SparkNotes Plus subscription is $4.99/month or $24.99/year as selected above. The free trial period is the first 7 days of your subscription. TO CANCEL YOUR SUBSCRIPTION AND AVOID BEING CHARGED, YOU MUST CANCEL BEFORE THE END OF THE FREE TRIAL PERIOD. You may cancel your subscription on your Subscription and Billing page or contact Customer Support at custserv@bn.com. Your subscription will continue automatically once the free trial period is over. Free trial is available to new customers only.
Choose Your Plan
Payment Details
Payment Summary
Your Free Trial Starts Now!
For the next 7 days, you'll have access to awesome PLUS stuff like AP English test prep, No Fear Shakespeare translations and audio, a note-taking tool, personalized dashboard, & much more!
Thanks for creating a SparkNotes account! Continue to start your free trial.
Please wait while we process your payment
Your PLUS subscription has expired
We’d love to have you back! Renew your subscription to regain access to all of our exclusive, ad-free study tools.
Figure %: Integrated rate laws for reaction orders zero through two
As you can see, each order of reaction has a unique input and output
variable that produces a straight
line. For example, if we graph the following rate data for
the decomposition of
H2O2 assuming that it could be zero, first, or
second order, we find
that only the graph for a 2nd order reaction (1/[A] versus t) gives a
straight line. Therefore, the
reaction has the rate law rate = k [H2O2]2.
Figure %: Rate data for the decomposition of hydrogen peroxide
Assuming that we did not know that the decomposition is second order, we
will make a series of three
graphs to determine the order of the reaction:
Figure %: Plot of Hydrogen Peroxide Concentration versus Time
If the graph was linear, we would conclude that the rate law is zero
Order, but it is not.
Therefore, let's see if it is first order by plotting ln
[H2O2] versus time in
the following graph:
Figure %: Plot of the natural log of hydrogen peroxide concentration versus
time
Seeing that the reaction is not first order due to the non-linearity of the
above graph, we move to
plotting 1/[H2O2] versus time to test
whether the
decomposition reaction is second order.
Figure %: Plot of inverse hydrogen peroxide concentration versus time
Because the above graph is linear, we know that the reaction is
second order. The
slope of the line is twice the rate constant, k from the rate law.
What you should learn from the above discussion is that you can use
integrated rate laws to determine
both the rate constant for a reaction and the form of the rate law. It may
seem more complicated to use integrated rate laws rather than the method of
initial rates to
determine the rate law, but it really
does take much longer to perform and analyze the several reactions needed
for the method of initial
rates than it does to produce the necessary graphs for the integrated rate
law method--especially with
a good graphing program.
You may have noticed that we only list three integrated rate laws,
ignoring rate laws like rate = k [A]
[B]. That doesn't mean we can't use integrated rate laws to determine the
rate law for those types of
reactions. We simply must be more clever about how we do it. For a two-
component
second order reaction
with a rate law rate = k [A] [B], we can make the concentration of B so
large as compared to A that
the concentration of B is almost constant. Assuming that the concentration
of B is constant, the
reaction becomes pseudo-first order--that is, the reaction will behave as
if it was first order. The
kinetics data for this reaction will give a graph of ln [A] versus time
that is linear, telling us that the
reaction is first order in A. If the reaction happened to be second order
in A, in the present example,
then a graph of 1/[A] versus time would be linear. We can, similarly,
determine the order of B by
making the concentration of A large. For multiple component rate laws, you
can simply make the
concentration of all reagents but the one of interest large to iteratively
determine the order of the
reaction in each component. (Chemists are fond of using the word 'iterative' to
say 'repetitive'.)
Half-Lives
An earlier and less common way to measure rate is by the half-life of
a reaction. A half-life is
the time it takes for one half of the starting material to be transformed
into its products. Often you
will hear half-life associated with radioactive decay phenomena (which
follow first order kinetics),
but the term can be applied to any reaction.
The half-life of a reaction not only depends on the rate constant of the
reaction (those with larger k's
have shorter half-lives) but also on the integrated rate law for the
reaction. To derive the form of the
half-life expression for a first order reaction, we start with its
integrated rate law, and then substitute the
value 0.5 for the ratio of [A] to [A]o:
Figure %: Derivation of the half-life for a first order reaction
Using the same techniques with the different integrated rate laws, you can
derive the half-life
expression for a reaction of any order. Summarized below are the
half-lives for reactions of orders
zero through two.
Figure %: Half-lives of reactions with orders zero through two