chemical kinetics chapter 13 dr. ali bumajdad. chapter 13 topics rate of a reaction reaction rates...
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Chemical KineticsChapter 13Dr. Ali Bumajdad
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Chapter 13 Topics
• Rate of a Reaction• Reaction Rates and Stoichiometry• The Rate Law• Relationship between Reactant Concentration and Time1)First-order reactions2)Second-ordered reactions3)Zero-ordered reactions• The Effect of Activation Energy and Temperature on
Rate• Collision Theory• The Arhenius Equation • Finding Ea from graph and from equation• Catalysis
Chemical Kinetics
Dr. Ali Bumajdad
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Chemical Kinetics = Chemistry and Time
•Thermodynamics – does a reaction take place?
•Kinetics – how fast does a reaction proceed?
- Why it is important:
To answer questions like:1) How quickly a medicine is able to work?2) How fast the depletion of Ozone?3) How rapidly food spoils?4) How fast steel rust?
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•Reaction rate is the change in the concentration of a reactant or a product with time (M/s).
A B
rate = -[A]t
rate = [B]t
[A] = change in concentration of A over time period t
[B] = change in concentration of B over time period t
Because [A] decreases with time, [A] is negative.
Rate of a reaction
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A B
rate = -[A]t
rate = [B]t
time
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•Kinetics can be studied by following a change in property as a function of time
This property could be:
1) UV absorption2) Pressure3) Conductivity4) Concentration5) Gas volume
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Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
time
393 nmlight
Detector
[Br2] Absorption
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Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
slope oftangent
slope oftangent slope of
tangent
average rate = -[Br2]t
= -[Br2]final – [Br2]initial
tfinal - tinitial
instantaneous rate = rate for specific instance in time(slope of a tangent)
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rate [Br2]
rate = k [Br2]
k = rate[Br2]
= rate constant
= 3.50 x 10-3 s-1
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2H2O2 (aq) 2H2O (l) + O2 (g)
PV = nRT
P = RT = [O2]RTnV
[O2] = PRT1
rate = [O2]t RT
1 Pt=
measure P over time
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Reaction Rates and Stoichiometry
2A B
Two moles of A disappear and one mole of B that is formed.
= [B]t
rate = -[A]t
12
aA + bB cC + dD
•In this case, the rate will be different if we consider the disappearance of A than if we consider the formation of B
•In order to make it the same we should divide by the coefficient
• In general for:
rate = -[A]t
1a
= -[B]t
1b
=[C]t
1c
=[D]t
1d
(1)
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Q) Write the rate expression for the following reaction:
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
rate = -[CH4]
t= -
[O2]t
12
=[H2O]
t12
=[CO2]
tSa. Ex. a) How is the rate of disappearance of ozone related to the rate of appearance of oxygen in the following equation:
2O3 (g) 3O2 (g)
b) If at certain time the rate of appearance of O2, [O2]/t = 6.0 10-5 M/s what is the value of the rate of disappearance of O3, - [O3]/t at the same time.
4.0 10-5 M/s
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The Rate Law•Rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers that can be determined by experiment only.
•Rate Law is different than the rate of reaction
aA + bB cC + dD
reaction is xth order in A
reaction is yth order in B
reaction is (x +y)th order overall
rate = -[A]t
1a
Rate = k [A]x[B]y (2)
Rate = k [A]x[B]y
Reactant Reactant not productnot product
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F2 (g) + 2ClO2 (g) 2FClO2 (g)
rate = k [F2]x[ClO2]y
When Double [F2] with [ClO2] constant
x = 1
Quadruple [ClO2] with [F2] constant
y = 1
rate = k [F2][ClO2]
Rate doubles
Rate quadruples
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F2 (g) + 2ClO2 (g) 2FClO2 (g)
rate = k [F2][ClO2]
Rate Laws
1
• Rate laws are always determined experimentally.
• Reaction order is always defined in terms of reactant (not product) concentrations.
• The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.
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Q) Determine the rate law and calculate the rate constant for the following reaction from the following data:S2O8
2- (aq) + 3I- (aq) 2SO42- (aq) + I3
- (aq)
Experiment [S2O82-] [I-]
Initial Rate (M/s)
1 0.08 0.034 2.2 x 10-4
2 0.08 0.017 1.1 x 10-4
3 0.16 0.017 2.2 x 10-4
rate = k [S2O82-]x[I-]y
Double [I-], rate doubles (experiment 1 & 2)
y = 1
Double [S2O82-], rate doubles (experiment 2 & 3)
x = 1
k = rate
[S2O82-][I-]
=2.2 x 10-4 M/s
(0.08 M)(0.034 M)= 0.08/M•s
rate = k [S2O82-][I-]
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First-Order Reactions
• For reaction of the type A product
rate = -[A]t
and rate = k [A]
[A] is the concentration of A at any time t[A]0 is the concentration of A at time t=0
Relationship between Reactant Concentration and Time
and [A]t= k-
[A]
[A]t
= k [A]-
Reactions with first overall order
Using Calculus
ln[A] = ln[A]0 - kt (3a) (3b)[A] = [A]0exp(-kt)
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2N2O5 4NO2 (g) + O2 (g)
k = rate[A]
= 1/s or s-1M/sM
=Unit of first order rate constant
ln[A] = ln[A]0 - kt (3a)
Y = b + mX
Where m = slope b = intercept
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Q) The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ?
kt = ln[A]0 – ln[A]
t =ln[A]0 – ln[A]
k= 66 s
[A]0 = 0.88 M
[A] = 0.14 M
ln[A]0
[A]
k=
ln0.88 M
0.14 M
2.8 x 10-2 s-1=
ln[A] = ln[A]0 - kt(3a)
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• half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2
ln[A]0
[A]0/2
k=t½
ln2k
=0.693
k=
0.693k
=ln2k
t1/2 =(4) Independent of
initial concentration
First-Order Reactions
1) It take the same time for the concentration of reactant to decrease from 0.1M to 0.05M as it does from 0.05M to 0.025M
2) If you know t1/2 you know k
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A product
First-order reaction
# of half-lives [A] = [A]0/n
1
2
3
4
2
4
8
16
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Q) What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?
t½ln2k
=0.693
5.7 x 10-4 s-1= = 1200 s = 20 minutes
How do you know decomposition is first order?
units of k (s-1)
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• For reactions of the type A product
rate = -[A]t
and rate = k [A]2
[A]t
= k [A]2and -
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
t½ = t when [A] = [A]0/2
Second-Order Reactions Reactions with second overall order
k = Unit of second order rate constantrate[A]2 = 1/M•sM/s
M2=
Using Calculuswhere1
[A]=
1[A]0
+ kt(5)
t½ =1
k[A]0
(6)
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A product rate = -[A]t
rate = k [A]0 = k
k = rate[A]0
= M/s[A]t
= k-
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
t½ = t when [A] = [A]0/2
Zero-Order Reactions
[A] = [A]0 - kt (7)(7)
t½ =[A]0
2k(8)(8)
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order Zero First Second
Rate law Rate = k Rate = k[A]
Rate = k[A]2
Integrated rate law [A] = -kt + [A]o
ln[A] = -kt +ln[A]o
1 1
[A] [A]o
kt
Plot needed to give
a straight line [A] versus t In[A] versus t 1
[A]versus t
Relationship of k to the slope of straight
line
Slope = -k Slope = -k Slope = k
Half-life 1
2
[ ]
2oA
tk
1
2
0.693t
k 1
2
1
[ ]o
tk A
Summary of the Kinetics for reactions of type aA products
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Summary of the Kinetics of Zero-Order, First-Orderand Second-Order Reactions
Order Rate LawConcentration-Time
Equation Half-Life
0
1
2
rate = k
rate = k [A]
rate = k [A]2
ln[A] = ln[A]0 - kt
1[A]
=1
[A]0
+ kt
[A] = [A]0 - kt
t½ln2k
=
t½ =[A]0
2k
t½ =1
k[A]0
k Unit
s-1
M-1 s-1
M s-1
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ln[A] = ln[A]0 - kt (3a)
ln [A] = - kt (3c)
[A]0
(a)(a)
(b,c)(b,c)
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• Plot of ln [A] versus t should be linear for first order reactionPlot of ln [A] versus t should be linear for first order reaction
• Since [A] Since [A] P (PV=nRT), then lnP =-kt +ln P P (PV=nRT), then lnP =-kt +ln P00
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0.693k
=ln2k
t1/2 =(4)
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1[A]
=1
[A]0
+ kt(5)
t½ =1
k[A]0
(6)
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- is the minimum amount of energy
required to initiate a chemical reaction
The Effect of Activation Energy (Ea) and Temperature on Rate
-Usually rate increase by increasing T
- Rate increase by decreasing Ea
- Unit: J/mol
- Unit: K
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Collision Theory
1) How reaction Occurs?By collisions between molecules or atoms
2) Does all collisions results in a reactions? No, a) The collision should be strong enough (more than the
activation energy or the reaction) b) The collision should be with the right orientation
3) What factor affect the speed (rate) of the reactions?
a) Collision frequency (A)b) Temperaturec) Activation energy (Ea)
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Orientation effect
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Exothermic Reaction Endothermic Reaction
A + B AB C + D++
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Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature (K)
A is the frequency and orientation factor
(Arrhenius equation)
The Arhenius Equation k = A • exp( -Ea / RT ) (9)
lnk = -Ea
R1T
+ lnA(10)
Y = m X + b
k T
k 1/Ea
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lnk = -Ea
R1T
+ lnA
Y = m X + b
1) Finding Ea from graph
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lnk1 =Ea
R1T1
lnA -
lnk2 =Ea
R1T2
lnA -
2) Finding Ea from Equation
Subtracting 2 from 1
(1)
(2)
1T2
lnk1 - lnk2 =Ea
R1T1
- Ea
R+
1T2
lnk1 - lnk2 =Ea
R1T1
-
1T2
ln k1 =Ea
R1T1
- k2
(11)
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Sa. Ex.Sa. Ex.
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lnk = -Ea
R1T
+ lnA(10)
Y = m X + b
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1T2
ln k1 =Ea
R1T1
- k2
(11)
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•Catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.
k = A • exp( -Ea / RT ) Ea k
ratecatalyzed > rateuncatalyzed
Ea < Ea‘
Uncatalyzed Catalyzed
Catalysis
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In heterogeneous catalysis, the reactants and the catalysts are in different phases.
In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid.
• Haber synthesis of ammonia
• Ostwald process for the production of nitric acid
• Catalytic converters
• Acid catalysis
• Base catalysis
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N2 (g) + 3H2 (g) 2NH3 (g)Fe/Al2O3/K2O
catalyst
Haber Process
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Ostwald Process
Hot Pt wire over NH3 solutionPt-Rh catalysts used
in Ostwald process
4NH3 (g) + 5O2 (g) 4NO (g) + 6H2O (g)Pt catalyst
2NO (g) + O2 (g) 2NO2 (g)
2NO2 (g) + H2O (l) HNO2 (aq) + HNO3 (aq)
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Catalytic Converters
CO + Unburned Hydrocarbons + O2 CO2 + H2Ocatalytic
converter
2NO + 2NO2 2N2 + 3O2
catalyticconverter
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Enzyme Catalysis: very selective biological catalyst
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uncatalyzedenzyme
catalyzed
rate = [P]t
rate = k [ES]