Upload
rhoda-melton
View
218
Download
1
Tags:
Embed Size (px)
Citation preview
13-1CHEM 102, Spring 2012, LA TECH
CTH 328 9:30-10:45 am
Instructor: Dr. Upali Siriwardane
e-mail: [email protected]
Office: CTH 311 Phone 257-4941
Office Hours: M,W 8:00-9:00 & 11:00-12:00 am;
Tu,Th,F 8:00 - 10:00 am..
Exams: 9:30-10:45 am, CTH 328.
March 21 , 2012 (Test 1): Chapter 13
April 18 , 2012 (Test 2): Chapter 14 &15
May 14 , 2012 (Test 3): Chapter 16 &18
Optional Comprehensive Final Exam: May 17, 2012 :
Chapters 13, 14, 15, 16, 17, and 18
Chemistry 102(01) Spring 2012
13-2CHEM 102, Spring 2012, LA TECH
Chapter 13. Chemical Kinetics
13.1 Reaction Rate 13.2 Effect of Concentration on Reaction Rate 13.3 Rate Law and Order of Reaction
13.4 A Nanoscale View: Elementary Reactions
13.5 Temperature and Reaction Rate: The
Arrhenius Equation13.6 Rate Laws for Elementary Reactions13.7 Reaction Mechanisms13.8 Catalysts and Reaction Rate13.9 Enzymes: Biological Catalysts13‑10 Catalysis in Industry
13-3CHEM 102, Spring 2012, LA TECH
How do you measure rates?Rates are related to the time it required to
decay reactants or form products. The rate reaction = change in concentration of reactants/products per unit time
Average rate
rate of reaction = – D[reactant]/DtInstantaneous rate
rate of reaction = – d[reactant]/dt
13-4CHEM 102, Spring 2012, LA TECH
Rate of Appearance & Disappearance2 N2O5(g) -----> 4 NO2 (g) + O2 (g)
Disappearance is based on reactants
rate = -(D[N2O5]/ D t
Appearance is based on products
rate = D[NO2]/ D t
rate = D[O2]/ D t
Converting rates of Appearance.
rate = (D[NO2]/ D t = - 4/2 D[N2O5]/ D t
D[O2]/ D t = - 1/2 D[N2O5]/ D t
13-5CHEM 102, Spring 2012, LA TECH
Measuring Rate a A --> b B
Based on reactants
rate = -(1/a) D[A]/ D t
Based on products
rate = +(1/b) D[B]/ D t
D[A]= [A]f - [A]I Change in A
D t= tf - ti Change in t
13-6CHEM 102, Spring 2012, LA TECH
Reaction of cis-platin with Water
13-7CHEM 102, Spring 2012, LA TECH
Disappearance of Color
13-8CHEM 102, Spring 2012, LA TECH
Gas
buret
Constant temperature bath
An example reaction where gas is produced
13-9CHEM 102, Spring 2012, LA TECH
Time (s) Volume STP O2, mL
0 0
300 1.15
600 2.18
900 3.11
1200 3.95
1800 5.36
2400 6.50
3000 7.42
4200 8.75
5400 9.62
6600 10.17
7800 10.53
Here are the results for
our experiment.
Here are the results for
our experiment.
Time vs. volume of gas
13-10CHEM 102, Spring 2012, LA TECH
2 N2O5(g) -----> 4 NO2 (g) + O2 (g)
13-11CHEM 102, Spring 2012, LA TECH
Graph of 2 N2O5(g) ---> 4 NO2 (g) + O2 (g)
13-12CHEM 102, Spring 2012, LA TECH
Graph
13-13CHEM 102, Spring 2012, LA TECH
a) Temperature
b) Concentration
c) Catalysts
d) Particle size of solid reactants
Factors that affect rates of chemical reactions
13-14CHEM 102, Spring 2012, LA TECH
Effect of Particle Size on Rate
13-15CHEM 102, Spring 2012, LA TECH
Chemical Kinetics Definitions and Concepts
a) rate law
b) rate constant
c) order
d) differential rate law
c) integral rate law
13-16CHEM 102, Spring 2012, LA TECH
Every chemical reaction has a Rate Law
The rate law is an expression that relates
the rate of a chemical reaction to a constant
(rate constant-k) and concentration of
reactants raised to a power.
The power of a concentration is called the order
with respect to a particular reactant.
Rate Law
13-17CHEM 102, Spring 2012, LA TECH
Rate LawE.g. A + B -----> C
rate a [A]l[B]m
rate = k [A]l[B]m; k = rate constant
[A] = concentration of A
[B] = concentration of B
l = order with respect to A
m = order with respect to B
l & m have nothing to do with stoichiometric coefficients
13-18CHEM 102, Spring 2012, LA TECH
Rate Constant
E.g. A + B -----> C
rate a [A]l[B]m
rate = k [A]l[B]m;
k = rate constant
proportionality constant of the rate law
Larger the k faster the reaction
It is related inversely to t½
13-19CHEM 102, Spring 2012, LA TECH
Decomposition Reaction
13-20CHEM 102, Spring 2012, LA TECH
Rate Law E.g.
2 N2O5(g) -----> 4 NO2 (g) + O2 (g)
rate a [N2O5]1
rate = k [N2O5]1 ;k = rate constant
[N2O5] = concentration of N2O5
1 = order with respect to N2O5
Rate and the order are obtained by experiments
13-21CHEM 102, Spring 2012, LA TECH
Order The power of the concentrations is the order with
respect to the reactant.
E.g. A + B -----> C
If rate law: rate = k [A]1[B]2
The order of the reaction with respect to A is one (1).
The order of the reaction with respect to B is two (2).
Overall order of a chemical reaction is equal to the
sum of all orders (3).
13-22CHEM 102, Spring 2012, LA TECH
Method of initial rates
The order for each reactant is found by:
•Changing the initial concentration of that reactant.
•Holding all other initial concentrations and conditions constant.
•Measuring the initial rates of reaction
The change in rate is used to determine the order for that specific
reactant. The process is repeated for each reactant.
Finding rate laws
13-23CHEM 102, Spring 2012, LA TECH
Initial rate
13-24CHEM 102, Spring 2012, LA TECH
How do you find order? A + B -----> C
rate = k [A]l[B]m;
Hold concentration of other reactants constant
If [A] doubled, rate doubled• 1st order, [2A]1 = 2 1 x [A]1 , 2 1 = 2
b) If [A] doubled, rate quadrupled• 2nd order, [2A]2 = 2 2 x [A]2 , 2 2 = 4
c) If [A] doubled, rate increased 8 times • 3rd order, [2A]3 = 2 3 x [A]3 , 2 3 = 8
13-25CHEM 102, Spring 2012, LA TECH
Rate data
13-26CHEM 102, Spring 2012, LA TECH
Determining order
13-27CHEM 102, Spring 2012, LA TECH
Determining K, Rate Constant
13-28CHEM 102, Spring 2012, LA TECH
Overall order
13-29CHEM 102, Spring 2012, LA TECH
Units of the Rate Constant (k) 1first order: k = ─── = s-1
s L second order k = ─── mol s
L2 third order k = ─── mol2 s
13-30CHEM 102, Spring 2012, LA TECH
First order reactions
13-31CHEM 102, Spring 2012, LA TECH
Rate Law Differential Rate Law Integral Rate
rate = k [A]0 - D [A]/Dt =k ; ([A]0=1) [A]f-[A]i = -kt
rate = k [A]1 - D [A]/Dt = k [A] ln [A]o/[A]t = kt
rate = k [A]2 - D [A]/Dt = k [A]2 1/ [A]f = kt + 1/[A]i
Differential and Integral Rate Law
13-32CHEM 102, Spring 2012, LA TECH
Integrated Rate Laws
13-33CHEM 102, Spring 2012, LA TECH
Graphical methodOrder
RateLaw
Integrated Rate Law GraphX vs. time
Slope
0 rate = k [A]t = -kt + [A]0 [A]t -k
1 rate = k[A]ln[A]t = -kt + ln[A]0
ln[A]t -k
2 rate=k[A]2 = kt + k1
[A]t
1
[A]0
1
[A]t
13-34CHEM 102, Spring 2012, LA TECH
Graphical Ways to get Order
13-35CHEM 102, Spring 2012, LA TECH
First-order, Second-order,and Zeroth-order Plots
13-36CHEM 102, Spring 2012, LA TECH
Finding rate laws
0
0.05
0.1
0.15
0.2
0 2000 4000 6000 8000
-4.5
-4
-3.5
-3
-2.5
-2
-1.50 2000 4000 6000 8000
0
20
40
60
80
100
0 1000 2000 3000 4000 5000 6000 7000 8000
0 order plot
1st order plot
2nd order plot
As you can see from these
plots of the N2O5 data,
only a first order plot
results in a straight line.
As you can see from these
plots of the N2O5 data,
only a first order plot
results in a straight line.
Time (s)
Time (s)
Time (s)
[N2O
5]
1/[
N2O
5]
ln[N
2O
5]
13-37CHEM 102, Spring 2012, LA TECH
This plot of ln[cis-platin] vs.
time produces a straight line,
suggesting that the reaction
is first-order.
Comparing graphs
13-38CHEM 102, Spring 2012, LA TECH
First Order ReactionsA ----> B
13-39CHEM 102, Spring 2012, LA TECH
t1/2 equation 0.693 = k t1/2
0.693 t1/2 =---- k
13-40CHEM 102, Spring 2012, LA TECH
The half-life and the rate constant are related.
t1/2 =
Half-life can be used to calculate the first order rate constant.
For our N2O5 example, the reaction took 1900 seconds to react half way so:
k = = = 3.65 x 10-4
s-1
0.693
k
0.693
t1/2
0.693
1900 s
Half-life
13-41CHEM 102, Spring 2012, LA TECH
A Nanoscale View:Elementary Reactions
Most reactions occur through a series of simple
steps or elementary reactions.
Elementary reactions could be
unimolecular - rearrangement of a molecule
bimolecular - reaction involving the collision of
two molecules or particles
termolecular - reaction involving the collision of
three molecules or particles
13-42CHEM 102, Spring 2012, LA TECH
2NO2 (g) + F2 (g) 2NO2F (g)
If the reaction took place in a single step the rate law would be: rate = k
[NO2]2 [F2]
Observed: rate = k1 [NO2] [ F2]
If the observed rate law is not the same as if the reaction took place in a single
step that more than one step must be involved
Elementary Reactions and Mechanism
13-43CHEM 102, Spring 2012, LA TECH
Elementary ReactionsA possible reaction mechanism might be:
Step one NO2 + F2 NO2F + F (slow)
Step two NO2 + F NO2F (fast)
Overall 2NO2 + F2 2NO2F
slowest step in a multi-step mechanismthe step which determines the overall rate of the reaction
rate = k1 [NO2] [ F2]
Rate Determining Step
13-44CHEM 102, Spring 2012, LA TECH
This type of plot
shows the energy
changes during
a reaction.
This type of plot
shows the energy
changes during
a reaction.
Reaction profile of rate determining step
DH
activation
energy
Pote
nti
al
En
erg
y
Reaction coordinate
13-45CHEM 102, Spring 2012, LA TECH
What Potential Energy Curves ShowExothermic Reactions
Endothermic Reactions
Activation Energy (Ea) of reactant or the minimum
energy required to start a reaction
Effect of catalysts
Effect of temperature
13-46CHEM 102, Spring 2012, LA TECH
Exothermic reaction
Endothermic reaction
Examples of reaction profiles
13-47CHEM 102, Spring 2012, LA TECH
High activation energy (kinetic)
Low heat of reaction (thermodynamic)
Low activation energy (kinetic)
High heat of reaction (thermodynamic)
Examples of reaction profiles
13-48CHEM 102, Spring 2012, LA TECH
Unimolecular Reactioncis-2-butene trans-2-butrne
13-49CHEM 102, Spring 2012, LA TECH
Bimolecular Reaction
I- + CH3Br ICH3 + Br
-
13-50CHEM 102, Spring 2012, LA TECH
Orientation Probability: Some Unsuccessful Collisions
I- + CH3Br ICH3 + Br
-
13-51CHEM 102, Spring 2012, LA TECH
Arrhenius Equation: Dependence of Rate Constant (k) on T Rate constant (k)
k = A e-Ea/RT
A = frequency factor: A = p x z
Ea = Activation energyR = gas constantT = Kelvin temperaturep = collision factorz = Orientation factor
13-52CHEM 102, Spring 2012, LA TECH
Energy Distribution Curves:Activation Energy
13-53CHEM 102, Spring 2012, LA TECH
An alternate form of the Arrhenius equation:
k = A e-Ea/RT
ln k = + ln A
If ln k is plotted against 1/T, a straight line of slope -Ea/RT is obtained.
Activation energy - Ea
The energy that molecules must have in order to react.
( ) ( )1
T
Ea
R
-
Arrhenius Equation: ln form
13-54CHEM 102, Spring 2012, LA TECH
Calculation of Eak = A e-
Ea/RT
ln k = ln A - Ea/RT
log k = log A - Ea/ 2.303 RT
using two set of values
log k1 = log A - Ea/ 2.303 RT1
log k2 = log A - Ea/ 2.303 RT2
log k1 - log k2 = - Ea/ 2.303 RT2 + Ea/ 2.303 RT1
log k1/ k2 = Ea/ 2.303 R[ 1/T1 - 1/T2 ]
13-55CHEM 102, Spring 2012, LA TECH
Reaction rates are temperature dependent.
0
1
2
3
4
5
6
7
20 25 30 35 40 45 50
Here are rate constants
for N2O5 decomposition
at various temperatures.
T, oC k x 10
4, s
-1
20 0.235
25 0.469
30 0.933
35 1.82
40 3.62
45 6.29
k x
10
4 (
s-1
)
Temperature (o
C)
Rate vs Temperature plot
13-56CHEM 102, Spring 2012, LA TECH
y = - 1 2 3 9 2 x + 4 0 . 8 0 9
S l o p e = - 1 2 3 9 2
R = 8 . 3 5 J / m ol K
E a = 1 0 3 k J / m ol
- 2
- 1
0
1
2
3
0 . 0 0 3 1 0 . 0 0 3 2 0 . 0 0 3 3 0 . 0 0 3 4 0 . 0 0 3 5
ln k
T-1
Calculation of Ea from N2O5 data
13-57CHEM 102, Spring 2012, LA TECH
Collision ModelThree conditions must be met at the nano-scale
level if a reaction is to occur:
the molecules must collide;
they must be positioned so that the reacting
groups are together in a transition state between
reactants and products;
and the collision must have enough energy to
form the transition state and convert it into
products.
13-58CHEM 102, Spring 2012, LA TECH
Effect of Concentrationon Frequency ofBimolecular Collisions
13-59CHEM 102, Spring 2012, LA TECH
Transition State: Activated Complex or Reaction Intermediatesan unstable arrangement of atoms that has the
highest energy reached during the rearrangement
of the reactant atoms to give products of a reaction
13-60CHEM 102, Spring 2012, LA TECH
Catalyst
A substance which speeds up the rate of a
reaction while not being consumed
Homogeneous Catalysis - a catalyst which is in
the same phase as the reactants
Heterogeneous Catalysis- a catalyst which is
in the different phase as the reactants
catalytic converter• solid catalyst working on gaseous materials
13-61CHEM 102, Spring 2012, LA TECH
Catalysts Lowers Ea
13-62CHEM 102, Spring 2012, LA TECH
Catalyzed & Uncatalyzed Reactions
13-63CHEM 102, Spring 2012, LA TECH
Conversion of NO to N2 + O2
13-64CHEM 102, Spring 2012, LA TECH
Catalytic Converter catalyst
H2O(g) + HCs CO(g) + H2(g) (unbalanced)
catalyst
2 H2(g) + 2 NO(g) N2(g) + 2 H2O(g)
catalyst
HCs + O2(g) CO2(g) + H2O(g) (unbalanced)
catalyst
CO(g) + O2(g) CO2(g) (unbalanced)
catalyst = Pt-NiOHCs = unburned hydrocarbons
13-65CHEM 102, Spring 2012, LA TECH
Enzymes: Biological catalystsBiological catalysts
Typically are very large proteins.
Permit reactions to ‘go’ at conditions that the body
can tolerate.
Can process millions of molecules every second.
Are very specific - react with one or only a few types
of molecules (substrates).
13-66CHEM 102, Spring 2012, LA TECH
The active site
Enzymes are typically HUGE proteins, yet only a
small part is actually involved in the reaction. The active site has two
basic components.
catalytic site
binding site
Model of
trios-phosphate-isomerase
Model of
trios-phosphate-isomerase
13-67CHEM 102, Spring 2012, LA TECH
Relationship of Enzyme to Substrate
13-68CHEM 102, Spring 2012, LA TECH
Enzyme Catalyzed Reaction
13-69CHEM 102, Spring 2012, LA TECH
Maximum Velocity for an Enzyme Catalyzed Reaction
13-70CHEM 102, Spring 2012, LA TECH
Enzyme Activity Destroyed by Heat
13-71CHEM 102, Spring 2012, LA TECH
Reaction Mechanism
A set of elementary reactions which represent
the overall reaction
13-72CHEM 102, Spring 2012, LA TECH
Mechanism Oxidation ofIodide Ion by Hydrogen Peroxide
13-73CHEM 102, Spring 2012, LA TECH
Rate Law of Oxidation ofIodide Ion by Hydrogen Peroxide
Step 1.
HOOH + I- HOI + OH-
slow step - rate determining step, suggests that
the reaction is first order with regard to hydrogen
peroxide and iodide ion
rate = k[HOOH][I-]
13-74CHEM 102, Spring 2012, LA TECH
Mechanisms with a Fast Initial Step
2 NO(g) + Br2(g) 2NOBr(g)
rateexperimental = k[NO]2[Br2]
13-75CHEM 102, Spring 2012, LA TECH
Mechanism of NO + Br2
Rate = k[NOBr2][NO]
13-76CHEM 102, Spring 2012, LA TECH
Rate Constants for NO + Br2
Step +1(forward), rate constant k1
Step -1(backward), rate constant k-1
Step 2, rate constant k2
rateStep+1 = rateStep-1 + rateStep2
k1[NO][Br2] = k-1[NOBr2] - k2[NOBr2]
13-77CHEM 102, Spring 2012, LA TECH
Relationships of Rate Constants
k1[NO][Br2] ~ k-1[NOBr2]
thus
[NOBr2] = (k1/k-1)[NO][Br2]
substituting into
rate = k2[NOBr2][NO]
rate = k2((k1/k-1)[NO][Br2])[NO]
rate = (k2k1/k-1)[NO]2[Br2]
13-78CHEM 102, Spring 2012, LA TECH
Chain Mechanismschain initiating step • - the step of a mechanism which • starts the chain chain
propagating step(s) • the step or steps which keeps the chain going
chain terminating step(s) • the step or steps which break the chain
13-79CHEM 102, Spring 2012, LA TECH
Chain Mechanismscombustion of gasoline in an internal
combustion
engine
chain initiating step - additives which generate
free radicals, particles with unpaired electrons
chain propagating step(s) - steps which generate
new free radicals
chain terminating step(s)
- steps which do not generate new free radicals