AP Chemistry - Notes - Chapter 14 - Kinetics
Chapter 14 outline : Chemical kinetics
A. Chemical Kinetics - chemistry of reaction rates
1. Reaction Rates
a. Reaction rate- the change in concentration of reactant or product per unit of time
- always positive (absolute change per unit of time)
- units are mol/L • s (or Ms-1)
- the instantaneous rate (rate at a particular time) can be determined by calculating the slope of a
line tangent to that point in time
For the reaction A  B, a reaction progress curve might look like :
[A]
a.
b.
c.
t1
t2
Time (s)
The rate of the reaction is -Δ[A] / Δt (negative Δ[A] because A is being used up as it changes to B)
- final value will be positive (absolute value)
- the slope of the line tangent to point a. is the initial rate t ≈ 0
- the slope of the line tangent to point b. is the rate at time = t1
- the slope of the line from a. to c. is the average rate from t = 0 to t = t2
- the stoichiometry of the equation determines the relative rates of consumption of reactants or
production of products (e.g. in the reaction 2H2O  2 H2 + O2, water is used up at the same
rate hydrogen is produced, per mole, and oxygen is produced at half the rate of hydrogen)
Sample problems :
1. The reaction for the Haber process used for the production of ammonia is
N2(g) + 3 H2(g)  2 NH3(g).
a. Compare the rate at which the ammonia will be produced to the rate at which nitrogen and
hydrogen will be consumed.
Answer : Ammonia will be produced at twice the rate at which nitrogen is consumed due to the
mole ratio of N2 to NH3 in the balanced equation which equals 1:2. Ammonia will be produced
at two-thirds the rate at which hydrogen is consumed due to the mole ratio of H2 to NH3 in the
balanced equation which equals 3:2
b. If hydrogen gas is used up at a rate of 24.0 L/s, what is the rate at which ammonia is produced?
Answer : 16.0 L/s
Solution : Ammonia is produced at a 2/3 rate of hydrogen so : 2/3 x 24.0 L/s = 16.0 L/s
AP Chemistry - Notes - Chapter 14 - Kinetics
2. Rate Laws : An Introduction
a. Reactions are reversible, therefore the rate of the forward reaction equals the difference in the
forward and reverse rates. If we examine a reaction in which only the reactants are present, the
reverse reaction can be ignored. For the following reaction : 2NO2  2NO + O2, then becomes :
Rate = k[NO2]n , where is the rate constant and n is the order of the reactant (n must be determined
experimentally and may be an integer (including zero) or a fraction).
- Care must be taken on how the rate is defined. In the reaction above is the rate determined by
how much nitrogen dioxide is used up?, How much oxygen is produced? These will be different
due to the mole ratio in the balanced equation.
- rate is dependent on concentration
- k is independent of concentration (k is temperature dependent)
b. Types of rate laws :
- Differential rate law - expresses how the rate depends on concentration
- Integrated rate law - expresses how the concentrations of species depend on time
- which rate law is used is based on the types of information obtained from a reaction or what
types of information are needed
- since the two types of rate laws are related in a defined way, only one of the rate laws for a
reaction need to be determined
-The importance of the rate law is that it reveals information about the steps in which a reaction
occurs
3. Determining the Form of the Rate Law
a. Determining the order of a reaction :
- The order of a reactant is the power to which the concentration of the reactant must be raised to
produce the amount of product indicated by the reaction. observe the following data for the
reaction :
aA + bB ------> cC + dD
Initial [A]
.02
.04
.04
Initial [B]
.01
.01
.02
Initial rate(mol/L-s)
.003
.012
.024
- Comparing the first two lines of data we see that [A] is doubled (.02 to .04) and the reaction rate
is increased by a factor of four(.003 to .012).
line 1 =
line 2
.003
k(.04)x(.01)y = .04x
=
.012
k(.02)x(.01)y
.02x
, 4 = 2x , x = 2
The order of reactant A is therefore 2, or second order (22=4). If [B] is doubled the reaction rate
is doubled and it is therefore a first order reactant (21=2). the overall reaction order is the sum of
each of the orders of the reactants and is therefore a third order reaction. The rate law equation
for this reaction can then be written as
rate = k[A]2[B]
- Here is another example: Determining the orders of each reactant and the overall reaction rate for
the following reaction.
aA + bB + cC -----> dD + eE
Data :
Initial [A]
Initial [B]
Initial [C]
Initial rate(mol/L-s)
.10
.06
.20
6.400
.05
.06
.20
3.200
.05
.03
.20
.800
AP Chemistry - Notes - Chapter 14 - Kinetics
.05
.03
.10
.200
Order of A is 1 : [A] is cut in half (lines 1-2) and the reaction rate is cut in half (½ )1=½
Order of B is 2 : [B] is cut in half (lines 2-3) and the reaction rate is cut to one-fourth (½)2=¼
Order of C is 2 : [C] is cut in half (lines 3-4) and the reaction rate is cut to one-fourth (½)2=¼
Overall order is 5 : equals the sum of the orders of each reactant
- Writing the equation for the rate law for this reaction we get : rate = k[A][B]2[C]2
b. Calculation of k : k can be determined by rearranging the rate law to solve for k using each
line of data and then averaging the values
- units for k : Units of the rate constant depend on the overall order of the reaction. These are
the units necessary to cancel properly and give the rate in units of Ms–1 ( mol/L • s).
Overall Order of
Reaction
0
1
2
3
Units of k
Ms –1 (or mol/L • s)
s–1 ( or 1/s)
M –1s–1 (or L/mol • s)
M –2s–1 ( or L2/mol2 • s)
4. The Integrated Rate Law - expresses the concentration of the reactant as a function of time
a. first order reactions : for the reaction : aA  products, the integrated first-order rate law is :
ln[A] = -kt +ln[A]0
where [A] is the concentration of A at time t, [A]0 is the initial concentration of A
- this allows us to calculate [A] at any time if t and k are known
- it is an equation for a straight line (y= mx + b) where the slope = -k
To determine which
- a plot of ln[A] vs t gives a straight line
order a reactant is the
- a useful derivation is ln ([A]0/[A]) = kt
data has to be plotted to
- half-life of a first-order reaction (t½) (half life depends on k alone)
see which type of data
- equation : t½ = 0.693/k
gives a straight line.
b. Second-order rate laws
-second-order rate law :
1/[A] = kt + 1/[A]0
- a plot of 1/[A] will produce a straight line with the slope equaling k
- t½=1/k[A]0 (half life depends on k and[A]0 and the length of each half life is twice the half-life
before it)
Sample problem : The following data were obtained for the reaction of the decomposition of nitrogen
dioxide in the gas phase at 300 ºC, NO2(g)  NO(g) + ½O2(g)
Time (s)
[NO2] (M)
0.0
0.01000
50.0
0.00787
100.0
0.00649
200.0
0.00481
300.0
0.00380
Determine the rate law for this reaction and the value of k.
Solution : To determine the order of NO2, we need to plot ln[NO2] and 1/[NO2] vs t to determine if it is
first or second order. Calculating the values we get :
AP Chemistry - Notes - Chapter 14 - Kinetics
(Sample problem cont'd)
Time
0.0
50.0
100.0
200.0
300.0
[NO2]
0.01000
0.00787
0.00649
0.00481
0.00380
ln[NO2]
-4.610
-4.845
-5.038
-5.337
-5.573
1/[NO2]
100
127
154
208
263
Constructing the plots we would get :
ln[NO2]
1/[NO2]
Time (s)
Time (s)
Since the plot of 1/[NO2] is linear the reaction is second order with regards to [NO2] giving the rate
law of Rate = k[NO2]2. Calculating the slope would give a value of 0.543. For a second
order reaction k = slope and therefore k = 0.543 L/mol • s
Sample problems : Half-life
1. A certain first-order reaction is 35.0% complete in 45.0 s. What are the values of the rate constant and
half-life for this reaction?
Solution : Use the derivation ln ([A]0/[A]) = kt to solve for k . Assume 100.0 for [A]0 , and then at 45.0
s when the reaction is 35.0% complete [A] = 65.0.
ln (100.0/65.0) = k(45.0s), k = 9.57 x 10-3L/mol • s
The equation for the half-life of a first-order reaction is t½ = 0.693/k
Solving t½ = 0.693/9.57 x 10-3L/mol • s = 72.4 s
2. In a second-order reaction it took 112 s for 50.0% of a substance to decompose. If the initial
concentration was .20 M, what is the rate constant for this reaction?
Solution : For a second-order reaction t½ = 1/k[A] 0 or k = 1/t½[A]0
Solving : k = 1/(112s • .20 M) = 4.5 x 10-2 L/mol • s
c. Zero order rate laws : - rate is constant and does not change with changes in concentration
-zero-order rate law
[A] = -kt + [A]0
- zero-order reactions usually involve a metal surface as a catalyst or an enzyme (these factors
control reaction rate rather than concentration)
- a plot of [A] vs t would give a straight line with the slope = -k
- t½ = [A] 0 /2k
AP Chemistry - Notes - Chapter 14 - Kinetics
d. Integrated Rate Laws for Reactions with More than One Reactant
- can be studied one reactant at a time by keeping the reactant concentration of the reactant in
question small compared to other reactants much larger concentrations can be considered
constant in comparison to the small concentration of the reactant in question)
5. Summary of rate laws
Summary of Kinetics for a zero, first or second order reaction of [A] in which A  Products
Order :
0
Rate Law
Integrated Rate Law
Plot needed for a
straight line
Relationship of rate
constant to the slope
of straight line
Half-life
1
2
Rate = k
[A]0 - [A] = kt
[A] vs t
Rate = k[A]
ln[A] 0 - ln[A] = kt
ln[A] vs t
Rate = k[A]2
1/[A] - 1/[A] 0 = kt
1/[A] vs t
Slope = k
Slope = k
Slope = k
t½ = [A] 0 /(2k)
t½ = 0.693/k
t½ = 1/(k[A] 0)
6. Reaction Mechanisms - the series of steps by which a reaction occurs
- the balanced equation tells us what we start and end with, the reaction mechanism tells us what's
in between
- the reaction mechanism must agree with the experimentally determined rate law
- the sum of the elementary steps must give the overall balanced equation for the reaction
a. intermediate - a species that is formed and then changed in a reaction mechanism - it is neither a
reactant nor a product
b. elementary step - a reaction whose rate law can be written from its molecularity (the number of
species that must collide to produce the reaction indicated by that step)
- unimolecular step - involves only one molecule
- bimolecular step - involves two species - always second order (e.g. k[A]2 for a single reactant or
k[A][B] for two reactants)
- termolecular step - involves three species (rare)
Elementary step
A  products
2A products
A + B  products
2A + B  products
A + B + C  products
Molecularity
Unimolecular
Bimolecular
Bimolecular
Termolecular
Termolecular
Rate Law
Rate = k[A]
Rate = k[A]2
Rate = k[A][B]
Rate = k[A]2[B]
Rate =k [A][B][C]
Sample Problem : Write the rate laws for the following elementary reactions :
a. A(g)  B(g)
b. A(g) + B(g)  C(g)
c. 2A + B  2C(g)
Solution : In elementary reactions the rate law can be determined using the coefficients in the balanced
equation as the order of each reactant.
a. Rate = k[A]
b. Rate = k[A][B]
c. rate = k[A]2[B]
AP Chemistry - Notes - Chapter 14 - Kinetics
- the rate determining step - the slowest step of the reaction mechanism which determines the
overall speed of the reaction
- reaction mechanisms can never be absolutely proven
Sample problem : From the equations below which describe the two elementary steps of a process by
which a certain chemical reaction occurs, write the rate law and the balanced chemical equation for
the overall reaction.
Step 1 :
Step 2 :
A + B  C + D (slow step)
C  E + F (fast step)
Solution The rate law can be written from the rate determining step (slow step) : Rate = k[A][B]
The balanced chemical equation can be determined by adding the two reaction and canceling out like
terms on the reactant and product sides.
A + B  C + D
+
C  E + F____
= A + B  D + E + F
7. A Model for Chemical Kinetics
a. Collision Model - molecules must collide in order to react
- Factors affecting the effectiveness of collisions :
- temperature
- concentrations of reactants
- nature of reaction (e.g. states of reactants)
- presence of catalysts
b. activation energy - energy needed to overcome the energy barrier - the energy needed to break the
bonds of the reactants
- kinetic energy becomes the potential energy - energy stored in bonds
c. activated complex (transition state) - the temporary combination of reactant species at the top of the
energy barrier
- number of collisions with activation energy = (total number of collisions)e-Ea/RT
Ea = activation energy
R- universal gas constant
e-Ea/RT - fraction of collisions with energy Ea or greater at temperature T
d. molecular orientation - molecules must collide with proper orientation to each in order for the
collision to be effective (assuming enough energy in collision)
d. In order for a collision to be effective it must meet both of the above criteria (enough energy and
proper orientation)
- Arrhenius equation : k = Ae-Ea/RT
where : A is the frequency factor
- taking the natural log of the Arrhenius equation : ln(k) = (- Ea/R)(1/T) + ln(A)
- to determine the activation energy :
ln(k2/k1)=(Ea/R)(1/T1-1/T2)
8. Catalysis
a. enzymes -organic catalysts (substances which speed up a reaction without being changed or used
up)
- catalysts work by lowering activation energy (provide alternate pathway (mechanism))
- homogeneous catalysts - same phase as reacting molecules
- heterogeneous catalysts - different phase than reacting molecules (usually a solid)
AP Chemistry - Notes - Chapter 14 - Kinetics
b. Heterogeneous catalysts - catalysts is in a different phase (state) as reactants
- adsorption - the collecting of one substance on the surface of another substance
- common heterogeneous catalysts -platinum, palladium, and transition metal oxides
- typical heterogeneous catalysis involves four steps :
- adsorption and activation of reactants
- migration of adsorbed reactants on the surface
- reaction of adsorbed substances
- escape or desorption of products
- examples of heterogeneous catalysis -hydrogenation of oils, conversion of sulfur dioxide to
sulfur trioxide, catalytic converters
c. Homogeneous catalysis - catalyst is in the same phase as reactants
-example- NO and freons (chlorofluorocarbons) in the upper atmosphere causes the destruction of
ozone