Writing Rate Laws from Reaction Orders
An experiment shows that the reaction of nitrogen dioxide with carbon monoxide:

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:

The reaction is second order in NO₂; thus m = 2. The reaction is zero order in CO; thus n = 0. The rate law is:

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 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
The rate law for the reaction:

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?

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:

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

What is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of reaction?

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 ([link]). One such reaction is the combination of nitric oxide, NO, with ozone, O₃:

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

Determine the rate law and the rate constant for the reaction at 25 °C.

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:

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. 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:

   \(\text{rate}=k{\left[\text{NO}\right]}^{1}{\left[{\text{O}}_{3}\right]}^{1}=k\left[\text{NO}\right]\left[{\text{O}}_{3}\right]\)

3. Determine the value of k from one set of concentrations and the corresponding rate.

   \(\begin{array}{cc}\hfill k& =\phantom{\rule{0.1em}{0ex}}\frac{\text{rate}}{\left[\text{NO}\right]\left[{\text{O}}_{3}\right]}\hfill \\ & =\phantom{\rule{0.1em}{0ex}}\frac{6.60\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}\overline{){\text{mol L}}^{-1}}{\text{s}}^{-1}}{\left(1.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\overline{){\text{mol L}}^{-1}}\right)\phantom{\rule{0.2em}{0ex}}\left(3.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{-1}\right)}\hfill \\ & =\text{2}.\text{2}0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}\phantom{\rule{0.2em}{0ex}}\text{L}\phantom{\rule{0.2em}{0ex}}{\text{mol}}^{-1}{\text{s}}^{-1}\hfill \end{array}\)

   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.

Check Your Learning
Acetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation:

Determine the rate law and the rate constant for the reaction from the following experimental data:

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:

Solution
The rate law for this reaction will have the form:

As in [link], 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:

1. 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:

   \(\frac{{\text{rate}}_{x}}{{\text{rate}}_{y}}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{k{\text{[}\text{NO}\text{]}}_{x}^{m}{\text{[}{\text{Cl}}_{2}\text{]}}_{x}^{n}}{k{\text{[}\text{NO}\text{]}}_{y}^{m}{\text{[}{\text{Cl}}_{2}\text{]}}_{y}^{n}}\)

   Using the third trial and the first trial, in which [Cl₂] does not vary, gives:

   \(\frac{\text{rate 3}}{\text{rate 1}}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{0.00675}{0.00300}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{k\left(0.15{\right)}^{m}\left(0.10{\right)}^{n}}{k{\text{(0.10)}}^{m}\left(0.10{\right)}^{n}}\)

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

   \(\frac{0.00675}{0.00300}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{{\left(0.15\right)}^{m}}{{\left(0.10\right)}^{m}}\)

   which simplifies to:

   \(2.25={\left(1.5\right)}^{m}\)

   We can use natural logs to determine the value of the exponent m:

   \(\begin{array}{ccc}\hfill \text{ln}\left(2.25\right)& =& m\text{ln}\left(1.5\right)\hfill \\ \hfill \frac{\text{ln}\left(2.25\right)}{\text{ln}\left(1.5\right)}& =& m\hfill \\ \hfill 2& =& m\hfill \end{array}\)

   We can confirm the result easily, since:

   \({1.5}^{2}=2.25\)

2. Determine the value of n from data in which [Cl₂] varies and [NO] is constant.

   \(\frac{\text{rate 2}}{\text{rate 1}}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{0.00450}{0.00300}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{k\left(0.10{\right)}^{m}\left(0.15{\right)}^{n}}{k\left(0.10{\right)}^{m}\left(0.10{\right)}^{n}}\)

   Cancelation gives:

   \(\frac{0.0045}{0.0030}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{{\left(0.15\right)}^{n}}{{\left(0.10\right)}^{n}}\)

   which simplifies to:

   \(1.5={\left(1.5\right)}^{n}\)

   Thus n must be 1, and the form of the rate law is:

   \(\text{Rate}=k{\left[\text{NO}\right]}^{m}{\left[{\text{Cl}}_{2}\right]}^{n}=k{\left[\text{NO}\right]}^{2}\left[{\text{Cl}}_{2}\right]\)

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:

   \(\begin{array}{ccc}\hfill 0.00300\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{-1}\phantom{\rule{0.2em}{0ex}}{\text{s}}^{-1}& =& k{\left(0.10\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{-1}\right)}^{2}{\left(0.10\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{-1}\right)}^{1}\hfill \\ \hfill k& =& 3.0\phantom{\rule{0.2em}{0ex}}{\text{mol}}^{-2}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{2}\phantom{\rule{0.2em}{0ex}}{\text{s}}^{-1}\hfill \end{array}\)

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

Determine the rate law expression and the value of the rate constant k with appropriate units for this reaction.