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Chemistry: Atoms First

17.3 Rate Laws

Chemistry: Atoms First17.3 Rate Laws
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Table of contents
  1. Preface
  2. 1 Essential Ideas
    1. Introduction
    2. 1.1 Chemistry in Context
    3. 1.2 Phases and Classification of Matter
    4. 1.3 Physical and Chemical Properties
    5. 1.4 Measurements
    6. 1.5 Measurement Uncertainty, Accuracy, and Precision
    7. 1.6 Mathematical Treatment of Measurement Results
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  3. 2 Atoms, Molecules, and Ions
    1. Introduction
    2. 2.1 Early Ideas in Atomic Theory
    3. 2.2 Evolution of Atomic Theory
    4. 2.3 Atomic Structure and Symbolism
    5. 2.4 Chemical Formulas
    6. 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 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:
  • Explain the form and function of a rate law
  • Use rate laws to calculate reaction rates
  • Use rate and concentration data to identify reaction orders and derive rate laws

As described in the previous module, the rate of a reaction is affected by the concentrations of reactants. Rate laws or rate equations are mathematical expressions that describe the relationship between the rate of a chemical reaction and the concentration of its reactants. In general, a rate law (or differential rate law, as it is sometimes called) takes this form:

rate=k[A]m[B]n[C]p…rate=k[A]m[B]n[C]p…
17.23

in which [A], [B], and [C] represent the molar concentrations of reactants, and k is the rate constant, which is specific for a particular reaction at a particular temperature. The exponents m, n, and p are usually positive integers (although it is possible for them to be fractions or negative numbers). The rate constant k and the exponents m, n, and p must be determined experimentally by observing how the rate of a reaction changes as the concentrations of the reactants are changed. The rate constant k is independent of the concentration of A, B, or C, but it does vary with temperature and surface area.

The exponents in a rate law describe the effects of the reactant concentrations on the reaction rate and define the reaction order. Consider a reaction for which the rate law is:

rate=k[A]m[B]nrate=k[A]m[B]n
17.24

If the exponent m is 1, the reaction is first order with respect to A. If m is 2, the reaction is second order with respect to A. If n is 1, the reaction is first order in B. If n is 2, the reaction is second order in B. If m or n is zero, the reaction is zero order in A or B, respectively, and the rate of the reaction is not affected by the concentration of that reactant. The overall reaction order is the sum of the orders with respect to each reactant. If m = 1 and n = 1, the overall order of the reaction is second order (m + n = 1 + 1 = 2).

The rate law:

rate=k[H2O2]rate=k[H2O2]
17.25

describes a reaction that is first order in hydrogen peroxide and first order overall. The rate law:

rate=k[C4H6]2rate=k[C4H6]2
17.26

describes a reaction that is second order in C4H6 and second order overall. The rate law:

rate=k[H+][OH−]rate=k[H+][OH−]
17.27

describes a reaction that is first order in H+, first order in OH−, and second order overall.

Example 17.3

Writing Rate Laws from Reaction Orders

An experiment shows that the reaction of nitrogen dioxide with carbon monoxide:
NO2(g)+CO(g)⟶NO(g)+CO2(g)NO2(g)+CO(g)⟶NO(g)+CO2(g)
17.28

is second order in NO2 and zero order in CO at 100 °C. What is the rate law for the reaction?

Solution

The reaction will have the form:
rate=k[NO2]m[CO]nrate=k[NO2]m[CO]n
17.29

The reaction is second order in NO2; thus m = 2. The reaction is zero order in CO; thus n = 0. The rate law is:

rate=k[NO2]2[CO]0=k[NO2]2rate=k[NO2]2[CO]0=k[NO2]2
17.30

Remember that a number raised to the zero power is equal to 1, thus [CO]0 = 1, which is why we can simply drop the concentration of CO from the rate equation: the rate of reaction is solely dependent on the concentration of NO2. When we consider rate mechanisms later in this chapter, we will explain how a reactant’s concentration can have no effect on a reaction despite being involved in the reaction.

Check Your Learning

The rate law for the reaction:
H2(g)+2NO(g)⟶N2O(g)+H2O(g)H2(g)+2NO(g)⟶N2O(g)+H2O(g)
17.31

has been determined to be rate = k[NO]2[H2]. What are the orders with respect to each reactant, and what is the overall order of the reaction?

Answer:

order in NO = 2; order in H2 = 1; overall order = 3

Check Your Learning

In a transesterification reaction, a triglyceride reacts with an alcohol to form an ester and glycerol. Many students learn about the reaction between methanol (CH3OH) and ethyl acetate (CH3CH2OCOCH3) as a sample reaction before studying the chemical reactions that produce biodiesel:
CH3OH+CH3CH2OCOCH3⟶CH3OCOCH3+CH3CH2OHCH3OH+CH3CH2OCOCH3⟶CH3OCOCH3+CH3CH2OH
17.32

The rate law for the reaction between methanol and ethyl acetate is, under certain conditions, determined to be:

rate=k[CH3OH]rate=k[CH3OH]
17.33

What is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of reaction?

Answer:

order in CH3OH = 1; order in CH3CH2OCOCH3 = 0; overall order = 1

It is sometimes helpful to use a more explicit algebraic method, often referred to as the method of initial rates, to determine the orders in rate laws. To use this method, we select two sets of rate data that differ in the concentration of only one reactant and set up a ratio of the two rates and the two rate laws. After canceling terms that are equal, we are left with an equation that contains only one unknown, the coefficient of the concentration that varies. We then solve this equation for the coefficient.

Example 17.4

Determining a Rate Law from Initial Rates

Ozone in the upper atmosphere is depleted when it reacts with nitrogen oxides. The rates of the reactions of nitrogen oxides with ozone are important factors in deciding how significant these reactions are in the formation of the ozone hole over Antarctica (Figure 17.9). One such reaction is the combination of nitric oxide, NO, with ozone, O3:
A view of Earth’s southern hemisphere is shown. A nearly circular region of approximately half the diameter of the image is shown in shades of purple, with Antarctica appearing in a slightly lighter color than the surrounding ocean areas. Immediately outside this region is a narrow bright blue zone followed by a bright green zone. In the top half of the figure, the purple region extends slightly outward from the circle and the blue zone extends more outward to the right of the center as compared to the lower half of the image. In the upper half of the image, the majority of the space outside the purple region is shaded green, with a few small strips of interspersed blue regions. The lower half however shows the majority of the space outside the central purple zone in yellow, orange, and red. The red zones appear in the lower central and left regions outside the purple zone. To the lower right of this image is a color scale that is labeled “Total Ozone (Dobsone units).” This scale begins at 0 and increases by 100’s up to 700. At the left end of the scale, the value 0 shows a very deep purple color, 100 is indigo, 200 is blue, 300 is green, 400 is a yellow-orange, 500 is red, 600 is pink, and 700 is white.
Figure 17.9 Over the past several years, the atmospheric ozone concentration over Antarctica has decreased during the winter. This map shows the decreased concentration as a purple area. (credit: modification of work by NASA)
NO(g)+O3(g)⟶NO2(g)+O2(g)NO(g)+O3(g)⟶NO2(g)+O2(g)
17.34

This reaction has been studied in the laboratory, and the following rate data were determined at 25 °C.

Trial [NO] (mol/L) [O3] (mol/L) Δ[NO2]Δt(molL−1s−1)Δ[NO2]Δt(molL−1s−1)
1 1.00 ×× 10−6 3.00 ×× 10−6 6.60 ×× 10−5
2 1.00 ×× 10−6 6.00 ×× 10−6 1.32 ×× 10−4
3 1.00 ×× 10−6 9.00 ×× 10−6 1.98 ×× 10−4
4 2.00 ×× 10−6 9.00 ×× 10−6 3.96 ×× 10−4
5 3.00 ×× 10−6 9.00 ×× 10−6 5.94 ×× 10−4

Determine the rate law and the rate constant for the reaction at 25 °C.

Solution

The rate law will have the form:
rate=k[NO]m[O3]nrate=k[NO]m[O3]n
17.35

We can determine the values of m, n, and k from the experimental data using the following three-part process:

  1. Step 1.

    Determine the value of m from the data in which [NO] varies and [O3] is constant. In the last three experiments, [NO] varies while [O3] remains constant. When [NO] doubles from trial 3 to 4, the rate doubles, and when [NO] triples from trial 3 to 5, the rate also triples. Thus, the rate is also directly proportional to [NO], and m in the rate law is equal to 1.

  2. Step 2.

    Determine the value of n from data in which [O3] varies and [NO] is constant. In the first three experiments, [NO] is constant and [O3] varies. The reaction rate changes in direct proportion to the change in [O3]. When [O3] doubles from trial 1 to 2, the rate doubles; when [O3] triples from trial 1 to 3, the rate increases also triples. Thus, the rate is directly proportional to [O3], and n is equal to 1.The rate law is thus:

    rate=k[NO]1[O3]1=k[NO][O3]rate=k[NO]1[O3]1=k[NO][O3]
    17.36
  3. Step 3.

    Determine the value of k from one set of concentrations and the corresponding rate.

    k=rate[NO][O3]=6.60×10−5mol L−1s−1(1.00×10−6mol L−1)(3.00×10−6molL−1)=2.20×107Lmol−1s−1k=rate[NO][O3]=6.60×10−5mol L−1s−1(1.00×10−6mol L−1)(3.00×10−6molL−1)=2.20×107Lmol−1s−1
    17.37

    The large value of k tells us that this is a fast reaction that could play an important role in ozone depletion if [NO] is large enough.

Check Your Learning

Acetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation:
CH3CHO(g)⟶CH4(g)+CO(g)CH3CHO(g)⟶CH4(g)+CO(g)
17.38

Determine the rate law and the rate constant for the reaction from the following experimental data:

Trial [CH3CHO] (mol/L) −Δ[CH3CHO]Δt(molL−1s−1)−Δ[CH3CHO]Δt(molL−1s−1)
1 1.75 ×× 10−3 2.06 ×× 10−11
2 3.50 ×× 10−3 8.24 ×× 10−11
3 7.00 ×× 10−3 3.30 ×× 10−10

Answer:

rate=k[CH3CHO]2rate=k[CH3CHO]2 with k = 6.73 ×× 10−6 L/mol/s

Example 17.5

Determining Rate Laws from Initial Rates

Using the initial rates method and the experimental data, determine the rate law and the value of the rate constant for this reaction:
2NO(g)+Cl2(g)⟶2NOCl(g)2NO(g)+Cl2(g)⟶2NOCl(g)
17.39
Trial [NO] (mol/L) [Cl2] (mol/L) −Δ[NO]Δt(molL−1s−1)−Δ[NO]Δt(molL−1s−1)
1 0.10 0.10 0.00300
2 0.10 0.15 0.00450
3 0.15 0.10 0.00675

Solution

The rate law for this reaction will have the form:
rate=k[NO]m[Cl2]nrate=k[NO]m[Cl2]n
17.40

As in Example 17.4, we can approach this problem in a stepwise fashion, determining the values of m and n from the experimental data and then using these values to determine the value of k. In this example, however, we will use a different approach to determine the values of m and n:

  1. Step 1.

    Determine the value of m from the data in which [NO] varies and [Cl2] is constant. We can write the ratios with the subscripts x and y to indicate data from two different trials:

    ratexratey=k[NO]xm[Cl2]xnk[NO]ym[Cl2]ynratexratey=k[NO]xm[Cl2]xnk[NO]ym[Cl2]yn
    17.41

    Using the third trial and the first trial, in which [Cl2] does not vary, gives:

    rate 3rate 1=0.006750.00300=k(0.15)m(0.10)nk(0.10)m(0.10)nrate 3rate 1=0.006750.00300=k(0.15)m(0.10)nk(0.10)m(0.10)n
    17.42

    After canceling equivalent terms in the numerator and denominator, we are left with:

    0.006750.00300=(0.15)m(0.10)m0.006750.00300=(0.15)m(0.10)m
    17.43

    which simplifies to:

    2.25=(1.5)m2.25=(1.5)m
    17.44

    We can use natural logs to determine the value of the exponent m:

    ln(2.25)=mln(1.5)ln(2.25)ln(1.5)=m2=mln(2.25)=mln(1.5)ln(2.25)ln(1.5)=m2=m
    17.45

    We can confirm the result easily, since:

    1.52=2.251.52=2.25
    17.46
  2. Step 2.

    Determine the value of n from data in which [Cl2] varies and [NO] is constant.

    rate 2rate 1=0.004500.00300=k(0.10)m(0.15)nk(0.10)m(0.10)nrate 2rate 1=0.004500.00300=k(0.10)m(0.15)nk(0.10)m(0.10)n
    17.47

    Cancelation gives:

    0.00450.0030=(0.15)n(0.10)n0.00450.0030=(0.15)n(0.10)n
    17.48

    which simplifies to:

    1.5=(1.5)n1.5=(1.5)n
    17.49

    Thus n must be 1, and the form of the rate law is:

    Rate=k[NO]m[Cl2]n=k[NO]2[Cl2]Rate=k[NO]m[Cl2]n=k[NO]2[Cl2]
    17.50
  3. Step 3.

    Determine the numerical value of the rate constant k with appropriate units. The units for the rate of a reaction are mol/L/s. The units for k are whatever is needed so that substituting into the rate law expression affords the appropriate units for the rate. In this example, the concentration units are mol3/L3. The units for k should be mol−2 L2/s so that the rate is in terms of mol/L/s.

    To determine the value of k once the rate law expression has been solved, simply plug in values from the first experimental trial and solve for k:

    0.00300molL−1s−1=k(0.10molL−1)2(0.10molL−1)1k=3.0mol−2L2s−10.00300molL−1s−1=k(0.10molL−1)2(0.10molL−1)1k=3.0mol−2L2s−1
    17.51

Check Your Learning

Use the provided initial rate data to derive the rate law for the reaction whose equation is:
OCl−(aq)+I−(aq)⟶OI−(aq)+Cl−(aq)OCl−(aq)+I−(aq)⟶OI−(aq)+Cl−(aq)
17.52
Trial [OCl−] (mol/L) [I−] (mol/L) Initial Rate (mol/L/s)
1 0.0040 0.0020 0.00184
2 0.0020 0.0040 0.00092
3 0.0020 0.0020 0.00046

Determine the rate law expression and the value of the rate constant k with appropriate units for this reaction.

Answer:

rate 2rate 3=0.000920.00046=k(0.0020)x(0.0040)yk(0.0020)x(0.0020)yrate 2rate 3=0.000920.00046=k(0.0020)x(0.0040)yk(0.0020)x(0.0020)y
 2.00 = 2.00y
 y = 1
 rate 1rate 2=0.001840.00092=k(0.0040)x(0.0020)yk(0.0020)x(0.0040)yrate 1rate 2=0.001840.00092=k(0.0040)x(0.0020)yk(0.0020)x(0.0040)y
 2.00=2x2y2.00=2x214.00=2xx=22.00=2x2y2.00=2x214.00=2xx=2
 Substituting the concentration data from trial 1 and solving for k yields:
 rate=k[OCl−]2[I−]10.00184=k(0.0040)2(0.0020)1k=5.75×104mol−2L2s−1rate=k[OCl−]2[I−]10.00184=k(0.0040)2(0.0020)1k=5.75×104mol−2L2s−1

Reaction Order and Rate Constant Units

In some of our examples, the reaction orders in the rate law happen to be the same as the coefficients in the chemical equation for the reaction. This is merely a coincidence and very often not the case.

Rate laws may exhibit fractional orders for some reactants, and negative reaction orders are sometimes observed when an increase in the concentration of one reactant causes a decrease in reaction rate. A few examples illustrating these points are provided:

NO2+CO⟶NO+CO2rate=k[NO2]2CH3CHO⟶CH4+COrate=k[CH3CHO]22N2O5⟶2NO2+O2rate=k[N2O5]2NO2+F2⟶2NO2Frate=k[NO2][F2]2NO2Cl⟶2NO2+Cl2rate=k[NO2Cl]NO2+CO⟶NO+CO2rate=k[NO2]2CH3CHO⟶CH4+COrate=k[CH3CHO]22N2O5⟶2NO2+O2rate=k[N2O5]2NO2+F2⟶2NO2Frate=k[NO2][F2]2NO2Cl⟶2NO2+Cl2rate=k[NO2Cl]
17.53

It is important to note that rate laws are determined by experiment only and are not reliably predicted by reaction stoichiometry.

Reaction orders also play a role in determining the units for the rate constant k. In Example 17.4, a second-order reaction, we found the units for k to be Lmol−1s−1,Lmol−1s−1, whereas in Example 17.5, a third order reaction, we found the units for k to be mol−2 L2/s. More generally speaking, the units for the rate constant for a reaction of order (m+n)(m+n) are mol1−(m+n)L(m+n)−1s−1.mol1−(m+n)L(m+n)−1s−1. Table 17.1 summarizes the rate constant units for common reaction orders.

Rate Constants for Common Reaction Orders
Reaction Order Units of k
(m+n)(m+n) mol1−(m+n)L(m+n)−1s−1mol1−(m+n)L(m+n)−1s−1
zero mol/L/s
first s−1
second L/mol/s
third mol−2 L2 s−1
Table 17.1

Note that the units in the table can also be expressed in terms of molarity (M) instead of mol/L. Also, units of time other than the second (such as minutes, hours, days) may be used, depending on the situation.

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