13.4 The Integrated Rate Law: The
Dependence of Concentration on Time
The Integrated Rate Law
• Integrated rate law for a chemical reaction is a relationship between the
concentration of the reactant(s) and time. The following equations are on
your reference table!
The first order integrated rate law is:
ln[A]t = -kt + ln[A]0
OR
[A]t
ln ------ = -kt
[A]0
Where [A]t is the concentration at any time t, k is the rate
constant, and [A]0 is the initial concentration of A.
The rate law above has the form of an equation of a straight line with
decreasing slope:
y = mx + b. This is because a plot for a first order reaction is a straight line, and
even though the slope is negative the rate constant is always positive.
Let’s Try a Practice Problem!
Cyclopropane rearranges to form propene in the gas
phase.
The reaction is first order in cyclopropene and has a
measured rate constant of 3.36X10-5s-1 at 720 K. If the
initial cyclopropane concentration is 0.0445 M, what will
the cyclopropane concentration be after 235.0 min?
ln[A]t = -kt + ln[A]0
ln[cyclopropane]t = -(3.36X10-5s-1)((235 min X (60 s / 1 min)) + ln(0.0445 M)
ln[cyclopropane]t = -3.586, so [cyclopropane]t = e-3.586 = 0.0277 M
Second Order Integrated Rate Law
1
1
---- = kt + ---[A]t
[A]0
This equation is also in the form of a straight line, but
has an increasing slope because the plot is of the
inverse of concentration of the reactant with respect
to time.
Zero-Order Integrated Rate Law
[A]t = -kt + [A]0
Again here, the equation is in the form of a straight
line.
These straight lines that can be produced by these
integrated rate law equations can be used to aid in
interpreting data, and extrapolating points on the
graph.
Half-Life of a Reaction
•
The half-life (t½) of a reaction – the time required for the concentration of a reactant to fall
to one-half of its initial value.
First order reaction half-life: (here t½ is independent of the initial concentration. This causes the
half-life to be constant, making the concept of half-life particularly useful for first-order
reactions).
0.693
t½ = -------k
This formula is
on your reference
table
From the College Board:
Let’s Try a Practice Problem!
Molecular iodine dissociates at 625 K with a first-order rate
constant of 0.271 s-1. What is the half-life of this reaction?
0.693
t½ = ---------k
0.693
k = -------------- = 2.56 s
0.271s-1
Let’s Try Another!
A first-order reaction has a half-life of 26.4 s. How long
does it take for the concentration of the reactant in a
reaction to fall to one-eighth of its initial value?
Half-life
Time
Amount
0
0
Initial value
1
26.4 s
½ initial value
2
52.8 s
¼ initial value
3
79.2 s
1/8 initial value
If you wanted to use an equation instead of a table here,
you would need to use: Time = half-life X log2(1/y) where
y= fraction of material remaining… so the table may seem
easier.
Summarizing Basic Kinetic
Relationships
• The reaction order and the rate law must be determined
experimentally.
• The rate law relates the rate of the reaction to the
concentration of the reactants.
• The integrated rate law (which is mathematically derived
from the rate law) relates the concentration of the reactant(s)
to time.
• The half-life is the time it takes for the concentration of the
reactant to fall to one-half of its initial value.
• The half-life of a first order reaction is independent of the
initial concentration.
(We will only worry about half-lives of first order reactions in
this course)
13.4 pgs. 640-641 #’s 45, 48, 53 & 56
Read 13.5 pgs. 615-622