Transcript Chapter 13 - Richsingiser.com

Daniel L. Reger
Scott R. Goode
David W. Ball
http://academic.cengage.com/chemistry/reger
Chapter 13
Chemical Kinetics
Kinetics
• Kinetics is the study of the rates of
chemical reactions.
• Rate is the change of concentration
(c) per unit time (t):
c
rate 
t
Rate of Reaction
•
•
Square brackets are used to denote
molar concentration.
Rate is expressed either as rate of
appearance of product or rate of
disappearance of reactant.
c products 
reactants 
Rate 


t
t
t
•
Rate has units of M/s or molar/s or
mol/(L·s).
Reaction Rate
Average Rates
•
•
An average rate is a change in
concentration measured over a
non-zero time interval.
Average rates are not very useful
because they depend on the
starting and ending times.
Instantaneous Reaction Rate
•
The instantaneous rate of the reaction
is equal to the slope of the line drawn
tangent to the curve at time t.
These graphs show how to determine the instantaneous
rate at 10.0 s.
Rate and Reaction Stoichiometry
•
•
The relative rates of consumption of
reactants and formation of products
depend on the reaction stoichiometry.
For the reaction
2HBr (g)  H2 (g) + Br2 (g)
two moles of HBr are consumed for every
one mole of H2 that is formed so rate of
change of [HBr] is double that of [H2].
1 HBr  H2 
 

2
t
t
Rate and Reaction Stoichiometry
•
For any reaction
aA+bBcC+dD
the reaction rate is given by:
1 A 
1 B
rate   
 
a t
b t
1 C
1 D




c t
d t
•
Note the signs as well as the
coefficients.
Example Problem
For the reaction
5H2O2 + 2MnO4- + 6H+ → 2Mn2+ + 5O2 + 8H2O
the experimentally determined rate of
disappearance of MnO4- is 2.2 x 10-3 M/s.
(a) Calculate the reaction rate.
(b) What is the rate of appearance of O2?
Relating Rate and Concentration
• Experimental rate law: analysis of
many experiments shows that the
rate of a reaction is proportional to
the product of the concentrations of
the reactants raised to some power.
• For a reaction aA + bB  products
the rate law is the equation:
x
y
rate = k[A] [B]
Rate Law
rate = k[A]x[B]y
• x and y are the orders of the
reaction in [A] and [B] respectively.
• The overall order of the reaction is
x + y.
• x and y are usually small integers,
but may be zero, negative, or
fractions.
• k is the specific rate constant.
Relating Rate and Concentration
• The reaction orders are determined by
noting the effect that changing the
concentration of each reactant have on
the rate.
• The rate constant, k, is evaluated once
the orders in the rate law are known.
Relating Rate and Concentration
• Most often you will be given initial
concentrations and rates and asked to
determine the order, which is the
exponent to which concentration is
raised.
• One way to learn to predict the order is
to predict the rate given the order and
concentration and see the relationship.
Dependence of Rate on Order
First order rate law
rate = k[conc]1
Second order rate law
rate = k[conc]2
Zero order rate law
rate = k[conc]0
[Concentration]
[Rate]
1
2
3
1
2
3
1
2
3
1
2
3
1
4
9
1
1
1
Initial Rate Method
• The initial rate method measures the
time during which a known small fraction
of the reactants are consumed.
• Experiments are performed in which initial
concentrations of [A] and [B] are individually
varied.
• The time period of measurement is small
enough that the measured rate is
approximately equal to the instantaneous
rate.
Initial Rates
•
Initial rates are given below for the
reaction
F2 + 2ClO2  2FClO2
Determine the rate law and rate constant
Trial
Init. conc. Init. conc.
[F2], M
[ClO2], M
Init. Rate
Ms
1
0.10
0.010
1.2x10-3
2
0.10
0.040
4.8x10-3
3
0.20
0.010
2.4x10-3
Initial Rates
• It is helpful to express concentrations
and rates on a relative scale, by
dividing the entries in each column by
the smallest value.
Trial
[F2]
Rel.
conc.
[ClO2]
Rel.
conc.
Initial
rate
Rel.
rate
1
0.10
1
0.010
1
1.2x10-3
1
2
0.10
1
0.040
4
1.8x10-3
4
3
0.20
2
0.010
1
2.4x10-3
2
Initial Rates
• In trials 1 and 2, the concentration of F2
does not change, so the 4-fold change in
[ClO2] causes the 4-fold change in rate;
the order of ClO2 must be 1.
• In trials 1 and 3, [ClO2] is the same, so
the doubling of rate was caused by
doubling [F2]; therefore the order of F2 is
also 1.
Initial Rates
• The rate law is first order in both
[F2] and [ClO2].
rate = k[F2][ClO2]
• Solve the equation for k, and
substitute the experimental
concentrations and rates.
3
rate
1.2 10 M / s
1 1
k

 1.2M s
F2 ClO 2  0.10M 0.010M 
Test Your Skill
• Write the rate law for the reaction
given the following data:
2NO + 2H2  N2 + 2H2O
Init conc. Init conc. Init. Rate,
Trial
[NO], M
[H2], M
MIs
1
0.00570
0.140
7.01x10-5
2
0.00570
0.280
1.40x10-4
3
0.0114
0.140
2.81x10-4
Concentration-Time Dependence
• A zero order rate law
rate = k
means that the reaction rate is
independent of reactant concentration.
Zero-Order Rate Laws
• A plot of concentration
vs. time yields a straight
line
•
A plot of rate vs. time
yields a straight line with
a slope of zero
Concentration-Time Dependence
• First order rate law
• Differential rate law
rate = k[R]
• Integrated rate law
R  Ro e
 kt
or
In R  In Ro  kt
Concentration-Time Dependence
•
First order rate law: [R] = [R]o e -kt
• A plot of concentration vs. time
yields a curve.
Concentration-Time Dependence
• First order rate law: ln[R] = -kt + ln[R]o
• A plot of ln(concentration) vs. time
yields a straight line.
First Order Rate Law
• C12H22O11 + H2O  C6H12O6 + C6H12O6
sucrose + water  glucose + fructose
The reaction is 1st order, k = 6.2 x 10-5 s-1.
If [R]o = 0.40 M, what is [R] after 2 hr?
Test Your Skill
• How long did it take for the
concentration in the same
experiment to drop to 0.30 M?
Half-Life
• Half-life, t½, is the time required for the
initial concentration to decrease by ½.
Half-Life
• Half-life, t½, is the time required for
the initial concentration to decrease
by ½.
• For a first order rate law, the halflife is independent of the
concentration.
t1
2
[R] o
1
1
0.693
 ln
 ln2 
k 1/2[R] o k
k
Calculating Half-Life
• k = 6.2 x 10-5 s-1 for the reaction
C12H22O11 + H2O  C6H12O6 + C6H12O6
Calculate the half-life.
Radiocarbon Dating
• The age of objects that were once living
can be found by 14C dating, because:
• The concentration 14C is a constant in the
biosphere (the atmosphere and all living
organisms).
• When an organism dies, the 14C content
decreases with first order kinetics (t½ =
5730 years).
• Scientists calculate the age of an object
from the concentration of 14C in a sample.
Example: 14C Dating
• A sample of wood has 58% of the
14C originally present. What is the
age of the wood sample?
Second Order Rate Law
• For a second order rate law
rate = k[R]2
1
1

 kt
[R] [R] o
• A plot of 1/[R] vs. t is a straight line
for a system described by second
order kinetics.
Example: Second Order Rate Law
• The reaction
2NOCl  2NO + Cl2
obeys the rate law
rate = 0.020 M-1s-1 [NOCl]2
Calculate the concentration of
NOCl after 30 minutes, when the
initial concentration was 0.050 M.
Example: Order of Reaction
Given the experimental data for the
decomposition of 1,3-pentadiene
shown below, determine the order of
the reaction.
Example: Order of Reaction (cont.)
Review table 13.4 on page 534.
Influence of Temperature on k
• Reactions proceed at faster rates at
higher temperatures.
Collision Theory
• The reaction rate is proportional to the
collision frequency, Z, the number of
molecular collisions per second.
• Z depends on the temperature and the
concentration of the colliding molecules.
• Not all molecular collisions result in the
formation of products.
Collision Theory
• Activation energy (Ea): the minimum
collision energy required for reaction to
occur.
• Activated complex: the highest energy
arrangement of atoms that occurs in the
course of the reaction.
The Activated Complex
NO + O3 [activated complex]*  NO2 + O2
Influence of Temperature on Kinetic Energy
• The fraction of collisions with energy in
excess of Ea is given by:
fr  e
Ea / RT
• The collision frequency is proportional to
the concentrations of colliding species.
• The reaction rate is proportional to the rate
of collisions time the fraction of collisions
with energy in excess of Ea.
rate = Z × fr
Orientation of Reactants
The Steric Factor
• The steric factor, p, is a number
between 0 and 1 that is needed to
account for factors other than energy
before a reaction can occur.
• The reaction rate is proportional to
the steric factor times the collision
frequency times the fraction of
collisions with energy in excess of Ea:
rate = p x Z x fr
The Arrhenius Equation
rate = p x Z x fr
rate  pZo colliding speciese
Ea / RT
Combine p and Zo into a term A:
rate  Acolliding speciese
Ea / RT
Experiments show that
rate = k[colliding species], so
k  Ae
 Ea / RT
The Arrhenius Equation
k  Ae
 Ea / RT
• Take the natural log of both sides of
the equation:
Ea  1 
In k  In A 
 
R T 
• A plot of ln k vs. 1/T gives a straight
line with a slope of -Ea/R and an
intercept of ln A.
Measuring Activation Energy
• Determine Ea for the reaction,
2NO2 2NO + O2
given the data:
k (M-1·s-1)
T (°C)
In k
1/T (K-1)
0.003
500
-5.8
2.00x10-3
0.037
550
-3.30
1.82x10-3
0.291
500
-1.234
1.67x10-3
1.66
650
0.507
1.54x10-3
7.39
700
2.000
1.43x10-3
Solution
•
Prepare a plot of ln k vs. 1/T
slope  1.37  10 4
slope  Ea /R


Ea  1.37  10 4 8.314 
Ea  1.14  105 J
 1.14  10 2 kJ
Example: Arrhenius Equation
• The rate of a reaction exactly doubles,
when the temperature is changed from
25.0o C to 36.2o C. Calculate the
activation energy for this reaction.
Catalysis
• A catalyst is a substance that
increases the reaction rate but is not
consumed in the reaction.
• A catalyst provides an alternate
reaction path with a lower activation
energy.
Homogeneous Catalysis
• A homogeneous catalyst is one that is
present in the same phase as the reactants.
• Bromide ion is a homogeneous catalyst for the
decomposition of hydrogen peroxide.
2H2O2(aq)  2H2O(l) + O2(g)
• step 1:
H2O2(aq) + 2Br-(aq) + 2H+(aq)  Br2(aq) + 2H2O(l)
• step 2:
H2O2(aq) + Br2(aq)  2Br-(aq) + 2H+(aq) + O2(g)
Heterogeneous Catalysis
• A heterogeneous catalyst is one that is
present in a different phase from the
reactants.
• The gas phase reaction of hydrogen with
many organic compounds is catalyzed by
solid platinum.
Enzyme Catalysis
• Enzymes are large molecules
(macromolecules) which catalyze
specific biochemical reactions.
• Enzymes can increase the rates of
14
reactions by factors as large as 10 .
• Enzymes are very specific in the
reactions they catalyze.
• Enzymes are active under mild reaction
conditions.
Reaction Mechanisms
• A mechanism is a sequence of
molecular-level steps that lead from
reactants to products.
• An elementary step is an equation that
describes an actual molecular level
event.
• The concentration dependence in the
rate law for an elementary step is
given by the coefficients in the
equation.
Molecularity
• The molecularity of an elementary step
is the number of reactant species
involved in that step.
• Most elementary steps are either
unimolecular (involving a single
molecule) or bimolecular (collision of two
species).
Elementary Steps
• The reaction
2NO + O2  2NO2
is believed to occur by the following
sequence of elementary steps:
2NO  N2O2 bimolecular reaction
N2O2 + O2  2NO2 bimolecular reaction
Rate Laws for Elementary Reactions
• The rate of an elementary step is
proportional to the concentration of
each reactant species raised to the
power of its coefficient in the equation:
• step 1: 2NO  N2O2
rate1 = k1[NO]2
• step 2: N2O2 + O2 2NO2
rate2 = k2[N2O2][O2]
Test Your Skill
• Write the rate law for the elementary
step
H2 + Cl  H2Cl
Rate-Limiting Steps
• The overall rate of a multistep reaction is
determined by its slowest step, called the
rate-limiting step.
• The rates of fast steps which follow the ratelimiting step have no effect on the overall
rate law.
• The rates of fast steps that precede the
rate-limiting step usually affect the
concentrations of the reactant species in the
rate-determining step.
Complex Reaction Mechanisms
• R  P (2 steps)
• R  intermediates
• Intermediates  P
rate1 = k1[R]
rate2 = k2[intermediates]
• If step 1 is slow, then it determines rate
• If step 2 is slow, how can we measure
[intermediates]?
• Many fast steps prior to slow step are fast
and reversible
Complex Reaction Mechanisms
• Fast reversible steps help with [intermediate]
• Consider the following reaction
2NO + 2H2  N2 + 2H2O
• rate = k[NO]2[H2]
• step 1: 2NO  N2O2
fast and reversible
• step 2: N2O2 + H2  N2O + H2O
slow
• step 3: N2O + H2  N2 + H2O
fast
Work on the board
Complex Reaction Mechanisms
• Consider the two-step reaction
2NO + O2  2NO2
• step 1: 2NO  N2O2
rate1 = k1[NO]2
• step 2: N2O2 + O2  2NO2
rate2 = k2[N2O2][O2]
Complex Reaction Mechanisms
• If the first step is the rate-limiting step,
the rate law is:
rate = k1[NO]2
• If the first step is rapid and the second
step is the rate limiting step, the rate law
is:
rate = k2[N2O2][O2]
Complex Reaction Mechanisms
If the first step reaches equilibrium
rate1 (forward) = rate-1 (reverse):
rate1  k1[NO] 2
rate1  k 1[N2 O 2 ]
Because the rates at equilibriu m are equal :
rate1  rate-1
k1[NO]  k 1[N2 O 2 ]
2
k1
2
[N2 O 2 ] 
[NO]
k 1
Complex Reaction Mechanisms
• Substituting into the expression for step
2:
rate = k2[N2O2][O2]
k1
rate = k2
[NO]2[O2]
k -1
Combine all the rate constants:
• rate = k [NO]2[O2]
• The reaction is second order in NO and
first order in O2.
Test Your Skill
• For the reaction
2NO2 + O3  N2O5 + O2
the experimentally determined rate law is
rate= k[NO2][O3].
• Identify the rate limiting step in the
proposed two-step mechanism:
NO2 + O3  NO3 + O2
NO3 + NO2  N2O5
step 1
step 2
The Hydrogen-Iodine Reaction
• H2 + I2  2HI
rate = k[H2][I2]
• For many years this reaction was believed to
occur as a single bimolecular step.
• From more recent data, a very different
mechanism is likely.
I2  2I
Fast, reversible
I + H2  H2I
Fast, reversible
H2I + I  2HI Slow
• Both mechanisms give the same rate
law.
Enzyme Catalysis
• Most enzymes follow the
Michaelis-Menten mechanism.
• You will see this again in
biochemistry!!