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Chemistry

12.4 Integrated Rate Laws

Chemistry12.4 Integrated Rate Laws
  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. 2.5 The Periodic Table
    7. 2.6 Molecular and Ionic Compounds
    8. 2.7 Chemical Nomenclature
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  4. 3 Composition of Substances and Solutions
    1. Introduction
    2. 3.1 Formula Mass and the Mole Concept
    3. 3.2 Determining Empirical and Molecular Formulas
    4. 3.3 Molarity
    5. 3.4 Other Units for Solution Concentrations
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  5. 4 Stoichiometry of Chemical Reactions
    1. Introduction
    2. 4.1 Writing and Balancing Chemical Equations
    3. 4.2 Classifying Chemical Reactions
    4. 4.3 Reaction Stoichiometry
    5. 4.4 Reaction Yields
    6. 4.5 Quantitative Chemical Analysis
    7. Key Terms
    8. Key Equations
    9. Summary
    10. Exercises
  6. 5 Thermochemistry
    1. Introduction
    2. 5.1 Energy Basics
    3. 5.2 Calorimetry
    4. 5.3 Enthalpy
    5. Key Terms
    6. Key Equations
    7. Summary
    8. Exercises
  7. 6 Electronic Structure and Periodic Properties of Elements
    1. Introduction
    2. 6.1 Electromagnetic Energy
    3. 6.2 The Bohr Model
    4. 6.3 Development of Quantum Theory
    5. 6.4 Electronic Structure of Atoms (Electron Configurations)
    6. 6.5 Periodic Variations in Element Properties
    7. Key Terms
    8. Key Equations
    9. Summary
    10. Exercises
  8. 7 Chemical Bonding and Molecular Geometry
    1. Introduction
    2. 7.1 Ionic Bonding
    3. 7.2 Covalent Bonding
    4. 7.3 Lewis Symbols and Structures
    5. 7.4 Formal Charges and Resonance
    6. 7.5 Strengths of Ionic and Covalent Bonds
    7. 7.6 Molecular Structure and Polarity
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  9. 8 Advanced Theories of Covalent Bonding
    1. Introduction
    2. 8.1 Valence Bond Theory
    3. 8.2 Hybrid Atomic Orbitals
    4. 8.3 Multiple Bonds
    5. 8.4 Molecular Orbital Theory
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  10. 9 Gases
    1. Introduction
    2. 9.1 Gas Pressure
    3. 9.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law
    4. 9.3 Stoichiometry of Gaseous Substances, Mixtures, and Reactions
    5. 9.4 Effusion and Diffusion of Gases
    6. 9.5 The Kinetic-Molecular Theory
    7. 9.6 Non-Ideal Gas Behavior
    8. Key Terms
    9. Key Equations
    10. Summary
    11. 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 Kinetics
    1. Introduction
    2. 12.1 Chemical Reaction Rates
    3. 12.2 Factors Affecting Reaction Rates
    4. 12.3 Rate Laws
    5. 12.4 Integrated Rate Laws
    6. 12.5 Collision Theory
    7. 12.6 Reaction Mechanisms
    8. 12.7 Catalysis
    9. Key Terms
    10. Key Equations
    11. Summary
    12. 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 Thermodynamics
    1. Introduction
    2. 16.1 Spontaneity
    3. 16.2 Entropy
    4. 16.3 The Second and Third Laws of Thermodynamics
    5. 16.4 Free Energy
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  18. 17 Electrochemistry
    1. Introduction
    2. 17.1 Balancing Oxidation-Reduction Reactions
    3. 17.2 Galvanic Cells
    4. 17.3 Standard Reduction Potentials
    5. 17.4 The Nernst Equation
    6. 17.5 Batteries and Fuel Cells
    7. 17.6 Corrosion
    8. 17.7 Electrolysis
    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 Organic Chemistry
    1. Introduction
    2. 20.1 Hydrocarbons
    3. 20.2 Alcohols and Ethers
    4. 20.3 Aldehydes, Ketones, Carboxylic Acids, and Esters
    5. 20.4 Amines and Amides
    6. Key Terms
    7. Summary
    8. Exercises
  22. 21 Nuclear Chemistry
    1. Introduction
    2. 21.1 Nuclear Structure and Stability
    3. 21.2 Nuclear Equations
    4. 21.3 Radioactive Decay
    5. 21.4 Transmutation and Nuclear Energy
    6. 21.5 Uses of Radioisotopes
    7. 21.6 Biological Effects of Radiation
    8. Key Terms
    9. Key Equations
    10. Summary
    11. 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 an integrated rate law
  • Perform integrated rate law calculations for zero-, first-, and second-order reactions
  • Define half-life and carry out related calculations
  • Identify the order of a reaction from concentration/time data

The rate laws we have seen thus far relate the rate and the concentrations of reactants. We can also determine a second form of each rate law that relates the concentrations of reactants and time. These are called integrated rate laws. We can use an integrated rate law to determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For example, an integrated rate law is used to determine the length of time a radioactive material must be stored for its radioactivity to decay to a safe level.

Using calculus, the differential rate law for a chemical reaction can be integrated with respect to time to give an equation that relates the amount of reactant or product present in a reaction mixture to the elapsed time of the reaction. This process can either be very straightforward or very complex, depending on the complexity of the differential rate law. For purposes of discussion, we will focus on the resulting integrated rate laws for first-, second-, and zero-order reactions.

First-Order Reactions

An equation relating the rate constant k to the initial concentration [A]0 and the concentration [A]t present after any given time t can be derived for a first-order reaction and shown to be:

ln([A]t[A]0)=ktln([A]t[A]0)=kt

or

ln([A]0[A]t)=ktln([A]0[A]t)=kt

or

[A] = [A]0 ekt [A]=[A]0ekt

Example 12.6

The Integrated Rate Law for a First-Order Reaction

The rate constant for the first-order decomposition of cyclobutane, C4H8 at 500 °C is 9.2 ×× 10−3 s−1:
C4H82C2H4C4H82C2H4

How long will it take for 80.0% of a sample of C4H8 to decompose?

Solution

We use the integrated form of the rate law to answer questions regarding time:
ln([A]0[A])=ktln([A]0[A])=kt

There are four variables in the rate law, so if we know three of them, we can determine the fourth. In this case we know [A]0, [A], and k, and need to find t.

The initial concentration of C4H8, [A]0, is not provided, but the provision that 80.0% of the sample has decomposed is enough information to solve this problem. Let x be the initial concentration, in which case the concentration after 80.0% decomposition is 20.0% of x or 0.200x. Rearranging the rate law to isolate t and substituting the provided quantities yields:

t=ln[x][0.200x]×1k=ln0.100mol L−10.020mol L−1×19.2×10−3s−1=1.609×19.2×10−3s−1=1.7×102st=ln[x][0.200x]×1k=ln0.100mol L−10.020mol L−1×19.2×10−3s−1=1.609×19.2×10−3s−1=1.7×102s

Check Your Learning

Iodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation:
I-131Xe-131+electronI-131Xe-131+electron

The decay is first-order with a rate constant of 0.138 d−1. All radioactive decay is first order. How many days will it take for 90% of the iodine−131 in a 0.500 M solution of this substance to decay to Xe-131?

Answer:

16.7 days

We can use integrated rate laws with experimental data that consist of time and concentration information to determine the order and rate constant of a reaction. The integrated rate law can be rearranged to a standard linear equation format:

ln[A]=(k)(t)+ln[A]0y=mx+bln[A]=(k)(t)+ln[A]0y=mx+b

A plot of ln[A] versus t for a first-order reaction is a straight line with a slope of −k and an intercept of ln[A]0. If a set of rate data are plotted in this fashion but do not result in a straight line, the reaction is not first order in A.

Example 12.7

Determination of Reaction Order by Graphing

Show that the data in Figure 12.2 can be represented by a first-order rate law by graphing ln[H2O2] versus time. Determine the rate constant for the rate of decomposition of H2O2 from this data.

Solution

The data from Figure 12.2 with the addition of values of ln[H2O2] are given in Figure 12.10.
A graph is shown with the label “Time ( h )” on the x-axis and “l n [ H subscript 2 O subscript 2 ]” on the y-axis. The x-axis shows markings at 6, 12, 18, and 24 hours. The vertical axis shows markings at negative 3, negative 2, negative 1, and 0. A decreasing linear trend line is drawn through five points represented at the coordinates (0, 0), (6, negative 0.693), (12, negative 1.386), (18, negative 2.079), and (24, negative 2.772).
Figure 12.10 The linear relationship between the ln[H2O2] and time shows that the decomposition of hydrogen peroxide is a first-order reaction.
Trial Time (h) [H2O2] (M) ln[H2O2]
1 0 1.000 0.0
2 6.00 0.500 −0.693
3 12.00 0.250 −1.386
4 18.00 0.125 −2.079
5 24.00 0.0625 −2.772

The plot of ln[H2O2] versus time is linear, thus we have verified that the reaction may be described by a first-order rate law.

The rate constant for a first-order reaction is equal to the negative of the slope of the plot of ln[H2O2] versus time where:

slope=change inychange inx=ΔyΔx=Δln[H2O2]Δtslope=change inychange inx=ΔyΔx=Δln[H2O2]Δt

In order to determine the slope of the line, we need two values of ln[H2O2] at different values of t (one near each end of the line is preferable). For example, the value of ln[H2O2] when t is 6.00 h is −0.693; the value when t = 12.00 h is −1.386:

slope=−1.386(−0.693)12.00 h6.00 h =−0.6936.00 h =−1.155×10−1h−1 k=slope=(−1.155×10−1h−1)=1.155×10−1h−1slope=−1.386(−0.693)12.00 h6.00 h =−0.6936.00 h =−1.155×10−1h−1 k=slope=(−1.155×10−1h−1)=1.155×10−1h−1

Check Your Learning

Graph the following data to determine whether the reaction AB+CAB+C is first order.
Trial Time (s) [A]
1 4.0 0.220
2 8.0 0.144
3 12.0 0.110
4 16.0 0.088
5 20.0 0.074

Answer:

The plot of ln[A] vs. t is not a straight line. The equation is not first order:

A graph, labeled above as “l n [ A ] vs. Time” is shown. The x-axis is labeled, “Time ( s )” and the y-axis is labeled, “l n [ A ].” The x-axis shows markings at 5, 10, 15, 20, and 25 hours. The y-axis shows markings at negative 3, negative 2, negative 1, and 0. A slight curve is drawn connecting five points at coordinates of approximately (4, negative 1.5), (8, negative 2), (12, negative 2.2), (16, negative 2.4), and (20, negative 2.6).

Second-Order Reactions

The equations that relate the concentrations of reactants and the rate constant of second-order reactions are fairly complicated. We will limit ourselves to the simplest second-order reactions, namely, those with rates that are dependent upon just one reactant’s concentration and described by the differential rate law:

Rate=k[A]2Rate=k[A]2

For these second-order reactions, the integrated rate law is:

1[A]=kt+1[A]01[A]=kt+1[A]0

where the terms in the equation have their usual meanings as defined earlier.

Example 12.8

The Integrated Rate Law for a Second-Order Reaction

The reaction of butadiene gas (C4H6) with itself produces C8H12 gas as follows:
2C4H6(g)C8H12(g)2C4H6(g)C8H12(g)

The reaction is second order with a rate constant equal to 5.76 ×× 10−2 L/mol/min under certain conditions. If the initial concentration of butadiene is 0.200 M, what is the concentration remaining after 10.0 min?

Solution

We use the integrated form of the rate law to answer questions regarding time. For a second-order reaction, we have:
1[A]=kt+1[A]01[A]=kt+1[A]0

We know three variables in this equation: [A]0 = 0.200 mol/L, k = 5.76 ×× 10−2 L/mol/min, and t = 10.0 min. Therefore, we can solve for [A], the fourth variable:

1[A]=(5.76×10−2L mol−1min−1)(10min)+10.200mol−11[A]=(5.76×10−1L mol−1)+5.00L mol−11[A]=5.58L mol−1[A]=1.79×10−1mol L−11[A]=(5.76×10−2L mol−1min−1)(10min)+10.200mol−11[A]=(5.76×10−1L mol−1)+5.00L mol−11[A]=5.58L mol−1[A]=1.79×10−1mol L−1

Therefore 0.179 mol/L of butadiene remain at the end of 10.0 min, compared to the 0.200 mol/L that was originally present.

Check Your Learning

If the initial concentration of butadiene is 0.0200 M, what is the concentration remaining after 20.0 min?

Answer:

0.0195 mol/L

The integrated rate law for our second-order reactions has the form of the equation of a straight line:

1[A]=kt+1[A]0y=mx+b1[A]=kt+1[A]0y=mx+b

A plot of 1[A]1[A] versus t for a second-order reaction is a straight line with a slope of k and an intercept of 1[A]0.1[A]0. If the plot is not a straight line, then the reaction is not second order.

Example 12.9

Determination of Reaction Order by Graphing

The data below are for the same reaction described in Example 12.8. Test these data to confirm that this dimerization reaction is second-order.

Solution

Trial Time (s) [C4H6] (M)
1 0 1.00 ×× 10−2
2 1600 5.04 ×× 10−3
3 3200 3.37 ×× 10−3
4 4800 2.53 ×× 10−3
5 6200 2.08 ×× 10−3

In order to distinguish a first-order reaction from a second-order reaction, we plot ln[C4H6] versus t and compare it with a plot of 1[C4H6]1[C4H6] versus t. The values needed for these plots follow.

Time (s) 1[C4H6](M−1)1[C4H6](M−1) ln[C4H6]
0 100 −4.605
1600 198 −5.289
3200 296 −5.692
4800 395 −5.978
6200 481 −6.175

The plots are shown in Figure 12.11. As you can see, the plot of ln[C4H6] versus t is not linear, therefore the reaction is not first order. The plot of 1[C4H6]1[C4H6] versus t is linear, indicating that the reaction is second order.

Two graphs are shown, each with the label “Time ( s )” on the x-axis. The graph on the left is labeled, “l n [ C subscript 4 H subscript 6 ],” on the y-axis. The graph on the right is labeled “1 divided by [ C subscript 4 H subscript 6 ],” on the y-axis. The x-axes for both graphs show markings at 3000 and 6000. The y-axis for the graph on the left shows markings at negative 6, negative 5, and negative 4. A decreasing slightly concave up curve is drawn through five points at coordinates that are (0, negative 4.605), (1600, negative 5.289), (3200, negative 5.692), (4800, negative 5.978), and (6200, negative 6.175). The y-axis for the graph on the right shows markings at 100, 300, and 500. An approximately linear increasing curve is drawn through five points at coordinates that are (0, 100), (1600, 198), (3200, 296), and (4800, 395), and (6200, 481).
Figure 12.11 These two graphs show first- and second-order plots for the dimerization of C4H6. Since the first-order plot (left) is not linear, we know that the reaction is not first order. The linear trend in the second-order plot (right) indicates that the reaction follows second-order kinetics.

Check Your Learning

Does the following data fit a second-order rate law?
Trial Time (s) [A] (M)
1 5 0.952
2 10 0.625
3 15 0.465
4 20 0.370
5 25 0.308
6 35 0.230

Answer:

Yes. The plot of 1[A]1[A] vs. t is linear:

A graph, with the title “1 divided by [ A ] vs. Time” is shown, with the label, “Time ( s ),” on the x-axis. The label “1 divided by [ A ]” appears left of the y-axis. The x-axis shows markings beginning at zero and continuing at intervals of 10 up to and including 40. The y-axis on the left shows markings beginning at 0 and increasing by intervals of 1 up to and including 5. A line with an increasing trend is drawn through six points at approximately (4, 1), (10, 1.5), (15, 2.2), (20, 2.8), (26, 3.4), and (36, 4.4).

Zero-Order Reactions

For zero-order reactions, the differential rate law is:

Rate=k[A]0=kRate=k[A]0=k

A zero-order reaction thus exhibits a constant reaction rate, regardless of the concentration of its reactants.

The integrated rate law for a zero-order reaction also has the form of the equation of a straight line:

[A]=kt+[A]0y=mx+b[A]=kt+[A]0y=mx+b

A plot of [A] versus t for a zero-order reaction is a straight line with a slope of −k and an intercept of [A]0. Figure 12.12 shows a plot of [NH3] versus t for the decomposition of ammonia on a hot tungsten wire and for the decomposition of ammonia on hot quartz (SiO2). The decomposition of NH3 on hot tungsten is zero order; the plot is a straight line. The decomposition of NH3 on hot quartz is not zero order (it is first order). From the slope of the line for the zero-order decomposition, we can determine the rate constant:

slope=k=1.3110−6mol/L/sslope=k=1.3110−6mol/L/s
A graph is shown with the label, “Time ( s ),” on the x-axis and, “[ N H subscript 3 ] M,” on the y-axis. The x-axis shows a single value of 1000 marked near the right end of the axis. The vertical axis shows markings at 1.0 times 10 superscript negative 3, 2.0 times 10 superscript negative 3, and 3.0 times 10 superscript negative 3. A decreasing linear trend line is drawn through six points at the approximate coordinates: (0, 2.8 times 10 superscript negative 3), (200, 2.6 times 10 superscript negative 3), (400, 2.3 times 10 superscript negative 3), (600, 2.0 times 10 superscript negative 3), (800, 1.8 times 10 superscript negative 3), and (1000, 1.6 times 10 superscript negative 3). This line is labeled “Decomposition on W.” A decreasing slightly concave up curve is similarly drawn through eight points at the approximate coordinates: (0, 2.8 times 10 superscript negative 3), (100, 2.5 times 10 superscript negative 3), (200, 2.1 times 10 superscript negative 3), (300, 1.9 times 10 superscript negative 3), (400, 1.6 times 10 superscript negative 3), (500, 1.4 times 10 superscript negative 3), and (750, 1.1 times 10 superscript negative 3), ending at about (1000, 0.7 times 10 superscript negative 3). This curve is labeled “Decomposition on S i O subscript 2.”
Figure 12.12 The decomposition of NH3 on a tungsten (W) surface is a zero-order reaction, whereas on a quartz (SiO2) surface, the reaction is first order.

The Half-Life of a Reaction

The half-life of a reaction (t1/2) is the time required for one-half of a given amount of reactant to be consumed. In each succeeding half-life, half of the remaining concentration of the reactant is consumed. Using the decomposition of hydrogen peroxide (Figure 12.2) as an example, we find that during the first half-life (from 0.00 hours to 6.00 hours), the concentration of H2O2 decreases from 1.000 M to 0.500 M. During the second half-life (from 6.00 hours to 12.00 hours), it decreases from 0.500 M to 0.250 M; during the third half-life, it decreases from 0.250 M to 0.125 M. The concentration of H2O2 decreases by half during each successive period of 6.00 hours. The decomposition of hydrogen peroxide is a first-order reaction, and, as can be shown, the half-life of a first-order reaction is independent of the concentration of the reactant. However, half-lives of reactions with other orders depend on the concentrations of the reactants.

First-Order Reactions

We can derive an equation for determining the half-life of a first-order reaction from the alternate form of the integrated rate law as follows:

ln[A]0[A]=ktt=ln[A]0[A]×1kln[A]0[A]=ktt=ln[A]0[A]×1k

If we set the time t equal to the half-life, t1/2,t1/2, the corresponding concentration of A at this time is equal to one-half of its initial concentration. Hence, when t=t1/2,t=t1/2, [A]=12[A]0.[A]=12[A]0.

Therefore:

t1/2=ln[A]012[A]0×1k=ln2×1k=0.693×1kt1/2=ln[A]012[A]0×1k=ln2×1k=0.693×1k

Thus:

t1/2=0.693kt1/2=0.693k

We can see that the half-life of a first-order reaction is inversely proportional to the rate constant k. A fast reaction (shorter half-life) will have a larger k; a slow reaction (longer half-life) will have a smaller k.

Example 12.10

Calculation of a First-order Rate Constant using Half-Life

Calculate the rate constant for the first-order decomposition of hydrogen peroxide in water at 40 °C, using the data given in Figure 12.13.
A diagram of 5 beakers is shown, each approximately half-filled with colored substances. Beneath each beaker are three rows of text. The first beaker contains a bright green substance and is labeled below as, “1.000 M, 0 s, and ( 0 h ).” The second beaker contains a slightly lighter green substance and is labeled below as, “0.500 M, 2.16 times 10 superscript 4 s, and ( 6 h ).” The third beaker contains an even lighter green substance and is labeled below as, “0.250 M, 4.32 times 10 superscript 4 s, and ( 12 h ).” The fourth beaker contains a green tinted substance and is labeled below as, “0.125 M, 6.48 times 10 superscript 4 s, and ( 18 h ).” The fifth beaker contains a colorless substance and is labeled below as, “0.0625 M, 8.64 times 10 superscript 4 s, and ( 24 h ).”
Figure 12.13 The decomposition of H2O2 (2H2O22H2O+O2)(2H2O22H2O+O2) at 40 °C is illustrated. The intensity of the color symbolizes the concentration of H2O2 at the indicated times; H2O2 is actually colorless.

Solution

The half-life for the decomposition of H2O2 is 2.16 ×× 104 s:
t1/2=0.693kk=0.693t1/2=0.6932.16×104s=3.21×10−5s−1t1/2=0.693kk=0.693t1/2=0.6932.16×104s=3.21×10−5s−1

Check Your Learning

The first-order radioactive decay of iodine-131 exhibits a rate constant of 0.138 d−1. What is the half-life for this decay?

Answer:

5.02 d.

Second-Order Reactions

We can derive the equation for calculating the half-life of a second order as follows:

1[A]=kt+1[A]01[A]=kt+1[A]0

or

1[A]1[A]0=kt1[A]1[A]0=kt

If

t=t1/2t=t1/2

then

[A]=12[A]0[A]=12[A]0

and we can write:

112[A]01[A]0=kt1/22[A]01[A]0=kt1/21[A]0=kt1/2112[A]01[A]0=kt1/22[A]01[A]0=kt1/21[A]0=kt1/2

Thus:

t1/2=1k[A]0t1/2=1k[A]0

For a second-order reaction, t1/2t1/2 is inversely proportional to the concentration of the reactant, and the half-life increases as the reaction proceeds because the concentration of reactant decreases. Consequently, we find the use of the half-life concept to be more complex for second-order reactions than for first-order reactions. Unlike with first-order reactions, the rate constant of a second-order reaction cannot be calculated directly from the half-life unless the initial concentration is known.

Zero-Order Reactions

We can derive an equation for calculating the half-life of a zero order reaction as follows:

[A]=kt+[A]0[A]=kt+[A]0

When half of the initial amount of reactant has been consumed t=t1/2t=t1/2 and [A]=[A]02.[A]=[A]02. Thus:

[A]02=kt1/2+[A]0kt1/2=[A]02[A]02=kt1/2+[A]0kt1/2=[A]02

and

t1/2=[A]02kt1/2=[A]02k

The half-life of a zero-order reaction increases as the initial concentration increases.

Equations for both differential and integrated rate laws and the corresponding half-lives for zero-, first-, and second-order reactions are summarized in Table 12.2.

Summary of Rate Laws for Zero-, First-, and Second-Order Reactions
Zero-Order First-Order Second-Order
rate law rate = k rate = k[A] rate = k[A]2
units of rate constant M s−1 s−1 M−1 s−1
integrated rate law [A] = −kt + [A]0 ln[A] = −kt + ln[A]0 1[A]=kt+(1[A]0)1[A]=kt+(1[A]0)
plot needed for linear fit of rate data [A] vs. t ln[A] vs. t 1[A]1[A] vs. t
relationship between slope of linear plot and rate constant k = −slope k = −slope k = +slope
half-life t1/2=[A]02kt1/2=[A]02k t1/2=0.693kt1/2=0.693k t1/2=1[A]0kt1/2=1[A]0k
Table 12.2
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