Rate Laws
Rate Laws
As described in the previous module, the rate of a reaction is affected by the concentrations of reactants. Rate laws or rate equations are mathematical expressions that describe the relationship between the rate of a chemical reaction and the concentration of its reactants. In general, a rate law (or differential rate law, as it is sometimes called) takes this form:
in which [A], [B], and [C] represent the molar concentrations of reactants, and k is the rate constant, which is specific for a particular reaction at a particular temperature. The exponents m, n, and p are usually positive integers (although it is possible for them to be fractions or negative numbers). The rate constant k and the exponents m, n, and p must be determined experimentally by observing how the rate of a reaction changes as the concentrations of the reactants are changed. The rate constant k is independent of the concentration of A, B, or C, but it does vary with temperature and surface area.
The exponents in a rate law describe the effects of the reactant concentrations on the reaction rate and define the reaction order. Consider a reaction for which the rate law is:
\(\text{rate}=k\left[A\right]^{m}\left[B\right]^{n}\)
If the exponent m is 1, the reaction is first order with respect to A. If m is 2, the reaction is second order with respect to A. If n is 1, the reaction is first order in B. If n is 2, the reaction is second order in B. If m or n is zero, the reaction is zero order in A or B, respectively, and the rate of the reaction is not affected by the concentration of that reactant. The overall reaction order is the sum of the orders with respect to each reactant. If m = 1 and n = 1, the overall order of the reaction is second order (m + n = 1 + 1 = 2).
The rate law:
\(\text{rate}=k\left[{\text{H}}_{2}{\text{O}}_{2}\right]\)
describes a reaction that is first order in hydrogen peroxide and first order overall. The rate law:
\(\text{rate}=k{\left[{\text{C}}_{4}{\text{H}}_{6}\right]}^{2}\)
describes a reaction that is second order in C4H6 and second order overall. The rate law:
\(\text{rate}=k\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{−}}\right]\)
describes a reaction that is first order in H+, first order in OH−, and second order overall.
Example
Writing Rate Laws from Reaction Orders
An experiment shows that the reaction of nitrogen dioxide with carbon monoxide:
\({\text{NO}}_{\text{2}}\left(g\right)+\text{CO}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{NO}\left(g\right)+{\text{CO}}_{2}\left(g\right)\)
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:\(\text{rate}=k\left[{\text{NO}}_{2}\right]^{m}{\left[\text{CO}\right]}^{n}\)
The reaction is second order in NO2; thus m = 2. The reaction is zero order in CO; thus n = 0. The rate law is:
\(\text{rate}=k\left[{\text{NO}}_{2}\right]^{2}\left[\text{CO}\right]^{0}=k{\left[{\text{NO}}_{2}\right]}^{2}\)
Remember that a number raised to the zero power is equal to 1, thus [CO]0 = 1, which is why we can simply drop the concentration of CO from the rate equation: the rate of reaction is solely dependent on the concentration of NO2. When we consider rate mechanisms later in this tutorial, we will explain how a reactant’s concentration can have no effect on a reaction despite being involved in the reaction.
It is sometimes helpful to use a more explicit algebraic method, often referred to as the method of initial rates, to determine the orders in rate laws. To use this method, we select two sets of rate data that differ in the concentration of only one reactant and set up a ratio of the two rates and the two rate laws. After canceling terms that are equal, we are left with an equation that contains only one unknown, the coefficient of the concentration that varies. We then solve this equation for the coefficient.
Example
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 (see the figure below). One such reaction is the combination of nitric oxide, NO, with ozone, O3:
Over the past several years, the atmospheric ozone concentration over Antarctica has decreased during the winter. This map shows the decreased concentration as a purple area. (credit: modification of work by NASA)
\(\text{NO}\left(g\right)+{\text{O}}_{3}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{\text{2}}\left(g\right)+{\text{O}}_{2}\left(g\right)\)
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) | \(\cfrac{\text{Δ}\left[{\text{NO}}_{2}\right]}{\text{Δ}t}\phantom{\rule{0.4em}{0ex}}\left(\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{-1}\phantom{\rule{0.2em}{0ex}}{\text{s}}^{-1}\right)\) |
|---|---|---|---|
| 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:\(\text{rate}=k\left[\text{NO}\right]^{m}\left[{\text{O}}_{\text{3}}\right]^{n}\)
We can determine the values of m, n, and k from the experimental data using the following three-part process:
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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.
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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:
\(\text{rate}=k{\left[\text{NO}\right]}^{1}{\left[{\text{O}}_{3}\right]}^{1}=k\left[\text{NO}\right]\left[{\text{O}}_{3}\right]\)
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Determine the value of k from one set of concentrations and the corresponding rate.
\(\begin{array}{cc}\hfill k& =\phantom{\rule{0.1em}{0ex}}\cfrac{\text{rate}}{\left[\text{NO}\right]\left[{\text{O}}_{3}\right]}\hfill \\ & =\phantom{\rule{0.1em}{0ex}}\cfrac{6.60\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{{\text{mol L}}^{-1}}{\text{s}}^{-1}}{\left(1.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{{\text{mol L}}^{-1}}\right)\phantom{\rule{0.2em}{0ex}}\left(3.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{-1}\right)}\hfill \\ & =\text{2}.\text{2}0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}\phantom{\rule{0.2em}{0ex}}\text{L}\phantom{\rule{0.2em}{0ex}}{\text{mol}}^{-1}{\text{s}}^{-1}\hfill \end{array}\)
The large value of k tells us that this is a fast reaction that could play an important role in ozone depletion if [NO] is large enough.
Example
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:
\(\text{2NO}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2NOCl}\left(g\right)\)
| Trial | [NO] (mol/L) | [Cl2] (mol/L) | \(-\phantom{\rule{0.2em}{0ex}}\cfrac{\text{Δ}\left[\text{NO}\right]}{\text{Δ}t}\phantom{\rule{0.4em}{0ex}}\left(\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{-1}\phantom{\rule{0.2em}{0ex}}{\text{s}}^{-1}\right)\) |
|---|---|---|---|
| 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:\(\text{rate}=k\left[\text{NO}\right]^{m}{\left[{\text{Cl}}_{2}\right]}^{n}\)
As in the example above, 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:
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Determine the value of m from the data in which [NO] varies and [Cl2] is constant. We can write the ratios with the subscripts x and y to indicate data from two different trials:
\(\cfrac{{\text{rate}}_{x}}{{\text{rate}}_{y}}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\cfrac{k{\text{[}\text{NO}\text{]}}_{x}^{m}{\text{[}{\text{Cl}}_{2}\text{]}}_{x}^{n}}{k{\text{[}\text{NO}\text{]}}_{y}^{m}{\text{[}{\text{Cl}}_{2}\text{]}}_{y}^{n}}\)
Using the third trial and the first trial, in which [Cl2] does not vary, gives:
\(\cfrac{\text{rate 3}}{\text{rate 1}}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\cfrac{0.00675}{0.00300}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\cfrac{k\left(0.15\right)^{m}\left(0.10\right)^{n}}{k{\text{(0.10)}}^{m}\left(0.10\right)^{n}}\)
After canceling equivalent terms in the numerator and denominator, we are left with:
\(\cfrac{0.00675}{0.00300}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\cfrac{{\left(0.15\right)}^{m}}{{\left(0.10\right)}^{m}}\)
which simplifies to:
\(2.25={\left(1.5\right)}^{m}\)
We can use natural logs to determine the value of the exponent m:
\(\begin{array}{ccc}\hfill \text{ln}\left(2.25\right)& =& m\text{ln}\left(1.5\right)\hfill \\ \hfill \cfrac{\text{ln}\left(2.25\right)}{\text{ln}\left(1.5\right)}& =& m\hfill \\ \hfill 2& =& m\hfill \end{array}\)
We can confirm the result easily, since:
\({1.5}^{2}=2.25\)
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Determine the value of n from data in which [Cl2] varies and [NO] is constant.
\(\cfrac{\text{rate 2}}{\text{rate 1}}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\cfrac{0.00450}{0.00300}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\cfrac{k\left(0.10\right)^{m}\left(0.15\right)^{n}}{k\left(0.10\right)^{m}\left(0.10\right)^{n}}\)
Cancelation gives:
\(\cfrac{0.0045}{0.0030}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\cfrac{{\left(0.15\right)}^{n}}{{\left(0.10\right)}^{n}}\)
which simplifies to:
\(1.5={\left(1.5\right)}^{n}\)
Thus n must be 1, and the form of the rate law is:
\(\text{Rate}=k{\left[\text{NO}\right]}^{m}{\left[{\text{Cl}}_{2}\right]}^{n}=k{\left[\text{NO}\right]}^{2}\left[{\text{Cl}}_{2}\right]\)
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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:
\(\begin{array}{ccc}\hfill 0.00300\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{-1}\phantom{\rule{0.2em}{0ex}}{\text{s}}^{-1}& =& k{\left(0.10\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{-1}\right)}^{2}{\left(0.10\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{-1}\right)}^{1}\hfill \\ \hfill k& =& 3.0\phantom{\rule{0.2em}{0ex}}{\text{mol}}^{-2}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{2}\phantom{\rule{0.2em}{0ex}}{\text{s}}^{-1}\hfill \end{array}\)
This lesson is part of:
Chemical Kinetics