chemistry 102(60) summer 2001 n instructor: dr. upali siriwardane n e-mail:[email protected] n...
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Chemistry 102(60) Summer 2001 Instructor: Dr. Upali Siriwardane e-mail:[email protected] Office: CTH 311 Phone 257-4941 Office Hours: 8:30-10:30 a.m., M, W, Tu,Th, F
(Test 1): Chapter 12(Test 2): Chapter 13. (Test 3): Chapter 14 (Test 4): Chapter 15. (Test 5): Chapter 17.
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Chapter 13. Rates of Reactions Chemical Kinetics
the branch of chemistry dealing with the
rates of chemical reactions.
Using chemical kinetics we can find
time to complete a reaction, the effect of
temperature on the rate
Effect of other substances (catalysts or
inhibitors) on the reactions
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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
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An example reaction
Gasburet
Constant temperature bath
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An example reaction 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.
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Rate a A --> b B based on reactants rate = -(1/a) [A]/ t Based on products rate = +(1/b) [B]/ t [A]= [A]f - [A]I Change in A
t= tf - ti Change in t
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Rate of Reaction
2 N2O5(g) -----> 4 NO2 (g) + O2 (g) based on reactants
rate = -(1/2) [N2O5]/ t Based on products rate = +(1/4) [NO2]/ t
rate = +(1/1) [O2]/ t
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Rate of Appearance & disappearance
2 N2O5(g) -----> 4 NO2 (g) + O2 (g) Disappearance is based on reactants
rate = -([N2O5]/ t Appearance is based on products rate = [NO2]/ t
rate = [O2]/ t converting rates of Appearance.. rate = ([NO2]/ t = - 4/2 [N2O5]/ t
[O2]/ t = - 1/2 [N2O5]/ t
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Chemical Kinetics Definitions and Concepts
a) rate law b) rate constant c) order d) differential rate law c) integral rate law
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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 the particular reactant.
Rate Law
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Rate Law E.g. A + B -----> C rate [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
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Rate Constant
E.g. A + B -----> C
rate [A]l[B]m rate = k [A]l[B]m; k = rate constant proportionality constant of the rate law
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Rate Law E.g. 2 N2O5(g) -----> 4 NO2 (g) + O2 (g)
rate [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
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Order The power of the concentrations is the
order with respect to the reactant. E.g. A + B -----> C 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).
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Finding rate laws Method of initial rates.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.
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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
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Method of Initial Rates A + B ----> C The rate law : rate = k [A]x [B]y
[A],mol/L [B],mol/L rate, mol/LS 4.6 x 10-4 3.1 x 10-5 2.08 x 10-3 4.6 x 10 -4 6.2 x 10 -5 4.16 x 10-3 9.2 x 10 -4 6.2 x 10 -5 1.664 x 10-2
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Order wrt A
1.664 x 10-2 k (9.2 x 10-4)x (6.2 x 10-5)y --------------- = -------------------------------------- 4.16 x 10-3 k (4.6 x 10-4)x (6.2 x 10-5)y 1.664 x 10-2 (9.2 x 10-4)x --------------- = -------------- 4.16 x 10-3 (4.6 x 10-4)x
4 = (9.2 x 10-4)/(4.6 x 10-4)x = 2x 4 = 2x
x = 2
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Order wrt B 4.16 x 10-3 k (4.6 x 10-4)x (6.2 x 10-5)y
----------- = ------------------------------ 2.08 x 10-3 k (4.6 x 10-4)x (3.1 x 10-5)y 4.16 x 10 -3 (6.2 x 10-5) y ----------- = ------------- 2.08 x 10 -3 (3.1 x 10-5) y
2 = (6.2 x 10-5/3.1 x 10-5)y = 2y 2 = 2y
y = 1 The rate law : rate = k [A]2 [B]1
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Units of the Rate Constant
1 first order: k = ─── = s-1
s L second order k = ─── mol s
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Rate Law Differential Rate Law Integral Rate
rate k [A]0 [A]/t =k ; ([A]0=1) [A]f-[A]i = -kt
rate k [A]1 [A]/t = k [A] ln [A]o/[A]t = kt
rate k [A]2 [A]/t = k [A]2 1/ [A]f = kt + 1/[A]i
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Differential Rate Law
Normal form Differential form
zero order rate = k [A]0 rate = - [A]/ t = k ( [A]0=1) first order rate law rate = k [A]1 rate = - [A]/ t = k [A]1
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Integral Rate Law
Differential form Integral form
zero order rate = - [A]/ t = [A]f-[A]i = -kt = k ( [A]0=1) first order rate law rate = - [A]/ t ln [A]o/[A]t = kt = k [A]1
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Finding rate lawsGraphical method.Graphical method.
Rate integrated Graph Slope
Order law rate law vs. time
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]0
1[A]t
1[A]t
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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 2000 4000 6000 8000
0 order plot
1st order plot
2nd order plot
As you can see from theseplots of the N2O5 data, only a first order plotresults in a straight line.
As you can see from theseplots of the N2O5 data, only a first order plotresults in a straight line.
Time (s)
Time (s)
Time (s)
[N2O
5]
1/[
N2O
5]
ln[N
2O
5]
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First order reactions Reactions that are first order with respect
to a reactant are of great importance. Describe how many drugs pass into the
blood stream or used by the body. Often useful in geochemistry Radioactive decay Half-life (Half-life (tt1/21/2)) The time required for one-half of the
quantity of reactant originally present to react.
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First Order Reactions A ----> B Differential rate law [A] - ───── = k [A] t [A]t [A]0
ln ─── = - k t or ln ─── = k t [A]0 [A]t
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Half-life form of 1st order
t2 is defined as time for [A]0, the initial concentration to decay half the original value
ie 1/2 x [A]0 = [A]t.
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t2 equation
0.693 = k t2
0.693 t2 = ---- k
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Half-life
The half-life and the rate constant are related. tt1/21/2 = = 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
0.6930.693kk
0.693
t1/2
0.6931900 s
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Half-life
0
0.05
0.1
0.15
0.2
0 2000 4000 6000 8000
From our N2O5 data, we cansee that it takes about 1900seconds for the concentrationto be reduced in half.
It takes another 1900 secondsto reduce the concentration inhalf again.
From our N2O5 data, we cansee that it takes about 1900seconds for the concentrationto be reduced in half.
It takes another 1900 secondsto reduce the concentration inhalf again.
Time (s)
[N2O
5]
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Theories of reaction rates Collision theoryCollision theory Based on kinetic-molecular theory. It assumes that reactants must collide for
a reaction to occur. They must hit with sufficient energy and
with the proper orientation so as to break the original bonds and form new ones.
As temperature is increased, the average kinetic energy increases - so will the rate.
As concentration increases, the number of collisions will also increase, also increasing the rate.
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Effective collision
Reactants must have sufficient energy and the proper orientation for a collision to result in a reaction.
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Transition state theory As reactants collide, they initially form an activated
complex. The activated complex is in the transition state. It lasts for approximately 10-100 fs. It can then form products or reactants. Once products are formed, it is much harder to return
to the transition state, for exothermic reactions.
Reaction profiles can be used to show this process.
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What are the factors that affect rates of chemical reactions?
a) Temperature b) Concentration c) Catalysts d) Particle size of solid reactants
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This type of plotshows the energychanges during
a reaction.
This type of plotshows the energychanges during
a reaction.
Reaction profile
Hactivation
energyPote
nti
al
En
erg
y
Reaction coordinate
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Potential Energy Curves
Exothermic Reactions Endothermic Reactions Effect of catalysts Effect of temperature
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Examples of reaction profiles
Exothermic reaction
Endothermic reaction
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Examples of reaction profiles
High activation energyLow heat of reaction
Low activation energyHigh heat of reaction
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Arrhenius Equation
Rate constant (k) k = A e-Ea/RT
A = frequency factor Ea = Activation energy R = gas constant T = Kelvintemperature
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Rate and temperature
Reaction rates are temperature dependent.
0
1
2
3
4
5
6
7
20 25 30 35 40 45 50
0
1
2
3
4
5
6
7
20 25 30 35 40 45 50
Here are rate constantsfor N2O5 decompositionat various temperatures.
T, oC k x 104, 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 (oC)
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Rate and temperature
The relationship between rate constant and temperature is mathematically described by the Arrhenius equationArrhenius equation.
k = A e A constant Ea activation energy T temperature, Kelvin R gas law constant
-Ea / RT
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Rate and temperature
An alternate form of the Arrhenius equation is:
ln k = + ln A
If ln k is plotted against 1/T, a straight line of slope -Ea/RT is obtained.
Activation energy - Activation energy - EEaa The energy that molecules must have in
order to react.
( )( ) 1
T
Ea
R-
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Calculation of Ea
k = 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 ]
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Calculation of Ea from N2O5 data
y = -12392x + 40.809
Slope = -12392R = 8.35 J/mol KEa = 103 kJ / mol
-2
-1
0
1
2
3
0.0031 0.0032 0.0033 0.0034 0.0035
ln k
T-1
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Reaction mechanisms
A detailed molecular-level picture of how a reaction might take place.
ON
OO
O+ O
NO
OO
ON
OO
O
activatedcomplex
= bonds in the process of breaking or being formed
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Reaction mechanisms
Elementary processElementary process Each step in a mechanism. MolecularityMolecularity The number of particles that come
together to form the activated complex in an elementary process.
1 - unimolecular 2 - bimolecular 3 - termolecular
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Reaction Mechanisms
Consider the following reaction. 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] However, the experimentally observed
rate law is: rate = k [NO2] [F2]
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Reaction Mechanisms Since the observed rate law is not the same as
if the reaction took place in a single step, we know two things.• More than one step must be involved• The activated complex must be produced from two
species. A possible reaction mechanism might be: Step oneStep one NO2 + F2 NO2F + F
Step twoStep two NO2 + F NO2F
OverallOverall 2NO2 + F2 2NO2F
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Reaction Mechanism
Elementary Reactions: NO2 + F2 --> NO2F + F (slow)
F + NO2 --> NO2F (fast) Molecularity? Of Elementary Reactions unimolecular, bimolecular,
termolecular?
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Reaction Mechanisms
Rate-determining step.Rate-determining step. When a reaction occurs in a series
of steps, with one slow step, it is the slowslow step that determines the overall rate.
Step oneStep one NO2 + F2 NO2F + F Expected to be slow. It involves breaking
an F-F bond. Step twoStep two NO2 + F NO2F Expected to be fast. A fluorine atom is
very reactive.
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Reaction Mechanisms Since step one is slow, we can expect this
step to determine the overall rate of the reaction.
NO2 + F2 NO2F + F This would give a rate expression of: rate = k1 [NO2] [ F2] This agrees with the experimentally
observed results.
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Catalysis CatalystCatalyst
A substance that changes the rate of a reaction without being consumed in the reaction.
Provides an easier way to react.
Lower activation energy.
Still make the same products. EnzymesEnzymes are biological catalysts. InhibitorInhibitor
A substance that decreases the rate of reaction.
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Catalysts Lowers ECatalysts Lowers Eaa
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Catalysis
Types of catalystsTypes of catalysts
HomogeneousHomogeneous - same phase
Catalyst is uniformly distributed throughout the reaction mixture
Example - I- in peroxide.
HeterogeneousHeterogeneous - different phase
Catalyst is usually a solid and thereactants are gases or liquids
Example - Automobile catalytic converter.
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Heterogeneous catalysis
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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 (substratessubstrates).
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Classification of enzymes Based on type of reactionBased on type of reaction
OxireductaseOxireductase catalyze a redox reaction
TransferaseTransferase transfer a functional group
HydrolaseHydrolase cause hydrolysis reactions
LyaseLyase break C-O, C-C or C-N bonds
IsomerasesIsomerases rearrange functional groups
LigaseLigase join two molecules
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The active site Enzymes are typically HUGE proteins, yet only
a small part is actually involved in the reaction.
The active site has twobasic components.
catalytic sitecatalytic site
binding sitebinding site
Model oftrios-phosphate-isomerase
Model oftrios-phosphate-isomerase
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Characteristics of enzyme active sites
Catalytic siteCatalytic site Where the reaction actually occurs.
Binding siteBinding site Area that holds substrate in proper place. Enzymes uses weak, non-covalent interactions to hold
the substrate in place based on R groups of amino acids.
Shape is complementary to the substrate and determines the specificity of the enzyme.
Sites are pockets or clefts on the enzyme surface.