Skip to main content
Chemistry LibreTexts

7.4 Kinetics

  • Page ID
    17142
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Rate Laws

    Typically, reaction rates decrease with time because reactant concentrations decrease as reactants are converted to products. You also learned that reaction rates generally increase when reactant concentrations are increased. We now examine the mathematical expressions called rate laws, which describe the relationships between reactant rates and reactant concentrations. Rate laws are laws that are mathematical descriptions of experimentally verifiable data.

    Rate laws may be written from either of two different but related perspectives. A differential rate law expresses the reaction rate in terms of changes in the concentration of one or more reactants (Δ[R]) over a specific time interval (Δt). In contrast, an integrated rate law describes the reaction rate in terms of the initial concentration ([R]0) and the measured concentration of one or more reactants ([R]) after a given amount of time (t); we will discuss integrated rate laws in Section 14.3. The integrated rate law can be found by using calculus to integrate the differential rate law, although the method of doing so is beyond the scope of this text. Whether you use a differential rate law or integrated rate law, always make sure that the rate law gives the proper units for the reaction rate, usually moles per liter per second (M/s).

    Reaction Orders

    For a reaction with the general equation

    Equation 14.8

    aA + bB → cC + dD

    the experimentally determined rate law usually has the following form:

    Equation 14.9

    rate = k[A]m[B]n

    The proportionality constant (k) is called the rate constant, and its value is characteristic of the reaction and the reaction conditions. A given reaction has a particular value of the rate constant under a given set of conditions, such as temperature, pressure, and solvent; varying the temperature or the solvent usually changes the value of the rate constant. The numerical value of k, however, does not change as the reaction progresses under a given set of conditions.

    Thus the reaction rate depends on the rate constant for the given set of reaction conditions and the concentration of A and B raised to the powers m and n, respectively. The values of m and n are derived from experimental measurements of the changes in reactant concentrations over time and indicate the reaction order, the degree to which the reaction rate depends on the concentration of each reactant; m and n need not be integers. For example, Equation 14.9 tells us that Equation 14.8 is mth order in reactant A and nth order in reactant B. It is important to remember that n and m are not related to the stoichiometric coefficients a and b in the balanced chemical equation and must be determined experimentally. The overall reaction order is the sum of all the exponents in the rate law: m + n.

    Note the Pattern

    Under a given set of conditions, the value of the rate constant does not change as the reaction progresses.

    Although differential rate laws are generally used to describe what is occurring on a molecular level during a reaction, integrated rate laws are used to determine the reaction order and the value of the rate constant from experimental measurements. (We present general forms for integrated rate laws) To illustrate how chemists interpret a differential rate law, we turn to the experimentally derived rate law for the hydrolysis of t-butyl bromide in 70% aqueous acetone. This reaction producest-butanol according to the following equation:

    \[(CH_3)_3CBr_{(soln)} + H_2O_{(soln)} \rightarrow (CH_3)_3COH_{(soln)} + HBr_{(soln)} \tag{14.10}\]

    Combining the rate expression in Equation 14.4 and Equation 14.9 gives us a general expression for the differential rate law:

    \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A]^m[\textrm B]^n \tag{14.11}\]

    Inserting the identities of the reactants into Equation 14.11 gives the following expression for the differential rate law for the reaction:

    \[\textrm{rate}=-\dfrac{\Delta[\mathrm{(CH_3)_3CBr}]}{\Delta t}=k[\mathrm{(CH_3)_3CBr}]^m[\mathrm{H_2O}]^n \tag{14.12}\]

    Experiments done to determine the rate law for the hydrolysis of t-butyl bromide show that the reaction rate is directly proportional to the concentration of \((CH_3)_3CBr\) but is independent of the concentration of water. Thus \(m\) and \(n\) in Equation 14.12 are 1 and 0, respectively, and

    \[\text{rate} = k[(CH_3)_3CBr]^1[H_2O]^0 = k[(CH_3)_3CBr] \tag{14.13}\]

    Because the exponent for the reactant is 1, the reaction is first order in (CH3)3CBr. It is zeroth order in water because the exponent for [H2O] is 0. (Recall that anything raised to the zeroth power equals 1.) Thus the overall reaction order is 1 + 0 = 1. What the reaction orders tell us in practical terms is that doubling the concentration of (CH3)3CBr doubles the reaction rate of the hydrolysis reaction, halving the concentration of (CH3)3CBr halves the reaction rate, and so on. Conversely, increasing or decreasing the concentration of water has no effect on the reaction rate. (Again, when you work with rate laws, there is no simple correlation between the stoichiometry of the reaction and the rate law. The values of k, m, and n in the rate law must be determined experimentally.) Experimental data show that k has the value 5.15 × 10−4 s−1 at 25°C. The rate constant has units of reciprocal seconds (s−1) because the reaction rate is defined in units of concentration per unit time (M/s). The units of a rate constant depend on the rate law for a particular reaction.

    Under conditions identical to those for the t-butyl bromide reaction, the experimentally derived differential rate law for the hydrolysis of methyl bromide (CH3Br) is as follows:

    \[\textrm{rate}=-\dfrac{\Delta[\mathrm{CH_3Br}]}{\Delta t}=k'[\mathrm{CH_3Br}] \tag{14.14}\]

    This reaction also has an overall reaction order of 1, but the rate constant in Equation 14.14 is approximately 106 times smaller than that for t-butyl bromide. Thus methyl bromide hydrolyzes about 1 million times more slowly than t-butyl bromide, and this information tells chemists how the reactions differ on a molecular level.

    Frequently, changes in reaction conditions also produce changes in a rate law. In fact, chemists often change reaction conditions to obtain clues about what is occurring during a reaction. For example, when t-butyl bromide is hydrolyzed in an aqueous acetone solution containing OH ions rather than in aqueous acetone alone, the differential rate law for the hydrolysis reaction does not change. For methyl bromide, in contrast, the differential rate law becomes rate =k″[CH3Br][OH], with an overall reaction order of 2. Although the two reactions proceed similarly in neutral solution, they proceed very differently in the presence of a base, which again provides clues as to how the reactions differ on a molecular level.

    Note the Pattern

    Differential rate laws are generally used to describe what is occurring on a molecular level during a reaction, whereas integrated rate laws are used for determining the reaction order and the value of the rate constant from experimental measurements.

    Example 3

    We present three reactions and their experimentally determined differential rate laws. For each reaction, give the units of the rate constant, give the reaction order with respect to each reactant, give the overall reaction order, and predict what happens to the reaction rate when the concentration of the first species in each chemical equation is doubled.

    1. \(\mathrm{2HI(g)}\xrightarrow{\textrm{Pt}}\mathrm{H_2(g)}+\mathrm{I_2(g)}
      \\ \textrm{rate}=-\frac{1}{2}\left (\frac{\Delta[\mathrm{HI}]}{\Delta t} \right )=k[\textrm{HI}]^2\)
    2. \(\mathrm{2N_2O(g)}\xrightarrow{\Delta}\mathrm{2N_2(g)}+\mathrm{O_2(g)}
      \\ \textrm{rate}=-\frac{1}{2}\left (\frac{\Delta[\mathrm{N_2O}]}{\Delta t} \right )=k\)
    3. \(\mathrm{cyclopropane(g)}\rightarrow\mathrm{propane(g)}
      \\ \textrm{rate}=-\frac{\Delta[\mathrm{cyclopropane}]}{\Delta t}=k[\mathrm{cyclopropane}]\)

    Given: balanced chemical equations and differential rate laws

    Asked for: units of rate constant, reaction orders, and effect of doubling reactant concentration

    Strategy:

    A Express the reaction rate as moles per liter per second [mol/(L·s), or M/s]. Then determine the units of each chemical species in the rate law. Divide the units for the reaction rate by the units for all species in the rate law to obtain the units for the rate constant.

    B Identify the exponent of each species in the rate law to determine the reaction order with respect to that species. Sum all exponents to obtain the overall reaction order.

    C Use the mathematical relationships as expressed in the rate law to determine the effect of doubling the concentration of a single species on the reaction rate.

    Solution:

    1. A [HI]2 will give units of (moles per liter)2. For the reaction rate to have units of moles per liter per second, the rate constant must have reciprocal units [1/(M·s)]:
    \(k\textrm M^2=\dfrac{\textrm M}{\textrm s}k=\dfrac{\textrm{M/s}}{\textrm M^2}=\dfrac{1}{\mathrm{M\cdot s}}=\mathrm{M^{-1}\cdot s^{-1}}\)

    B The exponent in the rate law is 2, so the reaction is second order in HI. Because HI is the only reactant and the only species that appears in the rate law, the reaction is also second order overall.

    C If the concentration of HI is doubled, the reaction rate will increase from k[HI]02 to k(2[HI])02 = 4k[HI]02. The reaction rate will therefore quadruple.

    1. A Because no concentration term appears in the rate law, the rate constant must have M/s units for the reaction rate to have M/s units.

    B The rate law tells us that the reaction rate is constant and independent of the N2O concentration. That is, the reaction is zeroth order in N2O and zeroth order overall.

    C Because the reaction rate is independent of the N2O concentration, doubling the concentration will have no effect on the reaction rate.

    1. A The rate law contains only one concentration term raised to the first power. Hence the rate constant must have units of reciprocal seconds (s−1) to have units of moles per liter per second for the reaction rate: M·s−1 = M/s.

    B The only concentration in the rate law is that of cyclopropane, and its exponent is 1. This means that the reaction is first order in cyclopropane. Cyclopropane is the only species that appears in the rate law, so the reaction is also first order overall.

    C Doubling the initial cyclopropane concentration will increase the reaction rate from k[cyclopropane]0 to 2k[cyclopropane]0. This doubles the reaction rate.

    Exercise

    Given the following two reactions and their experimentally determined differential rate laws: determine the units of the rate constant if time is in seconds, determine the reaction order with respect to each reactant, give the overall reaction order, and predict what will happen to the reaction rate when the concentration of the first species in each equation is doubled.

    1. \(\begin{align}\textrm{CH}_3\textrm N\textrm{=NCH}_3\textrm{(g)}\rightarrow\mathrm{C_2H_6(g)}+\mathrm{N_2(g)}\hspace{5mm}&\textrm{rate}=-\frac{\Delta[\textrm{CH}_3\textrm N\textrm{=NCH}_3]}{\Delta t}
      \\&=k[\textrm{CH}_3\textrm N\textrm{=NCH}_3]\end{align}\)
    2. \(\begin{align}\mathrm{2NO_2(g)}+\mathrm{F_2(g)}\rightarrow\mathrm{2NO_2F(g)}\hspace{5mm}&\textrm{rate}=-\frac{\Delta[\mathrm{F_2}]}{\Delta t}=-\frac{1}{2}\left ( \frac{\Delta[\mathrm{NO_2}]}{\Delta t} \right )
      \\&=k[\mathrm{NO_2}][\mathrm{F_2}]\end{align}\)

    Answer:

    1. s−1; first order in CH3N=NCH3; first order overall; doubling [CH3N=NCH3] will double the reaction rate.
    2. M−1·s−1; first order in NO2, first order in F2; second order overall; doubling [NO2] will double the reaction rate.

    Summary

    Reaction rates are reported either as the average rate over a period of time or as the instantaneous rate at a single time.

    The rate law for a reaction is a mathematical relationship between the reaction rate and the concentrations of species in solution. Rate laws can be expressed either as a differential rate law, describing the change in reactant or product concentrations as a function of time, or as an integrated rate law, describing the actual concentrations of reactants or products as a function of time.

    The rate constant (k) of a rate law is a constant of proportionality between the reaction rate and the reactant concentration. The power to which a concentration is raised in a rate law indicates the reaction order, the degree to which the reaction rate depends on the concentration of a particular reactant.

    Key Equations

    general definition of rate for A B

    Equation 14.4: \(\textrm{rate}=\frac{\Delta [\textrm B]}{\Delta t}=-\frac{\Delta [\textrm A]}{\Delta t}\)

    general form of rate law when A and B are reactants

    Equation 14.9: rate = k[A]m[B]n

    Conceptual Problems

    1. Explain why the reaction rate is generally fastest at early time intervals. For the second-order A + B → C, what would the plot of the concentration of C versus time look like during the course of the reaction?
    2. Explain the differences between a differential rate law and an integrated rate law. What two components do they have in common? Which form is preferred for obtaining a reaction order and a rate constant? Why?
    3. Diffusion-controlled reactions have rates that are determined only by the reaction rate at which two reactant molecules can diffuse together. These reactions are rapid, with second-order rate constants typically on the order of 1010 L/(mol·s). Would you expect the reactions to be faster or slower in solvents that have a low viscosity? Why? Consider the reactions H3O+ + OH → 2H2O and H3O+ + N(CH3)3 → H2O + HN(CH3)3+ in aqueous solution. Which would have the higher rate constant? Why?
    4. What information can you get from the reaction order? What correlation does the reaction order have with the stoichiometry of the overall equation?
    5. During the hydrolysis reaction A + H2O → B + C, the concentration of A decreases much more rapidly in a polar solvent than in a nonpolar solvent. How do you expect this effect to be reflected in the overall reaction order?

    Answers

    1. Reactant concentrations are highest at the beginning of a reaction. The plot of [C] versus t is a curve with a slope that becomes steadily less positive.
    1. Faster in a less viscous solvent because the rate of diffusion is higher; the H3O+/OH reaction is faster due to the decreased relative size of reactants and the higher electrostatic attraction between the reactants.

    Numerical Problems

    1. The reaction rate of a particular reaction in which A and B react to make C is as follows:
    \(\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=\dfrac{1}{2}\left ( \dfrac{\Delta[\textrm C]}{\Delta t} \right )\)

    Write a reaction equation that is consistent with this rate law. What is the rate expression with respect to time if 2A are converted to 3C?

    1. While commuting to work, a person drove for 12 min at 35 mph, then stopped at an intersection for 2 min, continued the commute at 50 mph for 28 min, drove slowly through traffic at 38 mph for 18 min, and then spent 1 min pulling into a parking space at 3 mph. What was the average rate of the commute? What was the instantaneous rate at 13 min? at 28 min?
    1. Why do most studies of chemical reactions use the initial rates of reaction to generate a rate law? How is this initial rate determined? Given the following data, what is the reaction order? Estimate.
    Time (s) [A] (M)
    120 0.158
    240 0.089
    360 0.062
    1. Predict how the reaction rate will be affected by doubling the concentration of the first species in each equation.
    1. C2H5I → C2H4 + HI: rate = k[C2H5I]
    2. SO + O2 → SO2 + O: rate = k[SO][O2]
    3. 2CH3 → C2H6: rate = k[CH3]2
    4. ClOO → Cl + O2: rate = k
    1. Cleavage of C2H6 to produce two CH3· radicals is a gas-phase reaction that occurs at 700°C. This reaction is first order, with k = 5.46 × 10−4 s−1. How long will it take for the reaction to go to 15% completion? to 50% completion?
    1. Three chemical processes occur at an altitude of approximately 100 km in Earth’s atmosphere.
    \(\mathrm{N_2^+}+\mathrm{O_2}\xrightarrow{k_1}\mathrm{N_2}+\mathrm{O_2^+}\)
    \(\mathrm{O_2^+}+\mathrm{O}\xrightarrow{k_2}\mathrm{O_2}+\mathrm{O^+}\)
    \(\mathrm{O^+}+\mathrm{N_2}\xrightarrow{k_3}\mathrm{NO^+}+\mathrm{N}\)

    Write a rate law for each elementary reaction. If the rate law for the overall reaction were found to be rate = k[N2+][O2], which one of the steps is rate limiting?

    1. The oxidation of aqueous iodide by arsenic acid to give I3 and arsenous acid proceeds via the following reaction:
    \(\mathrm{H_3AsO_4(aq)}+\mathrm{3I^-(aq)}+\mathrm{2H^+(aq)}\overset{k_{\textrm f}}{\underset{k_{\textrm r}}{\rightleftharpoons}}\mathrm{H_3AsO_3(aq)}+\mathrm{I_3^-(aq)}+\mathrm{H_2O(l)}\)

    Write an expression for the initial rate of decrease of [I3], Δ[I3]/Δt. When the reaction rate of the forward reaction is equal to that of the reverse reaction: kf/kr = [H3AsO3][I3]/[H3AsO4][I]3[H+]2. Based on this information, what can you say about the nature of the rate-determining steps for the reverse and the forward reactions?

    Answer

    1. 298 s; 1270 s

    Contributors

    • Anonymous

    Transition State Theory

    According to TST, between the state where molecules are reactants and the state where molecules are products, there is a state known as the transition state. In the transition state, the reactants are combined in a species called the activated complex. The theory suggests that there are three major factors that determine whether a reaction will occur:

    1. The concentration of the activated complex
    2. The rate at which the activated complex breaks apart
    3. The way in which the activated complex breaks apart: whether it breaks apart to reform the reactants or whether it breaks apart to form a new complex, the products.

    Collision theory proposes that not all reactants that combine undergo a reaction. However, assuming the stipulations of the collision theory are met and a successful collision occurs between the molecules, transition state theory allows one of two outcomes: a return to the reactants, or a rearranging of bonds to form the products.

    Template:ExampleStart

    The example reaction shown above involves the reactants, hydroxide and bromomethane, forming the products, methanol and bromide. The first part of the image shows the reactants. The second part of the image shows the transition state, in which the activated complex is formed, with rearranged molecules and different bonds. The dotted lines represent the transitory bonds that form during this stage. The activated complex is clearly different than the reactants or the products. A successful collision and reaction have clearly occurred because in the third part of the image products (methanol and bromide) are formed that are different than the reactants (hydroxide and bromomethane).

    Reaction Profiles

    Using a reaction profile, the energy necessary to complete a reaction can be determined by plotting energy values on the y-axis of a Cartesian plane. On the x-axis, the reaction progress is plotted. A graph such as the one below is the result:

    This graph illustrates where the molecules exist as reactants, products, or in the transition state. The relationship between potential energy and reaction rate is clear. Also illustrated is the amount of energy required to initiate a reaction—the activation energy (Ea). A reaction profile can also be used to find the enthalpy (ΔH) of the reaction, by subtracting the energy of the products from the energy of the reactants. This example reaction is exothermic. The Gibbs energy change of the reaction (ΔG) is equal to the activation energy of the forward reaction. The reaction profile for a two step reaction, shown below, displays all the activated complexes that occur during the reaction , as well as any intermediates formed.

    Applications

    The main application of the transition state theory is in reactions catalyzed by enzymes. The introduction of an enzyme catalyst into a reaction lowers the activation energy of that reaction. Research detailing enzymes' specific interactions with molecules, including those in the transition state, has shown that the enzymes actually increase the stability of the activated complex in the transition state. This is important in terms of enzyme activity such as the induced fit model and enzymes used in inhibitors. In more general terms, however, the transition state theory is useful in determining a reaction will proceed. By knowing which transition states will form, and what differing activation energies are, it is possible to map out the course of a reaction, and calculate quantities such as the reaction rate and the rate constant.

    Concept Questions

    1. Which two values in the Arrhenius equation are explained by transition state theory?
    2. Assuming a successful collision occurs, what two outcomes are possible for an activated complex?
    3. What is a reaction profile? What two values does it compare? What values can it be used to determine?
    4. Label the axes, reactants, products, transition state, Ea forward, Ea reverse, and ΔH for the blank reaction profile below.

    5. For a hypothetical exothermic reaction, the Ea reverse=600kJ and the Ea forward=200kJ. What is the value of ΔH?

    6. In the reaction depicted below, how many activated complexes are present? How many intermediates? Is the first step endothermic or exothermic? What about the reaction as a whole?

    Answers

    1. The pre-exponential factor (A) and the activation energy (Ea).
    2. The activated complex can either reform the reactants, or form products.
    3. A reaction profile is a graph that compares the potential energy of a reaction and the progression of the reaction. The activation energy (Ea) and enthalpy (ΔH) can be determined from it.

    4.

    5. Ea reverse=Ea forward + |ΔH|. Thus, 600kJ=200kJ + |ΔH|. |ΔH|=400kJ. The reaction is exothermic, thus ΔH must be negative. The answer is -400kJ.

    6. There are two activated complexes, and one intermediate. The first step is endothermic; however, the overall reaction is exothermic.

    References

    1. Gross, Dixie J., and Ralph H. Petrucci. General Chemistry Principles and Modern Applications, Ninth Editon : Study Guide. Upper Saddle River: Prentice Hall PTR, 2006.
    2. Pauling, Linus. General Chemistry. Minneapolis: Dover Publications, Incorporated, 1989.
    3. Petrucci, Ralph H., William S. Harwood, and Geoff E. Herring. General Chemistry : Principles and Modern Applications. Upper Saddle River: Prentice Hall PTR, 2006.
    4. Barker, Brett. Peterson's Master AP Chemistry. Lawrenceville, NJ: Peterson's, a Nelnet company, 2007.

    Further Reading


    7.4 Kinetics is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?