AP Chemistry Chapter 14 – Chemical Kinetics
Chemical Kinetics: Introduction


Chemical kinetics is the study of how fast chemical reactions occur
Four important factors affect rates of reactions:
o Concentration of reactants
o Temperature of reactions
o Presence or absence of a catalyst
o Surface area of solid or liquid reactants or catalysts
Reaction Rates







The speed of a reaction is defined as the change that occurs per unit time
o Often determined by measuring the change in concentration of a reactant or product with
time
o The speed of the reaction is its reaction rate
For a reaction A B
Average rate = change in number of moles of B
change in time
Here the change in moles of B is defined as:
ΔB(moles of B) = (moles of B at final time) – (moles of B at initial time)
Illustrated by the example:
o Suppose A reacts to form B. Let us begin with 1.00 mol A
o At t = 0 there is 1.00 mol A and no B
o At t = 10 min there is 0.74 mol A and 0.26 mol B
o At t = 20 min there is 0.54 mol A and 0.46 mol B
o Can use this data (and others determined at other times) to find the average rate
Average rate = Δ(mol B) = (mol B at t = 10 min) – (mol B at t = 0 min)
Δt
10 min – 0 min
Average rate = 0.26 mol – 0 mol = 0.026 mol
10.0 min – 0 min
min
For the reaction AB there are two ways of measuring rate:
o The rate of appearance of product B (as before)
o The rate of disappearance of reactant A
Average rate = - Δ(mol A)
Δt
Negative sign reminds us that rate is expressed in terms of disappearance of the reactant
*Resulting rates are ALWAYS positive
A plot of moles vs. time shows that as the reactants disappear, the products appear
See Fig. 14.4
Pract. Ex. 1
Rates in Terms of Concentrations
 In most chemical reactions we will determine the reaction rate by monitoring a change in
concentration (of a reactant or product)
o Most useful unit to use for rates is molarity (M)
o Since volume is constant, molarity and moles are directly proportional
 Consider the following reaction:
C4H9Cl + H2O  C4H9OH + HCl
o
o
o
o
We can calculate the average rate in terms of disappearance of C4H9Cl
The units for average rate are mol/L·s or M/s
The average rate decreases with time See Table 14.1 (use neg. sign for disappearance)
We can plot [C4H9Cl] vs. time
See Fig. 14.4
 The rate at any instant in time is called the instantaneous rate
 It is the slope of the straight line tangent to the curve at that instant
Pract. Ex. 2
 Instantaneous rate is different from average rate
 It is the rate at that particular instant in time
 For our purposes we will call the “instantaneous rate” the rate, unless otherwise
indicated
Reaction Rates and Stoichiometry
 For the reaction:
C4H9Cl + H2O  C4H9OH + HCl
o The rate of appearance of C4H9OH must equal the rate of disappearance of C4H9Cl
Rate = -Δ[C4H9Cl] = Δ[C4H9OH]
Δt
Δt
 What if the stoichiometric relationships are not 1 to 1?
o For the reaction:
2HI  H2 + I2
o The rate may be expressed as Rate = -Δ[HI] = 2Δ[H2] = 2Δ[I2] OR
Δt
Δt
Δt
Rate = -1 Δ[HI] = Δ[H2] = Δ[I2]
2 Δt
Δt
Δt
 For general reaction: aA + bB  cC + dD
o The rate may be expressed as:
Rate = -1 Δ[A] = -1 Δ[B] = 1 Δ[C] = 1 Δ[D]
a Δt
b Δt
c Δt
d Δt
Pract. Ex. 3
Concentration and Rate



In general, rates: Increase (decrease) when reactant concentration is increased (decreased)
Generally measure the way the reaction rate at the beginning of a reaction depends on the starting
conditions
o Consider the reaction:
NH4+ + NO2-  N2 + 2H2O
o We measure initial reaction rate, which is the instantaneous rate at t = 0
o We find this at various initial concentrations of each reactant See Table 14.2
o As [NH4+] doubles, with [NO2-] constant, the rate doubles
o So rate is proportional to [NH4+]
o As [NO2-] doubles, with [NH4+] constant, the rate doubles
o So rate is also proportional to [NO2-]
The overall concentration dependence of reaction rate is given by a rate law (rate expression)
o Example: rate law is
Rate = k[NH4+] [NO2-]
o Proportionality constant, k, is the rate constant (same for any set of data in example, but is
temperature dependent)
o Once the rate law and rate constant are known, they can be used to calculate initial reaction
rates under any set of initial concentrations
Exponents in the Rate Law (Reaction Order)
 For a general reaction with rate law
Rate = k[reactant 1]m [reactant 2]n
o The exponents m and n are called reaction orders
o The overall reaction order is the sum of the reaction orders (m + n +…)
o For the previous reaction: NH4+ + NO2-  N2 + 2H2O
o The reaction is said to be first order in [NH4+], first order in [NO2-], and second order
overall
o *Note: reaction orders are determined experimentally and do not necessarily correspond to
the stoichiometric coefficients in the balanced equation
o Common to have reaction orders of 0, 1, 2 or even fractional or negative values
Pract. Ex. 4
Units of Rate Constants (very important)
 Units of the rate constant depend on the overall reaction order
 Example: for a reaction that is overall second order
Units of rate = (units of k)(units of concentration)2
M/s
= (units of k)(M)2
Units of k = M/s
M2
Units of k = 1 .
or M-1s-1
Ms
 Units of rate are always M/s
Pract. Ex. 5
Using Initial Rates to Determine Rate Laws
 To determine the rate law, we observe the effect of changing initial concentrations
 For example:
o If changing the initial concentration of a reactant has no effect on rate, the reaction order
for that reactant is zero; the reactant will not appear in the rate law
o If doubling the concentration doubles the rate, the reaction is 1st order for that reactant
o If doubling the concentration causes a 22 increase in rate, the reaction is 2nd order for that
reactant
o Similarly, if tripling the concentration causes a 32 increase in rate, the reaction is 2nd order
for that reactant
 A quick way to figure out the reaction orders (exponents) for the rate laws is to use the following
equation: yx = z
where
y = what is done to concentration of reactant
x = reaction order for that reactant (unknown)
z = what happens to rate
o Examples:
If doubling the concentration is found to double the rate,
2x = 2 so x = 1 (1 becomes the exponent for the reactant in the rate law)
If doubling the concentration causes the rate to be cut in half,
2x = ½
so x = -1 (-1 becomes the exponent for the reactant in the rate law)
o To solve for x in any case: take the log of both sides
Example:
2x = ½
log 2x = log ½
x log 2 = log ½
x = log ½
log 2
so x = -1
o Note: the rate (not the rate constant) depends on concentration; both depend on
temperature and the presence/absence of catalysts (defined later)
Sample Ex. 6 and Pract. Ex. 6
The Change of Concentration with Time – need to express concentration as a function of time
First-Order Reactions
 For a first-order reaction, the rate is proportional to the concentration of a reactant, raised to the
first power
 For a reaction that is first order for a reactant, A: rate = -∆[A] = k[A]
∆t
 By using calculus integration (do not need to understand how) we get:
ln[A]t – ln[A]0 = -kt where [A]t is the concentration of A at time t
[A]0 is the initial concentration of A
k is the rate constant
t is the total time
rearranging:
ln[A]t = -kt + ln[A]0
(similar to y = mx +b)
 Plotting ln[A]t vs. t gives a straight line with slope –k and intercept ln[A]0 See Fig. 14.7
alternative form: ln [A]t = -kt
[A]0
may also see an equivalent form:
log [A]t = -kt .
[A]0 2.303
Pract. Ex. 7
Half-Life
 Half-life (t1/2) is the time required for the concentration of a reactant to decrease to half its original
value
 i.e. it is the time required for [A]0 to reach ½ [A]0
ln [A]t = -kt
[A]0
so, for t = t1/2 and [A]t = ½ [A]0
ln ½ [A]0 = -kt1/2
[A]0
ln ½ = -kt1/2
t1/2 = - ln ½ = 0.693
k
k
 Note: half-life of a first-order reaction does not depend of the initial concentration of the reactant
Pract. Ex. 9
Second-Order Reactions
 A second-order reaction is one whose rate depends on the reactant concentration to the second
power or on the concentration of two reactants, each raised to the first power
 For a reaction that is second-order for a particular reactant, A
Rate = - ∆[A] = k[A]2
∆t
Through integration:
1 = kt + 1 .


[A]t
[A]0
A plot of 1/[A]t vs. t is a straight line with slope k and intercept 1/[A]0
o For second-order reactions, a plot of ln[A]t vs. t is NOT linear
Can also show that half-life for a second-order reaction is:
t1/2 = 1 .
k[A]0
Temperature and Rate


Most reactions speed up as temperature increases
See Fig. Rate approx. doubles with each 10°C increase
Collision Model
 Rates are affected by concentration and temperature
 Molecules must collide to react
o More collisions  faster rate
o More molecules (i.e. higher concentration)  more collisions
o Higher temp  more frequent collisions
 Molecules must also collide in the correct orientation toward one another and with enough energy
to form products
 The collision model explains reaction rates in terms of frequency of collisions, the minimum
energy requirements of the collisions and the probability that collisions occur with suitable
orientations
Activation Energy
 Arrhenius: molecules must possess a minimum amount of energy to react. Why?
o To form products, bonds must be broken
o Bond breakage requires energy (can come from Ek)
 Activation energy (Ea) is the minimum energy required to cause a reaction See Fig.
 Consider the rearrangement of methyl isonitrile to form acetonitrile (p.)
o Energy is required to stretch the bond between the CH3 group and the N≡C group to allow
the N≡C to rotate
o The C−C bond begins to form
o The energy associated with the molecule drops
o The energy barrier between the starting molecule and the highest energy state found along
the reaction pathway is the activation energy
o The species at the top of the barrier is called the activated complex (transition state)
o The change in energy for the reaction is the difference in energy between CH3NC and
CH3CN
Fig. 14.
o ΔErxn has no effect on reaction rate
o The activation energy is the difference in energy between the reactants (CH3NC) and the
transition state
o The rate depends on the magnitude of Ea (lower Ea general means faster rate)
 Note: If a forward reaction is exothermic, then the reverse reaction is endothermic
 How does this relate to temperature?
o At any particular temperature, the molecules present have an average kinetic energy
associated with the population
o In the same distribution, some molecules have less energy than the average while others
have more
o The fraction of molecules with an energy equal to or greater than the ΔEa is given by:
f = e-Ea/RT
o Molecules that have an energy equal to or greater than ΔEa have sufficient energy to react
o As we increase the temperature, the fraction of the population that has an energy equal to
or greater than ΔEa increases; so more molecules react Fig. 14.
The Orientation Factor
 The orientation of a molecule during collision can have a profound effect on whether or not a
reaction occurs
 Consider the reaction: Cl + NOCl  NO + Cl2
o If the Cl collides with the Cl of NOCl, the products are Cl2 and NO
o If the Cl collides with the O of NOCl, no products are formed
The Arrhenius Equation
 Arrhenius discovered that most reaction-rate data obeyed an equation based on three factors:
number of collisions per time, fraction of collisions that occur with the correct orientation, fraction
of colliding molecules that have energy equal to or greater than ΔEa
 He developed the Arrhenius equation:
k = Ae-Ea/RT where k is the rate constant
Ea is the activation energy
R is 8.314 J/mol∙K
T is the Kelvin temperature
A is the frequency factor – relates the frequency of collisions and
the probability that a collision will have a favorable
orientation
o Both A and Ea are specific to a given reaction
o Taking the ln of both sides and rearranging the equation:
ln k = ln A + ln e -Ea/RT
ln k = - Ea + ln A
RT
o For two different temperatures:
ln k1 = - Ea + ln A
ln k2 = - Ea + ln A
RT1
RT2
ln k1 – ln k2 = (-Ea + ln A) – (-Ea + ln A)
RT1
RT2
ln k1/k2 = -Ea/RT1 + ln A +Ea/RT2 –ln A
ln k1/k2 = Ea/RT2 – Ea/RT1
ln k1/k2 = Ea/R (1/T2 – 1/T1)
Pract. Ex.
Reaction Mechanisms


The balanced chemical equation provides information about substances present at the beginning
and end of the reaction
The reaction mechanism is the process by which the reaction occurs (which bonds are broken and
formed)
Elementary Steps
 Elementary steps are any processes that occur in a single step
 The number of molecules present in an elementary step gives the molecularity of that step
o Unimolecular steps have one molecule
o Bimolecular steps have two molecules
o Termolecular steps have three molecules (very rare)
 A multistep mechanism consists of a sequence of elementary steps
o Elementary steps must add together to give the balanced chemical equation
o Some multistep mechanisms include intermediates
 These are species that appear in an elementary step but are neither reactants nor
products
 They are formed in one step and consumed in a subsequent step
 They do not appear in the overall balanced reaction
Rate Laws of Elementary Steps
 The rate laws of elementary steps determine the overall rate law of the reaction
 The rate law of an elementary step is determined by its molecularity (stoichiometry of the step)
o Unimolecular processes are first order
o Bimolecular processes are second order
o Termolecular steps are third order (rare)
Rate Laws of Multistep Mechanisms
 Most reactions occur via multistep mechanisms
 Often one step (rate-determining step) is slower than the others
 This step determines the rate law for the overall reaction Fig. 14.
 Consider the reaction:
NO2 + CO  NO + CO2
o The experimentally derived rate law is: rate = k[NO2]2
o A mechanism is given for the reaction:
Step 1:
NO2 + NO2 ---k1---> NO3 + NO slow step
Step 2:
NO3 + CO ---k2---> NO2 + CO2 fast step
o Note: NO3 is an intermediate
o If k2 >> k1 then the overall reaction rate will depend on the first step (the rate-determining
step)
o So, the rate law:
rate = k[NO2]2
would agree with the exp. rate law
o This supports (but does not prove) the mechanism
Mechanisms with an Initial Fast Step
 Consider the reaction:
2NO + Br2  2NOBr
o Exp. rate law found to be: rate = k[NO]2[Br2]
o A mechanism is proposed for the reaction:
Step 1:
NO + Br2 ---k1---> NOBr2
fast step
o
o
o
o
o
<---k-1--Step 2:
NOBr2 + NO ---k2---> 2NOBr slow step
The theoretical rate law for this mechanism based on the slow step is:
Rate = k2[NOBr2][NO]
Problem: this rate law depends on the concentration of an intermediate
Intermediates are usually unstable and have low/unknown concentrations
Can express [NOBr2] in terms of NOBr and Br2 by assuming that there is an equilibrium in
the fast step
Forward rate would equal the reverse rate:
K[NO][Br2] = k-1[NOBr2]
Rearranging:
[ NOBr2] = k1/k-1 [NO][Br2]
Overall rate law becomes:
rate = k2 k1/k-1 [NO][Br2][NO = k[NO]2[Br2]
Note: final rate law consistent with exp. rate law
Catalysis


A catalyst is a substance that changes the rate of a chemical reaction without itself undergoing a
permanent chemical change
Two types of catalysts: homogeneous and heterogeneous
Homogeneous Catalysis
 Homogeneous catalyst – one that is present in the same phase as the reacting molecules
 For example: decomp of hydrogen peroxide occurs very slowly
2H2O2  2H2O + O2
o With a bromide catalyst, it occurs rapidly in acidic solution
2Br- +H2O2 +2H+  Br2 + 2H2O
Br2 + H2O2  2 Br- +2H+ + O2
o Br- is a catalyst because it is regenerated at the end of the reaction
o Net reaction is still: 2H2O2  2H2O + O2
 How do catalysts increase reaction rates?
o Catalysts generally operate by lowering the Ea
o Can also operate by increasing the number of effective collisions
 i.e. increase k by increasing A or decreasing Ea (see Arrhenius eqn)
o A catalyst usually provides an alternate mechanism
o When a catalyst adds an intermediate, the activation energies for both steps must be lower
than the activation energy for the uncatalyzed reaction
Fig. 14.
Heterogeneous Catalysis
 Heterogeneous catalyst – one that is in a different phase than the reaction
 How do they work?
o First step is adsorption – the binding of reactant molecules to the catalyst surface
o Molecules are adsorbed to active sites on the catalyst surface
 Read about hydrogenation of ethylene to form ethane
Enzymes
 Enzymes are biological catalysts
 Usually large protein molecules
 Very specific for specific reactions
 Works in a lock-and-key-model
 Substances that undergo reaction at the active site on enzymes are called substrates

o Substrates fit into active site like a lock-and-key; forms the enzyme-substrate complex
Enzyme serves as a favorable environmental surface on which reactions can take place