJ} \\ \ce{C}+\ce{O2}&\rightarrow \ce{CO2} & \Delta H&=-393.5\text{ kJ} \\ \ce{2Ca}+\ce{O2}&\rightarrow \ce{2CaO} & \Delta H&=-1,270\text{ kJ} \end{align*}

Answers

  1. If chemical equations are combined, their energy changes are also combined.
  1. ΔH = 65.6 kJ
  1. ΔH = −131.2 kJ
  1. ΔH = 88 kJ
  1. ΔH = −220 kJ
  1. ΔH = 570 kJ

Annotate

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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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