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AP Chemistry
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14  Kinetics
NOTES
14.1 Factors That Affect Reaction Rates
 Chemical kinetics is the art of chemistry that deals with the speeds, or rates, of reactions. Reaction
rates can range from microseconds to millions of years.
 There are 5 main factors that affect reaction rates – the book leaves out the first to be understood
o Nature of the reactants: large, complex molecules tend to have slower reaction rates than smaller,
simpler molecules. They have more area and thusly could have collisions at spots other than the
reaction site.
o Physical state of the reactants: reactants with the same state react more quickly. Gases and
liquids have more surface area and thusly react faster than solids. Also, surface area in general is
factor as a powdered solid reacts faster than a big block of the same solid.
o Concentration of the reactants: the higher the concentration then the greater the amount of
collisions which leads to a faster the reaction rate. For gases, this means that the pressure and
reaction rate are directly related.
o Temperature at which the reaction occurs: reaction rate and temperature are directly related. An
increase in temperature means an increase in kinetic energy of molecules and this leads to more
collisions.
o Catalyst present: a catalyst increases the rate of a reaction by lowering the activation energy.
They affect the mechanism, or type of collisions, in the reaction.
 In general, reaction rates depend on the amount of collisions between the molecules, since they must
collide in order to react.
14.2 Reaction Rates
 Reaction rate is the speed of a chemical reaction
o It is the change in concentration of reactants, or products, per unit of time.
o It is a derived unit and is in molarity per seconds (M/s). You might also see it as
mol·L-1·s-1 or mol /L·s
*Fig 14.3 (Pg 559) shows a great example of a reaction rate. Notice how the number of moles is
consistent and that reactant disappears as product is made.
 Average rate can be given by the following equations
Average rate of appearance = Δ[Product] / Δt
OR
Average rate of disappearance = - Δ[Reactant] / Δt
where Δ[Reactant] is the change in concentration of the reactant, Δ[Product] is the change in
concentration of the product and Δt is the change in time
 The rate of a reaction decreases over time. As the concentration of the reactants decreases, they are
less likely to collide and react with each other.
*Table 14.1 shows the reaction rate of butyl chloride with water. Note that as the molarity of the butyl
chloride decreases over time, so does the reaction rate.
*Fig 14.4 shows the time versus concentration of this reaction graphically. The slope of this graph is the
reaction rate; the sloe gets less steep over time showing the decrease in rate.


Instantaneous rate is the rate at a specific moment during the reaction. This can easily be found
graphically.
o Instantaneous rate can be calculated through the average rate of appearance/disappearance
equations.
o Instantaneous rate at t = 0 is the initial rate
Reaction rates can be found stoichiometrically. If you consider that reactants disappear at the same
rate as each other and the same rate as the products appear you could assume that given the
following reaction
aA + bB → cC + dD
where the lower case letter represent moles and the capital letters represent the substances then, the
rate could be expressed as the following
 D 


 t 
*Sample Exercise 14.3 shows how stoichiometry can be used to determine rates
Rate 
1  A  1  B  1  C  1

  
 

a  t  b  t  c  t  d
14.3 The Rate Law: The Effect of Concentration on Rate
 Changing the concentration of a reactant changes the rate of the reaction
*Table 14.2 shows this for the reaction of Ammonium and Nitrate. Notice that if you leave one of the
reactants’ concentration steady, whatever factor you change the other concentration by, that is the factor
that the rate is changed by (double the concentration = double the rate, etc.0
 Rate Law is an equation that shows how the rate depends on the concentration of the reactants. If
you are given the reaction
aA + bB → cC + dD
then the rate law would be
Rate = k[A]m[B]n
o Notice that the products are not in the rate law
o k is the rate constant and is simply a constant for each chemical reaction at a given temperature.
It will be given unless you are solving for it.
o m and n are small whole number (usually 0,1, or 2) and are the reaction orders
 Reaction orders tell the effect the change in concentration of that reactant will have on the reaction
rate.
o Zero order reaction: m = 0 means that a change in the reactant concentration will have no effect
on the rate
o First order reaction: m = 1 means that the concentration will have a 1:1 effect on the rate (if the
concentration is doubled, then the rate will be doubled)
o Second order reaction: m = 2 means that concentration will have a 1:2 effect on the rate (if the
concentration is doubled, then the rate will be quadrupled)
o Overall reaction order is the sum of each reactants’ reaction order
o For example, consider this rate law
Rate = k[A]1[B]2
We would say that this reaction is first order with respect to A, second order with respect to B
and third order overall.
o Rate orders can only be found experimentally. Sometimes they are the same as the coefficients
from the balanced reactions but that is not where the rate orders come from.
*Sample Exercise 14.4 (Pg 566) shows how rate orders affect concentration and in turn affect the
reaction rate.
 The units of the rate constant for a particular reaction depend on the overall reaction order
Units of rate constant =
units of rate
(units of concentration)overall reaction order
So if a reaction was second order overall then
Units of rate constant = M/s = M-1s-1
M2
o
This equation can also be used to solve for the rate constant itself if you use the rate in
place of the top and the concentration of a reactant in place of the bottom
 We can determine the reaction orders of each reactant and then use those to determine the rate law
for a reaction.
o Keep in mind the relationship between rate orders and concentrations and their effect on rate
o The rate of a reaction depends on the concentration but the rate constant (k) does not
*Sample Exercise 14.6 goes through the entire process of finding a rate law based on experimental
initial rate data by comparing the concentrations to the reaction rates to determine the reaction orders.
The reaction orders were then used to determine the rate law.
14.4 The Change of Concentration with Time
 Overall reaction orders have their own equations
 First-order reactions are reactions where the rate depends on the concentration of a single reactant
raised to the first power, or
Rate = - Δ[A] = k[A]
Δt
This is called the differential rate law and it shows how rate depends on concentration.
o We can use a calculus process called integration to come up with an equation that will allow us
to use concentrations at time = 0, [A]0, up to the concentration at any other time, [A]t. This
creates the integrated rate law for a first-order reaction, which is
ln[A]t – ln[A]0 = -kt
OR
ln ([A]t / [A]0) = -kt
*Sample Exercise 14.7 (Pg 570) shows how to use the integrated first-order rate law to solve for
a concentration at a specific time.
o A first order reaction will always yield a straight line graphically when graphing time vs
concentration. The slope will always be –k.
*Fig 14.8b (Pg 581) shows the graph of a first order reaction
 Second-order reactions are reactions where the rate depends on the concentration of one reactant
raised to the second power OR the concentrations of two reactants with each raised to the first
power. If we consider the first type, we get the differential equation
Rate = - Δ[A] = k[A]2
Δt
o
The integrated rate law for a second-order reaction is
1
- 1 = kt
[A]t
[A]0
o A second-order reaction will also yield a straight line graphically but the slope will be +k
*The slope-intercept format, y = mx + b can be applied to all forms of rate law equations to produce a
graph
 Half-life, t½ , is the amount of time it takes to for concentration of a reactant to be half of its original
concentration
o Half-life of a first-order reaction, which is most half-life problems and radioactive decay, can be
found using
t½ = 0.693 / k
o Half-life of a second-order reaction can be found using
t½ = 1 / k[A]0
14.5 Temperature and Rate
 Temperature causes the rate of a chemical reaction to increase because the temperature increase
causes the rate constant to increase
 The collision model, which is based on the Kinetic Molecular Theory that we studied with gas laws,
states that molecules must collide in order to react. This doesn’t mean that every collision results in
a reaction.
*Fig 14.15 (Pg 576) shows how orientation of molecules can affect reactions during collisions
 Arrhenius concluded that not only should molecules be oriented properly, but they must collide with
a certain amount of energy so that bonds could be broken.
o Activation energy, Ea, is the minimum amount of energy required to start a chemical reaction
(“a” on below graph)
o Activated complex is the peak energy of a reaction (“c” on below graph)
+50
c
+35
PE
a
0
d
b
–35
–50
e
reaction coordinate

o The Arrhenius equation relates:
1) The fraction of molecules that have the activation energy or greater
2) Collisions occurring per second
3) The fraction of collisions that have the appropriate orientation
k = Ae-Ea/RT
where k is the rate constant, Ea is the activation energy, R is the gas constant (8.314 J/molK,
and T is the absolute temperature
The rate constant can be calculated by rearranging the equation to get
ln k1 = Ea 1 – 1
k2 R T2 T1
14.6 Reaction Mechanisms
 Reaction mechanism is the process by which a reaction occurs
o Many reactions actually take several steps and require a series of reactions that you don’t see
o Reaction mechanism is the entire sequence of steps
 Elementary reactions (also called elementary steps) involve only two reactant molecules colliding
o These are the individual steps in a reaction mechanism
 Molecularity is the amount of molecules that are reactants in an elementary step
o Unimoleculars are one molecule and usually involves rearrangement of that molecule or
decomposition
o Bimoleculars involve two reactants or 2 moles of one type of reactant
o Termolculars are very rare and involve three reactants, 3 moles of the same reactant, or 2 moles
of one reactant and 1 mole of the other
 Molecularity directly relates to the rate law
*Table 14.3 (Pg 584) shows this relationship. It would be best to memorize it.
 A multistep mechanism involves a series of elementary steps
o The elementary steps must always add up to get the equation of the overall process
 An intermediate (or reaction intermediate) is a substance that appears in the elementary steps but is
not in the overall equation
 Let’s consider the following reaction
NO2(g) + CO(g) → NO(g) + CO2(g)
This is the overall reaction and comes from two elementary reactions
Step 1:
NO2(g) + NO2(g) → NO3(g) + NO(g)
(slow)
Step 2:
NO3(g) + CO(g) → NO2 (g) + CO2(g) (fast)
The sum is
2 NO2(g) + NO3(g) + CO(g) → NO2(g) + NO3(g) + NO(g) + CO2(g)
Cancelling out the substances that appear on both sides produces the overall reaction
2 NO2(g) + NO3(g) + CO(g) → NO2(g) + NO3(g) + NO(g) + CO2(g)
o NO3(g) is the intermediate because it is completely cancelled out and although it is in the
elementary steps, it does not appear in the final overall reaction
 Each step in a multistep reaction mechanism has its own rate. The overall rate of the reaction can
not be faster than the rate of its slowest elementary step.
o Rate determining step (rate-limiting step) is the slowest elementary step in a mechanism and
controls the rate for the overall reaction
 Rate law for an overall reaction depends on the order of the elementary steps, or basically where the
rate determining step is located in the mechanism
 If the first elementary step is the slowest, then the rate law of the first step is equal to the overall rate
law
*Sample Exercise 14.14 shows this process.
 If the first step is not the rate-determining step, we must consider other issues.
o We can assume that the first step must be fast and therefore is reversible. That means it actually
has 2 processes – the forward reaction and the reverse reaction.
o This also means that the rate determining step probably contains an intermediate, which is not
good since they are usually unstable.


We can relate the fast step and the rate determining step by writing the intermediate in terms of its
reverse reaction, since they will have the same rate anyway. For example, given the following
mechanism
Step 1:
NO(g) + Br2(g)
↔ NOBr2(g)
(fast)
Step 2:
NOBr2(g) + NO(g) → 2 NOBr(g)
(slow)
The rate law would be based on the slow step and thusly be
Rate = k[NOBr2][NO]
But since NOBr2 is an intermediate, we do not want it in the rate law. Since a reversible reaction can
be assumed to be the same rate both ways, we can say that
k1[NO][Br2] = k-1[NOBr2]
Substituting the intermediate with its reverse reaction will give us the rate law
Rate = k[NO]2[Br2]
It is always safe to assume that if the initial step is fast, then it is reversible.
14.7 Catalysis
 A catalyst is a substance that speeds up the rate of a chemical reaction without actually being used in
the chemical reaction.
o It may take part in the reaction or even be changed during the reaction but it is recoverable in its
original form once the reaction has taken place.
 A homogeneous catalyst is a catalyst that is in the same phase of matter as the reactants. These are
mostly liquid solutions and gases
o They lower the activation energy of the reaction
 A heterogeneous catalyst is in a different phase of matter than the reactants. Most commonly, the
catalyst is a solid and the reactants are liquids or gases.
o Adsorption is the first step in heterogeneous catalysis and involves the reactants sticking to the
outer surface of the catalyst
*Fig. 14.24 (Pg 591) shows the adsorption method for a heterogeneous catalyst
 An enzyme is a biological catalyst and occurs in living things. It allows for the break down of very
specific substances called substrates
 A catalytic convertor is a piece of equipment in a car’s engine that allows fuel to burn cleaner by
o Oxidizing CO and any unburned hydrocarbon into CO2 and water and
o Reducing nitrogen oxides to nitrogen gas
Additional Notes: