Skip to ContentGo to accessibility pageKeyboard shortcuts menu
OpenStax Logo
Chemistry: Atoms First 2e

17.3 Rate Laws

Chemistry: Atoms First 2e17.3 Rate Laws
Menu
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 Ionic and Molecular 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

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 often affected by the concentrations of reactants. Rate laws (sometimes called differential 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. As an example, consider the reaction described by the chemical equation

aA+bB⟶productsaA+bB⟶products

where a and b are stoichiometric coefficients. The rate law for this reaction is written as:

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

in which [A] and [B] 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 and n are the reaction orders and are typically positive integers, though they can be fractions, negative, or zero. The rate constant k and the reaction orders m and n 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 reactant concentrations, but it does vary with temperature.

The reaction orders in a rate law describe the mathematical dependence of the rate on reactant concentrations. Referring to the generic rate law above, the reaction is m order with respect to A and n order with respect to B. For example, if m = 1 and n = 2, the reaction is first order in A and second order in B. The overall reaction order is simply the sum of orders for each reactant. For the example rate law here, the reaction is third order overall (1 + 2 = 3). A few specific examples are shown below to further illustrate this concept.

The rate law:

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

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

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

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

rate=k[H+][OH−]rate=k[H+][OH−]

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)

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

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

Remember that a number raised to the zero power is equal to 1, thus [CO]0 = 1, which is why the CO concentration term may be omitted from the rate law: the rate of reaction is solely dependent on the concentration of NO2. A later chapter section on reaction mechanisms will explain how a reactant’s concentration can have no effect on a reaction rate 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)

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

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

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

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

A common experimental approach to the determination of rate laws is the method of initial rates. This method involves measuring reaction rates for multiple experimental trials carried out using different initial reactant concentrations. Comparing the measured rates for these trials permits determination of the reaction orders and, subsequently, the rate constant, which together are used to formulate a rate law. This approach is illustrated in the next two example exercises.

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 17.8). 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 17.8 A contour map showing stratospheric ozone concentration and the “ozone hole” that occurs over Antarctica during its spring months. (credit: modification of work by NASA)
NO(g)+O3(g)⟶NO2(g)+O2(g)NO(g)+O3(g)⟶NO2(g)+O2(g)

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

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]
  3. Step 3.

    Determine the value of k from one set of concentrations and the corresponding rate. The data from trial 1 are used below:

    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

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)

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)
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

As in Example 17.4, 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, an explicit algebraic approach (vs. the implicit approach of the previous example) will be used 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. 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

    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

    Canceling equivalent terms in the numerator and denominator leaves:

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

    which simplifies to:

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

    Use logarithms 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

    Confirm the result

    1.52=2.251.52=2.25
  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

    Cancelation gives:

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

    which simplifies to:

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

    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]
  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

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)
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⟶NO2+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⟶NO2+O2rate=k[N2O5]2NO2+F2⟶2NO2Frate=k[NO2][F2]2NO2Cl⟶2NO2+Cl2rate=k[NO2Cl]

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

The units for a rate constant will vary as appropriate to accommodate the overall order of the reaction. The unit of the rate constant for the second-order reaction described in Example 17.4 was determined to be Lmol−1s−1.Lmol−1s−1. For the third-order reaction described in Example 17.5, the unit for k was derived to be L2mol−2s−1.L2mol−2s−1. Dimensional analysis requires the rate constant unit for a reaction whose overall order is x to be Lx−1mol1−xs−1.Lx−1mol1−xs−1. Table 17.1 summarizes the rate constant units for common reaction orders.

Rate Constant Units for Common Reaction Orders
Overall Reaction Order (x) Rate Constant Unit (Lx−1 mol1−x s−1)
0 (zero) mol L−1 s−1
1 (first) s−1
2 (second) L mol−1 s−1
3 (third) L2 mol−2 s−1
Table 17.1

Note that the units in this table were derived using specific units for concentration (mol/L) and time (s), though any valid units for these two properties may be used.

PreviousNext
Order a print copy

As an Amazon Associate we earn from qualifying purchases.

Citation/Attribution

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/chemistry-atoms-first-2e/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/chemistry-atoms-first-2e/pages/1-introduction
Citation information

© Dec 15, 2022 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.