1. Home
  2. Chemistry
  3. Kinetics
  4. Rate laws

12.3 Rate laws  (Page 2/10)

<< Chapter < Page
  Chemistry     Page 2 / 10
Chapter >> Page >

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.

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, O 3 :

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.
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)
NO( g ) + O 3 ( g ) ⟶ NO 2 ( g ) + O 2 ( g )

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

Trial [NO] (mol/L) [O 3 ] (mol/L) Δ [ NO 2 ] Δ t ( mol L −1 s −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.

Solution

The rate law will have the form:

rate = k [ NO ] m [ O 3 ] n

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 [O 3 ] is constant. In the last three experiments, [NO] varies while [O 3 ] 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 [O 3 ] varies and [NO]is constant. In the first three experiments, [NO] is constant and [O 3 ] varies. The reaction rate changes in direct proportion to the change in [O 3 ]. When [O 3 ] doubles from trial 1 to 2, the rate doubles; when [O 3 ] triples from trial 1 to 3, the rate increases also triples. Thus, the rate is directly proportional to [O 3 ], and n is equal to 1.The rate law is thus:

    rate = k [ NO ] 1 [ O 3 ] 1 = k [ NO ] [ O 3 ]

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

    k = rate [ NO ] [ O 3 ] = 6.60 × 10 −5 mol L −1 s − 1 ( 1.00 × 10 −6 mol L −1 ) ( 3.00 × 10 −6 mol L − 1 ) = 2 . 2 0 × 10 7 L mol − 1 s − 1

    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:

CH 3 CHO( g ) ⟶ CH 4 ( g ) + CO( g )

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

Trial [CH 3 CHO] (mol/L) − Δ[CH 3 CHO] Δ t ( mol L −1 s −1 )
1 1.75 × 10 −3 2.06 × 10 −11
2 3.50 × 10 −3 8.24 × 10 −11
3 7.00 × 10 −3 3.30 × 10 −10
<< Chapter < Page Page > Chapter >>
Practice Key Terms 5

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Chemistry' conversation and receive update notifications?

Ask