Chapter 14
Chemical Kinetics
*concerned with speed or rates of chemical
reactions
reaction rate- the speed at which a chemical
reaction occurs
reaction mechanism- step-by-step pathway from
reactants to products
Factors That Affect Reaction Rates
1. Physical state of the reactants
-if reactants are in different phases, the reaction
is limited by the area of contact
*inc surface area of solid to inc rate
2. Reactant concentrations
*inc conc. = inc collisions = inc rate
3. Reaction temp
*inc temp = inc KE/collisions = inc rate
4. Presence of a catalyst
*inc rate without being used up
rate of appearance of product = ∆ [product]
∆ time
[ ] = concentration
rate units = M/s
*conc. of reactants dec with time while the conc.
of product inc
*rate of disappearance of reactants = rate of
appearance of product
rate of disappearance of reactants = -∆ [reactant]
∆ time
Sample Exercise 14.1 page 560
Using data in figure 14.3 pg 559, calculate the
rate at which A disappears over the time interval
from 20s to 40s.
rate of disappearance = -∆ [reactant]
∆ time
rate = -(0.30M-0.54M)
(40s-20s)
= 0.012M/s
Practice Exercise
rate of product = (0.70M - 0M)
(40s - 0s)
= 0.018M/s
-it is typical for rates to dec. over time
WHY?
-the conc. of reactants dec.
page 561 Table 14.1 and Fig 14.4
instantaneous rate- rate at a particular instant during
a reaction
-determined from the slope of the curve at a
particular point in time
-tangent lines are drawn to find the rate
initial rate- instantaneous rate at t=0
Page 561 figure 14.4
*Problems page 562
Rate = -∆ [reactant] = ∆ [product]
∆t
∆t
aA + bB → cC + dD
-a,b,c,d are the coefficients from the balanced
equation
rate = -1 ∆[A] = -1 ∆[B] = +1 ∆[C] = +1 ∆[D]
a ∆t
b ∆t
c ∆t
d ∆t
Rate Law
-relationship between the rate of the reaction and
the conc. of the reactants
-consider this reaction: aA + bB → cC + dD
-the rate law would be:
rate = k[A]m[B]n
k = rate constant
m and n = reaction order (small whole numbers)
*the value of m and n determines how the rate
depends on the conc. of the reactant- can only
be found experimentally
-to find reaction order, look at two values of a
reactant that do not change
-see what happens to the rate when the other
reactant is changed
*if the rate change is the same as the conc. change,
it is first order (21, 31, 41)
ex: conc. doubles, rate doubles
*if the rate change is the square of the conc. change,
it is second order (22, 32, 42)
ex: conc. doubles, rate quadruples
*if the rate change does not change as the conc.
changes, it is zero order
-do the same holding each reactant constant
Example: look at page 563 Table 14.2
*find order of each reactant
*m and n are both first order based on data
rate = k [NH4+]1 [NO2-]1
*solve for k
*using experiment #1 data
5.7 x 10-1M = k(0.0100M)(0.200M)
k = 2.7x10-4/M∙s
*can use rate law and k to calculate rate for any
set of conc.
ex:
*find rate of same reaction if each conc is
0.100M
rate = 2.7x10-4/M∙s(0.100M)(0.100M)
= 2.7x10-6M/s
-a large value of k (≥ 109) means a fast reaction
-a small value of k (10 or lower) means a slow
reaction
-units of k change depending on overall order of
reaction
zero order = Ms-1
1st order = s-1
2nd order = M-1s-1
-overall order of reaction found by adding
together orders of each reactant
First Order Reactions
-rate is directly proportional to a single reactant
rate = -∆ [A] = k[A]
∆t
Integrated Rate Law
-relationship between initial conc. of reactant and
conc. at any other time
ln[A]t = -kt + ln[A]0
y = mx + b
*m= slope = -k
*has form of general equation for a straight line
*if when t vs. ln[A]t is graphed and a straight line is
produced then it is first order
Second-Order Reactions
-rate is proportional to the square of the [A] or to
reactants each raised to first power
rate = -∆ [A] = k[A] 2
∆t
-more sensitive to conc. of the reactants than first
order
Integrated Rate Law
1/[A]t = kt + 1/[A]0
[A]t = 1/ (kt + 1/[A]0)
*if when t vs. 1/[A]t is graphed and a straight line is
produced then it is second order
Zero-Order Reactions
-rate of reaction is independent of the [A]
rate = -∆ [A] = k
∆t
-rate is the same at any concentration
Integrated Rate Law
[A]t = -kt + [A]0
*if when t vs. [A]t is graphed and a straight line
is produced then it is zero order
Half-life
-time required for the conc. of a reactant to fall
to one half of its initial value
Zero-Order
t1/2 = [A]0 /2k
First-Order
t1/2 = 0.693/k
Second-Order
t1/2 = 1/ k[A]0
Temperature and Rate
*as temp inc. rate inc.
Ex- glow sticks
collision model- molecules must collide in order
to react
-the greater the # of collisions/sec the greater the
reaction rate
-inc temp and conc of reactants inc the collisions
which inc the rate
-for a reaction to occur more is required than just
collisions
-molecules must be oriented in a certain way to
make sure they are positioned to form new bonds
-page 576 figure 14.15
-molecules must also possess a certain amount of
energy to react
-this energy comes from collisions
-KE is used to stretch, bend and break bonds
activation energy/energy barrier- minimum
amount of energy required to initiate a chemical
reaction (Ea)
*as Ea inc. reaction rate dec.
activated complex/transition state- a high energy
intermediate step between reactants and products
-page 577 figure 14.17
-for reverse reaction Ea = ∆E + Ea
*must change sign on ∆E because going from right
*Ea will dec. with inc. temp
-page 578 figure 14.18
-the fraction of molecules that have an energy ≥ Ea
is given by:
f = e-Ea /RT
Arrhenius found that most reaction data obeyed
an equation based on:
a) the fraction of molecules possessing energy
Ea or greater
b) # of collisions per second
c) the fraction of collisions that have the
appropriate orientation
Arrhenius Equation: k = Ae-Ea /RT
A= frequency factor related to frequency of
collisions and the probability that the collisions
are favorably oriented
Sample Exercise 14.10 page 579
Rank slowest to fastest
2<3<1
Ea=
#2= 25kJ/mol #3= 20kJ/mol #1= 15kJ/mol
Practice Exercise
Reverse from slowest to fastest
2<1<3
Ea= ∆E + Ea *change sign on Ea b/c reversing
#2= 40kJ/mol #1= 25kJ/mol #3= 15kJ/mol
Reaction Mechanisms
-the steps by which a reaction occurs
Elementary Reactions
-when a reaction occurs in a single event or step
molecularity- defined by the # of reactants in an
elementary reaction
unimolecular- single reactant is involved
bimolecular- collision of two reactants
termolecular- collision between three molecules
(not as probable as uni or bi)
Multistep Mechanisms
-sequence of elementary equations
ex- page 582
-the elementary reactions in a multistep
mechanism must always add to give equation of
overall process
intermediate- not a reactant or a product, formed
in one elementary reaction and consumed in the
next
Sample Problem 14.12 page 582
a) molecularity = uni and bi
b) equation for overall reaction
2O3(g)  3O2(g)
c) identify intermediates
O(g)
Practice Exercise
a) yes it is consistent b/c both equations add to
give overall reaction
b) molecularity = uni and bi
c) intermediates = Mo(CO)5
Rate Laws for Elementary Reactions
-rate law is based on molecularity for elementary
reactions
*table 14.3 page 584
Sample Problem 14.13 page 584
Multistep Mechanisms
-each step in a mechanism has its own rate
constant and Ea
-often one step is much slower than the others
-the overall rate cannot exceed the rate of the
slowest elementary step
rate determining step/rate limiting step- limits
overall reaction rate
-determines the rate law
Mechanisms with slow initial step
*pg 585
-since step 1 is slow it is rate-determining step
-rate of overall reaction depends on rate of step 1
-rate law equals rate law of step 1 (k1)
-step 1 is bimolecular so using page 584 it is
second order
rate = k1[NO2]2
Mechanisms with fast initial step
*pg 586 and 587
-step 1 has forward (k1) and reverse (k-1) reaction
-step 2 is slow and determines overall rate
rate = k2[NOBr2][NO]
-NOBr2 is an intermediate (unstable, low conc.)
-rate law depends on unknown conc.
-not desirable
-want to express rate law in terms of reactants
and products