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By the end of this section, you will be able to:
  • Explain the form and function of an integrated rate law
  • Perform integrated rate law calculations for zero-, first-, and second-order reactions
  • Define half-life and carry out related calculations
  • Identify the order of a reaction from concentration/time data

The rate laws we have seen thus far relate the rate and the concentrations of reactants. We can also determine a second form of each rate law that relates the concentrations of reactants and time. These are called integrated rate laws . We can use an integrated rate law to determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For example, an integrated rate law is used to determine the length of time a radioactive material must be stored for its radioactivity to decay to a safe level.

Using calculus, the differential rate law for a chemical reaction can be integrated with respect to time to give an equation that relates the amount of reactant or product present in a reaction mixture to the elapsed time of the reaction. This process can either be very straightforward or very complex, depending on the complexity of the differential rate law. For purposes of discussion, we will focus on the resulting integrated rate laws for first-, second-, and zero-order reactions.

First-order reactions

An equation relating the rate constant k to the initial concentration [ A ] 0 and the concentration [ A ] t present after any given time t can be derived for a first-order reaction and shown to be:

ln ( [ A ] t [ A ] 0 ) = k t

or

ln ( [ A ] 0 [ A ] t ) = k t

or

[ A ] = [ A ] 0 e k t

The integrated rate law for a first-order reaction

The rate constant for the first-order decomposition of cyclobutane, C 4 H 8 at 500 °C is 9.2 × 10 −3 s −1 :

C 4 H 8 2C 2 H 4

How long will it take for 80.0% of a sample of C 4 H 8 to decompose?

Solution

We use the integrated form of the rate law to answer questions regarding time:

ln ( [ A ] 0 [ A ] ) = k t

There are four variables in the rate law, so if we know three of them, we can determine the fourth. In this case we know [ A ] 0 , [ A ], and k , and need to find t .

The initial concentration of C 4 H 8 , [ A ] 0 , is not provided, but the provision that 80.0% of the sample has decomposed is enough information to solve this problem. Let x be the initial concentration, in which case the concentration after 80.0% decomposition is 20.0% of x or 0.200 x. Rearranging the rate law to isolate t and substituting the provided quantities yields:

t = ln [ x ] [ 0.200 x ] × 1 k = ln 0.100 mol L −1 0.020 mol L −1 × 1 9.2 × 10 −3 s −1 = 1.609 × 1 9.2 × 10 −3 s −1 = 1.7 × 10 2 s

Check your learning

Iodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation:

I-131 Xe-131 + electron

The decay is first-order with a rate constant of 0.138 d −1 . All radioactive decay is first order. How many days will it take for 90% of the iodine−131 in a 0.500 M solution of this substance to decay to Xe-131?

Answer:

16.7 days

Got questions? Get instant answers now!

We can use integrated rate laws with experimental data that consist of time and concentration information to determine the order and rate constant of a reaction. The integrated rate law can be rearranged to a standard linear equation format:

Questions & Answers

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An atom is the smallest indivisible particle of an element that can take part in chemical reaction
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the study of matter
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energy can neither be created or distroyed it can only be transferred or converted from one form to another
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Graham's law of Diffusion
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aromaticity is a conjugated pi system specific to organic rings like benzene, which have an odd number of electron pairs within the system that allows for exceptional molecular stability
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sodium hydroxide (NaOH)
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a hydrocarbon contains 7.7 percent by mass of hydrogen and 92.3 percent by mass of carbon
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they are three 3
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TYPES OF COVALENT BOND-POLAR BOND-NON POLAR BOND-DOUBLE BOND-TRIPPLE BOND. There are three types of covalent bond depending upon the number of shared electron pairs. A covalent bond formed by the mutual sharing of one electron pair between two atoms is called a "Single Covalent bond.
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Practice Key Terms 2

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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