Chapter 18: Kinetics

18.3 Rate Laws

Learning Outcomes

  • Determine the rate law and overall order for a chemical reaction using initial rate data
  • Compare and contrast the effect of concentration or pressure on the rates of three common orders of reactions (zero-, first-, and second-order)

As described in the previous section, the rate of a reaction is affected by the concentrations of reactants. Rate law 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

[latex]aA+bB\rightarrow\text{ products}[/latex]

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


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:


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


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


describes a reaction that is first order in [latex]\ce{H+}[/latex], first order in [latex]\ce{OH-}[/latex], and second order overall.

Example 18.3.1: Writing Rate Laws from Reaction Orders

An experiment shows that the reaction of nitrogen dioxide with carbon monoxide:


is second order in [latex]\ce{NO2}[/latex] and zero order in [latex]\ce{CO}[/latex] at 100 °C. What is the rate law for the reaction?

Show Solution

The reaction will have the form:


The reaction is second order in [latex]\ce{NO2}[/latex]; thus m = 2. The reaction is zero order in [latex]\ce{CO}[/latex]; thus n = 0. The rate law is:


Remember that a number raised to the zero power is equal to 1, thus [latex]\ce{[CO]^0}[/latex] = 1, which is why we can simply drop the concentration of [latex]\ce{CO}[/latex] from the rate equation: the rate of reaction is solely dependent on the concentration of [latex]\ce{NO2}[/latex]. 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

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 18.3.2: 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.

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 18.3.1. 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: NASA)

One such reaction is the combination of nitric oxide, [latex]\ce{NO}[/latex], with ozone, [latex]\ce{O3}[/latex]:


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

Trial [latex]\ce{[NO]}[/latex] (mol/L) [latex]\ce{[O3]}[/latex] (mol/L) [latex]\dfrac{\Delta \ce{[NO2]}}{\Delta t}[/latex] 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.

Show Solution

The rate law will have the form:


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

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

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

[latex]\text{rate}=k\ce{[NO]^1[O3]^1} = k\ce{[NO][O3]}[/latex]

Step 3. Determine the value of k from one set of concentrations and the corresponding rate. The data from trial 1 are used below:

[latex]\begin{array}{cc}\hfill k& =\dfrac{\text{rate}}{\left[\ce{NO}\right]\left[{\ce{O}}_{3}\right]}\hfill \\ & =\dfrac{0.660\times {10}^{-4}\cancel{{\text{mol L}}^{-1}}{\text{s}}^{-1}}{\left(1.00\times {10}^{-6}\cancel{{\text{mol L}}^{-1}}\right)\left(3.00\times {10}^{-6}\text{mol}{\text{L}}^{-1}\right)}\hfill \\ & =\text{2}.\text{2}0\times {10}^{7}\text{ L}{\text{ mol}}^{-1}{\text{s}}^{-1}\hfill \end{array}[/latex]

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

Check Your Learning

Example 18.3.3: 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:


Trial [latex]\ce{[NO]}[/latex] (mol/L) [latex]\ce{[Cl2]}[/latex] (mol/L) [latex]-\dfrac{\Delta\left[\text{NO}\right]}{\Delta t}\left(\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}\right)[/latex]
1 0.10 0.10 0.00300
2 0.10 0.15 0.00450
3 0.15 0.10 0.00675
Show Solution

The rate law for this reaction will have the form:


As in Example 18.3.2, 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:

Step 1. Determine the value of m from the data in which [latex]\ce{[NO]}[/latex] varies and [latex]\ce{[Cl2]}[/latex] is constant: We can write the ratios with the subscripts x and y to indicate data from two different trials:


Using the third trial and the first trial, in which [latex]\ce{[Cl2]}[/latex] does not vary, gives:

[latex]\dfrac{\text{rate 3}}{\text{rate 1}}=\dfrac{0.00675}{0.00300}=\dfrac{k{\left(0.15\right)}^{m}{\left(0.10\right)}^{n}}{k{\left(0.10\right)}^{m}{\left(0.10\right)}^{n}}[/latex]

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


which simplifies to:


Use logarithms to determine the value of the exponent m:

[latex]\begin{array}{ccc}\hfill \text{ln}\left(2.25\right)& =& m\text{ln}\left(1.5\right)\hfill \\ \hfill \dfrac{\text{ln}\left(2.25\right)}{\text{ln}\left(1.5\right)}& =& m\hfill \\ \hfill 2& =& m\hfill \end{array}[/latex]

Confirm the result:



Step 2. Determine the value of n from data in which [latex]\ce{[Cl2]}[/latex] varies and [latex]\ce{[NO]}[/latex] is constant:

[latex]\dfrac{\text{rate 2}}{\text{rate 1}}=\dfrac{0.00450}{0.00300}=\dfrac{k{\left(0.10\right)}^{m}{\left(0.15\right)}^{n}}{k{\left(0.10\right)}^{m}{\left(0.10\right)}^{n}}[/latex]

Cancelation gives:


which simplifies to:

[latex]1.5 = (1.5)^{n}[/latex]

Thus [latex]n[/latex] must be 1, and the form of the rate law is:



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:

[latex]\begin{array}{ccc}\hfill 0.00300\text{mol}{\text{ L}}^{-1}{\text{s}}^{-1}& =& k{\left(0.10\text{mol}{\text{L}}^{-1}\right)}^{2}{\left(0.10\text{mol}{\text{ L}}^{-1}\right)}^{1}\hfill \\ \hfill k& =& 3.0{\text{mol}}^{-2}{\text{L}}^{2}{\text{s}}^{-1}\hfill \end{array}[/latex]

Check Your Learning

Use the provided initial rate data to derive the rate law for the reaction whose equation is:


Trial [latex]\ce{[OCl-]}[/latex] (mol/L) [latex]\ce{[I-]}[/latex] (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

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:

[latex]\begin{array}{l}\\ {\ce{NO}}_{2}+\ce{CO}\rightarrow\ce{NO}+{\ce{CO}}_{\text{2}}\qquad\text{rate}=\text{k}{\left[{\ce{NO}}_{2}\right]}^{2}\\ {\ce{CH}}_{3}\ce{CHO}\rightarrow{\ce{CH}}_{4}+\ce{CO}\qquad\text{rate}=\text{k}{\left[{\ce{CH}}_{3}\ce{CHO}\right]}^{2}\\ {\ce{2N}}_{2}{\ce{O}}_{5}\rightarrow{\ce{2NO}}_{2}+{\ce{O}}_{\text{2}}\qquad\text{rate}=\text{k}\left[{\ce{N}}_{2}{\ce{O}}_{5}\right]\\ {\ce{2NO}}_{2}+{\ce{F}}_{2}\rightarrow{\ce{2NO}}_{2}\ce{F}\qquad\text{rate}=\text{k}\left[{\ce{NO}}_{2}\right]\left[{\ce{F}}_{2}\right]\\ {\ce{2NO}}_{2}\ce{Cl}\rightarrow{\ce{2NO}}_{2}+{\ce{Cl}}_{2}\qquad\text{rate}=\text{k}\left[{\ce{NO}}_{2}\ce{Cl}\right]\end{array}[/latex]

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 18.3.2 was determined to be [latex]\text{L}{\text{mol}}^{-1}{\text{s}}^{-1}[/latex]. For the third-order reaction described in Example 18.3.3, the unit for [latex]k[/latex] was derived to be [latex]\text{L}^{2}\text{mol}^{-2}\text{s}^{-1}[/latex]. More generally speaking, dimensional analysis requires the units for the rate constant for a reaction whose overall order is [latex]x[/latex] to be ([latex]\text{L}^{x-1}\text{mol}^{1-x}\text{s}^{-1}[/latex]). Table 18.3.1 summarizes the rate constant units for common reaction orders.

Table 18.3.1. Rate Constants for Common Reaction Orders
Overall Reaction Order ([latex]x[/latex] Rate Constant Unit [latex]\text{L}^{x-1}\text{mol}^{1-x}\text{s}^{-1}[/latex]
0 (zero) [latex]\text{mol}\text{ L}^{-1}\text{s}^{-1}[/latex]
1 (first) [latex]\text{s}^{-1}[/latex]
2 (second) [latex]\text{L}{\text{ mol}}^{-1}{\text{s}}^{-1},[/latex]
3 (third) [latex]\text{L}^{2}\text{ mol}^{-2}\text{s}^{-1}[/latex]

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.


Key Concepts and Summary

Rate laws provide a mathematical description of how changes in the amount of a substance affect the rate of a chemical reaction. Rate laws are determined experimentally and cannot be predicted by reaction stoichiometry. The order of reaction describes how much a change in the amount of each substance affects the overall rate, and the overall order of a reaction is the sum of the orders for each substance present in the reaction. Reaction orders are typically first order, second order, or zero order, but fractional and even negative orders are possible.

Try It

  1. Doubling the concentration of a reactant increases the rate of a reaction four times. With this knowledge, answer the following questions:
    1. What is the order of the reaction with respect to that reactant?
    2. Tripling the concentration of a different reactant increases the rate of a reaction three times. What is the order of the reaction with respect to that reactant?
  2. How much and in what direction will each of the following affect the rate of the reaction: [latex]\ce{CO}(g)+\ce{NO2}(g)\rightarrow\ce{CO2}(g)+\ce{NO}(g)[/latex] if the rate law for the reaction is [latex]\text{rate}=k{\ce{[NO2]^2}}?[/latex]
    1. Decreasing the pressure of [latex]\ce{NO2}[/latex] from 0.50 atm to 0.250 atm.
    2. Increasing the concentration of [latex]\ce{CO}[/latex] from 0.01 M to 0.03 M.
Show Selected Solutions
    1. Since the concentration of the reactant doubled and the rate quadrupled, we can conclude that the order with respect to the reactant is 2, since 22 = 4.
      1. [latex]\begin{array}{l}\text{rate}=k{\left[\text{reactant}\right]}^{m}\\ \text{4}\left(\text{rate}\right)=k{\left[\text{2}\left(\text{reactant}\right)\right]}^{2}\end{array}[/latex]
    2. Since the concentration of the reactant and the rate both tripled, we can conclude that [latex]m=1,[/latex] and the order with respect to this reactant is 1.
    3. [latex]\begin{array}{l}\text{rate}=k{\left[\text{reactant}\right]}^{m}\\ \text{3}\left(\text{rate}\right)=k{\left[\text{3}\left(\text{reactant}\right)\right]}^{1}\end{array}[/latex]
    1. [latex]\frac{{\text{rate}}_{2}}{{\text{rate}}_{1}}=\frac{\text{k}{\text{[}0.25{\ce{NO}}_{2}\text{]}}^{2}}{\text{k}{\text{[}0.50{\ce{NO}}_{2}\text{]}}^{2}}=\frac{0.0675}{0.25}=\frac{1}{4}[/latex]
    2. Since Rate1 is four times as large as Rate2, the process reduces the rate by a factor of 4.
    3. Since [latex]\ce{CO}[/latex] does not appear in the rate law, the rate is not affected.


method of initial rates: use of a more explicit algebraic method to determine the orders in a rate law

overall reaction order: sum of the reaction orders for each substance represented in the rate law

rate constant (k): proportionality constant in the relationship between reaction rate and concentrations of reactants

rate law (also, rate equation): mathematical equation showing the dependence of reaction rate on the rate constant and the concentration of one or more reactants

reaction order: value of an exponent in a rate law, expressed as an ordinal number (for example, zero order for 0, first order for 1, second order for 2, and so on)

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