<< Chapter < Page Chapter >> Page >
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

Types of electrolytes
Treasure Reply
list 6 subatomic particles and their mass, speed and charges
Dubem Reply
combination of acid and base
Ayibiro Reply
that salt
Talhatu
calculate the mass in gram of NaOH present in 250cm3 of 0.1mol/dm3 of its solution
Omego Reply
The mass is 1.0grams. First you multiply the molecular weight and molarity which is 39.997g/mol x 0.1mol/dm3= 3.9997g/dm3. Then you convert dm3 to cm3. 1dm3 =1000cm3. In this case you would divide 3.9997 by 1000 which would give you 3.9997*10^-3 g/cm3. To get the mass you multiply 3.9997*10^-3 and
Kokana
250cm3 and get the mass as .999925, with significant figures the answer is 1.0 grams
Kokana
nitrogen, phosphorus, arsenic, antimony and Bismuth
faith Reply
What is d electronic configuration of for group 5
Miracle Reply
Can I know d electronic configuration of for group 5 elements
Miracle
2:5, 2:8:5, 2:8:8:5,...
Maxime
Thanks
Miracle
Pls what are d names of elements found in group 5
Miracle
define define. define
Muh Reply
what is enthalpy
Ayilaran Reply
total heat contents of the system is called enthalpy, it is state function.
Sajid
background of chemistry
Banji Reply
what is the hybridisation of carbon in formic acid?
Maham Reply
sp2 hybridization
Johnson
what is the first element
Josh Reply
HYDROGEN
Liklai
Element that has positive charge and its non metal Name the element
Liklai
helium
oga
sulphur
oga
hydrogen
Banji
account for the properties of organic compounds
mercy Reply
properties of organic compounds
mercy
what's the difference between molecules and compounds
Amha Reply
A compound can be a molecule however compounds must contain more than one element. For example ozone, O3 is a molecule but not a compound.
Justin
what is che? nd what is mistry?
Mukhtar
What's elixir?
EMEKA Reply
An Elixir is a substance held capable of changing base metals into Gold.
Nwafor
Give an example for each of the six groups of element
Francis Reply
Practice Key Terms 2

Get the best Chemistry course in your pocket!





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