# 12.4 Integrated rate laws

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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:

$\text{ln}\left(\frac{{\left[A\right]}_{t}}{{\left[A\right]}_{0}}\right)=\text{−}kt$

or

$\text{ln}\left(\frac{{\left[A\right]}_{0}}{{\left[A\right]}_{t}}\right)=kt$

or

$\left[A\right]={\left[A\right]}_{0}{e}^{-kt}$

## 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 :

${\text{C}}_{4}{\text{H}}_{8}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2C}}_{2}{\text{H}}_{4}$

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:

$\text{ln}\left(\frac{{\left[A\right]}_{0}}{\left[A\right]}\right)=kt$

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:

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:

$\text{I-131}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Xe-131}+\text{electron}$

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?

16.7 days

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:

what is a balanced equation 4 trioxonitrate (V)acid and sodium hydroxide?
proved ur Worth: If A is a of trioxonitrate(V)acid,HNO3' of unknown concentration .B is a standard solution of sodium hydroxide containing 4.00g per dm cube of solution.25cm cube portions solution B required an average of 24.00cm cube of solution A for neutralization,using 2 drops of methyl orange.
Marcel
calculate the concentration of solution B in moles per dm cube
Marcel
calculate the concentration of solution A and B in moles per DM cube
Marcel
finally calculate the concentration in g/dm cube of HNO3 in solution A (H=1,N=14,O=16,Na=23)
Marcel
calculate the standard enthalpy of formation for propane(C3H8) from the following data; 1), C3H8+5O2->3CO2+4H2O; -222.0kJ/mol 2), C+O2->CO2;-395.5kJ/mol 3),H2+O->H2O; 285.8kJ/mol
Josephine
let eventually of formation of propane = X X + (-222)=3×(-395.5)+4×(-286) rearrange to find X
Paul
wat is electrolysis?
it is the chemical decomposition of a substance when electric current is passed through it either in molten form or aqueous solution
Nuru
list the side effect of chemical industries
how do you ionise an atom
many ways ,but one of them is when the atom becomes heated to a certain temperature the surface electron becomes too energetic and leaves the atom because the attraction between the nucleus and the electron becomes overpowered by the energetic eletron
sunday
also hitting of two atoms can cause transfer of surface electrons
sunday
and when this transfers occur the atom becomes ionised
sunday
who is doing Cape chemistry tomorrow?
What is hybridization
the mix between different breeds of species in one
Jared
it is the blending of orbitals.
stanley
the mixing of orbital
caramel
are covalent bonds influenced by factors such as temperature and pressure?
what is catalyst used for mirror test
when an atom looses electron, what does it become?
it's oxidized and called an ion
Anora
thanks
Abdullahi
Now, I get it
Abdullahi
cation
Anora
can you give an example please, if you don't mind
Abdullahi
a positive ion,become positively charged/a cation.
Janis
sodium plus one is simple cation is exmpl
ajmal
Taking Sodium as example..... it carries a positive charge which means it is positively charged.....when it gains an electron, it is reduced cuz an electron is negatively charged.....also when an atom looses an electron, it becomes positively charged and when it gains, it becomes negatively charged.
Nuru
typically, ionization is the process where an atom looses or gains electron(s) to form ion(s) either a positively or negatively
Nuru
what is copper
just an element
Power
Cu
daniel
Why is water a single covalent bond?
nitrogen is a gas whereas phosphorus is solid .Explain.
can you explain what you are needing it now better than maybe I'm just not interpreting it what you're needing to know
Alex
cool nitrogen down to around negative 270 °F and it will be solid. now they are both solid
daniel
they are different elements and dats how they are pal.....check the periodic table
Nuru
Nitrogen is a diatomic molecule with relatively weak van de waals forces between the molecules. These forces are overcome when the solid melts or liquid evaporates. Phosphorus forms larger molecules consisting of four phosphorus atoms in a tetradedral shape. The intermolecular forces are stronger
Paul
whats a base
A base is a substance which will neutralize an acid to yield salt and water only
Zainab
base is a substance that produces OH(aq) ions in aqueous solution. Strong soluable bases are in water and are completely dislocated. Therefore weak base ionize slightly...
Roy
a base is a substance that neutralise and acid to form salt and water
Daksalma
write electrolysis of bright solution using either carbon or platinum and write the reaction at the anode or at the cathode
what is the H3O of a solution with the pH of 2.5
pH<7, therefore there are only H3O+HX3OX+particles in the solution. [H3O+]=10−pH=10−6.99=1.02⋅10−7[HX3OX+]=10−pH=10−6.99=1.02⋅10−7 When the pH is smaller than 6 or greater than 8, one will not notice the difference, but here it is logarithmically speaking  and I'll give you another one if this is ki
Alex
Alex
When the pH is smaller than 6 or greater than 8, one will not notice the difference, but here it is logarithmically speaking
Alex
sorry I don't know why that sent again
Alex
We have [H3O+]=10−pH=10−6.99=1.02⋅10−7[HX3OX+]=10−pH=10−6.99=1.02⋅10−7 and [OH−]=10−pOH=10−7.01=9.77⋅10−8[OHX−]=10−pOH=10−7.01=9.77⋅10−8.  Because of H3O++OH−⟶2H2OHX3OX++OHX−⟶2HX2O we are left with [H3O+]=1.02⋅10−7−9.77⋅10−8=4.6⋅10−9
Alex