AP CHEMISTRY CHAPTER REVIEW
CHAPTER 11: RATE OF REACTION
 You should understand the definition of reaction rate, as well as how rates might be measured in a laboratory
setting.
 You should know the difference between average rate, instantaneous rate, and initial rate.
 You should be able to determine the rate expression (reaction order) for a given reaction based on initial rate
data, concentration-time data, or the reaction mechanism.
 You should understand the relationship between concentration and time for a zero-, first-, and second-order
reaction.
 You should be able to calculate the order with respect to the reactants and the rate constant from a given set of
experimental data (i.e., initial rates at various reactant concentrations).
 Once the rate constant and the rate expression is known, you should be able to calculate the reaction rate at with
a given set of reactant concentrations.
 You should be able to use the Arrhenius equation to relate the rate constant to temperature. The activation
energy can be calculated from a given set of data.
 You should be able to decide if a proposed mechanism is consistent with the rate law for a given reaction.
This chapter deals with the rate at which a reaction proceeds. What are the factors that will affect the overall reaction
rate? These include the reactant concentration, the activation energy, the number of effective collisions between
reactant molecules, the temperature, the presence of a catalyst, and the reaction mechanism.
11.1
MEANING OF REACTION RATE
Consider the following chemical reaction:
N2O5(g)  2 NO2(g) + ½ O2(g)
As this reaction proceeds over time, the concentration of N2O5
decreases and the concentration of NO2 and O2 increases.
These three concentrations are constantly changing until eventually
they reach a plateau. Notice that each substance changes
concentration at a different rate.
When one mole of N2O5 decomposes, two moles of NO2 are
formed and one-half mole of O2 is formed.
We define the reaction rate with the following units:
concentration
moles per liter
rate = -------------------- = -------------------time
time
How do we measure the rate of the reaction? Depending on the properties of the reactants and products, you could
measure the disappearance of a reactant or the formation of a product.
Some ways of measuring the reaction rate include the following:
 absorption of visible light (using a spectrophotometer)
 change in pressure
Notice in the graph at right that the concentration of N2O5
(expressed with the N2O5 in brackets) is decreasing.
In a time span of 4 minutes, the concentration goes from
0.160 M to 0.040 M. The average rate would be calculated
as follows:
0.040 – 0.160
rate = ------------------- = –0.030 mol L-1 min-1
4 min
The instantaneous rate can also be calculated if you draw
a line tangent to the curve at a specific time, t.
For example, at t = 2 min, the instantaneous rate is
determined to be –0.028 mol L-1 min-1
11.2
REACTION RATE AND CONCENTRATION
It would follow logically that if you increase the concentration of the reactants, then you should increase the rate of the
reaction. This can be explained by the following theory:
Chemical reactions occur as the result of collisions between reactant molecules.
The higher the concentration of reactant molecules, the more frequently they will collide with each other. The more
often the reactant molecules collide, the faster the rate of reaction. As the reaction proceeds, the concentration of
reactant decreases, and consequently the rate will decrease.
Notice that the plot of rate versus [N2O5] is a straight line.
This means that the rate is directly proportional to the concentration.
The rate can be described by the following equation:
rate = k[N2O5]
This equation is called the rate expression or rate law.
and k is called the rate constant.
Suppose you have a single reactant, such as you would have in a decomposition reaction (A  products).
The rate expression will have the general form
rate = k[A]m
The letter m stands for the order of the reaction. It is very important to know that the order of a reaction must be
determined experimentally; it cannot be deduced from the coefficients in the balanced equation!
Most reactions that we will study in chemistry will fall into one of these categories:
zero order
1st order
2nd order
So how exactly is the order of a reactant determined by experiment?
The initial rate of the reaction (i.e., the rate at time zero) is measured for two different reactant concentrations.
Using two different reactant concentrations will produce two values for the rate:
rate1 = k[A]1m and
rate2 = k[A]2m
Then you calculate rate2 divided by rate1:
rate2 = [A]2m = [A]2
rate1 [A]1m
[A]1
m
Since all quantities are known except for m, the value of m can be calculated.
In Example 11.2 on p. 288, you can see how this calculation was done for the decomposition of CH3CHO.
Once you have calculated the order of the reaction, the rate constant can be calculated by simple “plug and chug,”
using the sample data from the experiment.
Once you have calculated the order of the reaction and the rate constant, the rate at any concentration can be
calculated.
What if there is more than one reactant?
Many reactions will involve more than one reactant. For a reaction that looks like this: aA + bB  products
The rate expression takes the following general form:
rate = k[A]m[B]n
m = the order of the reaction with respect to A, and n = the order of the reaction with respect to B
The overall order of the reaction is the sum of m + n.
With more than one reaction, the experimental set up and calculations can get a little tricky. If you want to determine
the order of the reaction with respect to A, it would be best to keep the concentration of B constant and only vary the
concentration of A. The initial rate of the reaction is measured for two different concentrations of A (while keeping [B]
constant.)
rate1 = k[A]1m[B]n and
rate2 = k[A]2m[B]n
Then you calculate rate2 divided by rate1:
rate2 = [A]2m[B]n = [A]2
rate1 [A]1m[B]n
[A]1
m
Since all quantities are known except for m, the value of m can be calculated.
Similarly, you can calculate the order of the reaction with respect to B if you keep [A] constant and only vary the
concentration of B.
In Example 11.2 on p. 289, you can see how this calculation was done for the reaction between (CH3)3CBr and OH11.3
REACTANT CONCENTRATION AND TIME
It is important for us to understand the relationship between reactant concentration and time. In other words, you might
need to figure out how much reactant is remaining after a certain period of time.
FIRST ORDER REACTIONS
For a first-order reaction, the rate expression is as follows:
rate = k[A]
The rate expression can be expressed as the rate of disappearance of [A] with respect to time:
d[A] = – k[A]
dt
1 d[A] = – k dt
[A]
Okay, now it is time for a little calculus. We must perform an integral on both sides of the equation.
On the left, the integral of 1/[A] with respect to d[A] will give us ln[A].
On the right, the integral of 1 with respect to dt will give us t.
We consider the integral from time zero to time t.
ln[A]t – ln[A]o = – kt
OR
ln[A]o – ln[A]t = kt
OR
ln [A]o = kt
[A]t
[A]o is the initial concentration of A, [A]t is the concentration of A at time t, and k is the first-order rate constant.
One of the questions that is very important for AP Chemistry students to answer is the following:
“What data can I plot versus time that will give me a straight-line (linear) plot?”
Think about the general equation for a straight line: y = mx + b
In the case of a first-order reaction, the rate expression can be written as ln[A]t = – kt + ln[A]o
So if ln[A] is plotted versus time, the slope will be equal to –k and the y-intercept will be ln[A]o
Another concept that is important to understand with respect to reaction rate is the half-life.
Half-life (t1/2) is defined as the time required for half of the original amount of A to decay (or react).
When the reaction has reached t = half-life, t1/2, then [A] would be equal to ½ [A]o
ln [A]o = kt
[A]
ln (1) = kt1/2
(0.5)
ln (2) = kt1/2
0.693 = t1/2
k
Note that with a first-order reaction, the half-life has a fixed value that is not dependent on the concentration of the
reactant. A first-order reaction is the only reaction type where [A] is not included in the half-life expression.
Radioactive decay is a first-order reaction. You should know and remember that for the AP Chemistry Exam!
ZERO ORDER REACTIONS
For a zero-order reaction, the rate expression would be given as follows: rate = k[A]0 = k
The rate of a zero-order reaction is constant, independent of the concentration of A.
Zero-order reactions are rare. One example is the decomposition of a molecule that is bound to a solid metal surface.
If the metal surface is completely covered with reactant molecules, then increasing the concentration of the reactant
will have no effect on the reaction rate.
The rate expression can be expressed as the rate of disappearance of [A] with respect to time:
d[A] = – k
dt
d[A] = – k dt
Okay, now it is time for a little calculus again. We must perform an integral on both sides of the equation.
On the left, the integral of 1 with respect to d[A] will give us [A].
On the right, the integral of 1 with respect to dt will give us t.
We consider the integral from time zero to time t.
[A]t – [A]o = – kt
OR
[A]t = [A]o – kt
[A]o is the initial concentration of A, [A]t is the concentration of A at time t, and k is the zero-order rate constant.
Think about the general equation for a straight line: y = mx + b
So if [A] is plotted versus time, the slope will be equal to –k and the y-intercept will be [A]o
SECOND ORDER REACTIONS
For a second-order reaction, the rate expression would be given as follows:
rate = k[A]2
The rate expression can be expressed as the rate of disappearance of [A] with respect to time:
d[A] = – k[A]2
dt
1 d[A] = – k dt
[A]2
Okay, Here is our last application of calculus. We must perform an integral on both sides of the equation.
On the left, the integral of 1/[A]2 with respect to d[A] will give us -1/[A].
On the right, the integral of 1 with respect to dt will give us t.
We consider the integral from time zero to time t.
(-1/[A]t) – (-1/[A]o) = – kt
1/[A]t – 1/[A]o = kt
1/[A]t = kt + 1/[A]o
Think about the general equation for a straight line: y = mx + b
So if 1/[A] is plotted versus time, the slope will be equal to +k and the y-intercept will be 1/[A]o
The following table is an important summary of the information that was discussed in this section.
Order
Rate Expression
Concentration-Time
Relationship
Half-Life
Linear Plot
Slope of Line
0
rate = k
[A]o – [A]t = kt
[A]o/2k
[A] vs. t
-k
1
rate = k[A]
ln[A]t – ln[A]o = -kt
0.693/k
ln[A] vs. t
-k
2
rate = k[A]2
1 - 1 = kt
[A]t
[A]o
1/k[A]o
1 vs. t
[A]
+k
You should be very familiar with zero, first, and second order reactions.
For each order, you should know what sort of plot will produce a straight line
If you are given the necessary equations, you should be able to calculate the concentration of A at any time t.
Here is what the Equation Sheet will give you on the AP Exam:
11.4
MODELS FOR REACTION RATE
In this section, we shift from experimental evidence and calculations to a theory that will explain our observations.
Why do some reactions occur rapidly while others occur very slowly? Consider the following reaction:
CO + NO2  CO2 + NO
For the reaction above, experimental evidence confirms that the reaction is first-order with respect to CO and firstorder with respect to NO2. How often do you think that the reactant molecules collide with each other? According to the
kinetic theory of gases, every CO molecule should collide with about 109 NO2 molecules in one second! If every
collision between CO and NO2 resulted in the formation of products, the reaction would be done in a fraction of a
second! So do you think that every time a CO molecule collides with an NO2 molecule, it will lead to the formation of
products? The answer is no. Here are a couple of reasons why every CO-NO2 collision may NOT be effective:


The molecules have to properly oriented with respect to one another when they collide.
The molecules have to have enough kinetic energy when they collide to break the required bonds.
The minimum energy that reactant molecules need in order to be successfully converted into products is called the
activation energy. It has the symbol Ea and it is expressed in kilojoules. The Ea depends on the nature of the reaction.
The higher the Ea, the slower the reaction rate. The lower the Ea, the faster the reaction rate.
At the top of the “hill” in an energy diagram is an intermediate known as the activated complex or transition state. At
this point, the bonds in the reactants are partially broken and the bonds in the products are partially formed. Reactants
have to have enough energy (and proper orientation when they collide) in order to get to this point. Then they will
either proceed to form the products or they will go back into reactants.
11.5
REACTION RATE AND TEMPERATURE
It should be no surprise to you that the rate of reaction increases as the temperature increases. As a general rule, the
approximation is that for every 10oC increase, the reaction rate doubles. Why does increasing the temperature speed
up the reaction? At higher temps, the reactant molecules will have higher kinetic energy. They will collide more often.
In short, a larger fraction of molecules will have the required activation energy to proceed to products.
There is an equation that you should be familiar with, called the Arrhenius equation. It relates the activation energy
(Ea), rate constant (k) and the temperature at which the reaction is carried out (T). It involves the form of the gas law
constant (R) that uses joules for its units (8.31 J mol-1 K-1)
– Ea
1
ln k = --------- ------- + ln A
R
T
where A is some constant (called a “frequency factor”, don’t worry about it)
If you plot the “ln k” versus “1/T” for a variety of temperatures, you should get a straight line with a negative slope. The
slope will be equal to the value “– Ea/R.” So basically the Arrhenius equation can be used to calculate the activation
energy of a reaction, if several rate constants at different temperatures are known.
Be careful of the units! R is in JOULES but activation energies are often listed in KILOJOULES.
11.6
CATALYSIS
You already know that a catalyst is something that increases the rate of a reaction without being consumed by it. How
does it do this? A catalyst changes the reaction pathway to one with lower activation energy. As mentioned earlier, the
top of the “hill” in an energy diagram represents the transition state or activated complex. A catalyst will do something
to stabilize this intermediate. If the intermediate has been stabilized, it will have lower energy. Therefore the reactants
will have a lower energy pathway to get to products. This will speed up the reaction.
Catalysts can be heterogeneous or homogeneous. In either case, they will not be consumed in the reaction. Enzymes
are a specific example of proteins that behave as catalysts.
11.7
REACTION MECHANISMS
A reaction mechanism is a set of steps that represents the pathway of the reaction. The simplest mechanism would
be one step in which reactants are instantly converted into products. Most reaction mechanisms involve two or more
steps. Some steps may be fast and other steps may be slow. The whole point that you must consider about
mechanisms is the following question:
Is the proposed reaction mechanism consistent with the experimentally observed rate law?
Now earlier in this packet I mentioned the following:
The order of a reaction must be determined experimentally; it cannot be deduced from the coefficients in the
balanced equation!
However, now that we are talking about reaction mechanisms, we will be breaking down the reaction into elementary
steps. In many cases, you WILL be able to figure out the order of the reaction by looking at the coefficients in these
elementary steps. Just keep in mind that the coefficients of these elementary steps will NOT NECESSARILY be the
same as the coefficients of the overall reaction. Is that clear? Great.
So why are we talking about mechanisms? The reason is very important. You should know that the rate expression
and the reaction order will depend on the mechanism by which the reaction takes place.
Elementary steps are the individual steps that constitute a reaction mechanism. These steps may be
unimolecular
AB+C
rate = k[A]
bimolecular
A+BC+D
rate = k[A][B]
termolecular
A+B+CD+E
rate = k[A][B][C] (In RARE cases)
The rate of an elementary step is equal to a rate constant k multiplied by the concentration of each reactant molecule.
If there is a coefficient, it would be in the rate law:
bimolecular
2AB
rate = k[A]2
Imagine a one lane road where you cannot pass any cars. If there is a car going really slowly on the road, a traffic jam
results, because everybody has to slow down behind that one car. In reaction mechanisms, the same kind of thing
happens. Often one step in the mechanism is the slow step. If this is the case, then that step is the rate-determining
step. For example:
AB
fast
BC
slow
CD
fast
AD
In this case, the rate at which A is converted to D will depend on the slow step in which B is converted
into C. Therefore the overall rate would include everything that happens up to and including the rate
determining step.
Important points to know about mechanisms include the following:



The overall rate of the reaction cannot exceed the rate of the slowest step
The rate of the slowest step will be approximately equal to the rate of the overall reaction.
The slowest step in a mechanism is usually the one that has the highest activation energy.
Don’t forget that rate expressions must be determined experimentally!
Once this has been done, then a plausible reaction mechanism can be derived that will be consistent with the
observed rate law.
Sometimes the mechanism will involve intermediates. If a chemical appears as a product in one step and then again
as a reactant in another step, and it can be crossed out, then it is an intermediate.
Sometimes the mechanism will involve a catalyst. If a chemical appears as a reactant in one step and then again as a
product in another step, and it can be crossed out, then it is a catalyst.
Intermediates and catalysts do NOT normally appear in the overall rate expression.