Chemical Kinetics
CHAPTER 14
Part B
Chemistry: The Molecular Nature of Matter, 6th edition
By Jesperson, Brady, & Hyslop
CHAPTER 14 Chemical Kinetics
CHAPTER 14 Chemical Kinetics
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Learning Objectives:
Factors Affecting Reaction Rate:
oConcentration
oState
oSurface Area
oTemperature
oCatalyst
Collision Theory of Reactions and Effective Collisions
Determining Reaction Order and Rate Law from Data
Integrated Rate Laws
Rate Law Concentration vs Rate
Integrated Rate Law Concentration vs Time
Units of Rate Constant and Overall Reaction Order
Half Life vs Rate Constant (1st Order)
Arrhenius Equation
Mechanisms and Rate Laws
Catalysts
CHAPTER 14 Chemical Kinetics
CHAPTER 14 Chemical Kinetics
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Lecture Road Map:
①Factors that affect reaction rates
②Measuring rates of reactions
③Rate Laws
④Collision Theory
⑤Transition State Theory & Activation Energies
⑥Mechanisms
⑦Catalysts
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Integrated RateLaws
Integrated RateLaws
CHAPTER 14 Chemical Kinetics
CHAPTER 14 Chemical Kinetics
IntegratedRate Laws
IntegratedRate Laws
Concentration & Time
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Rate law tells us how speed of reaction varies withconcentrations.
Sometimes want to know
oConcentrations of reactants and products atgiven time during reaction
oHow long for the concentration of reactants todrop below some minimum optimal value
Need dependence of rate on time
IntegratedRate Laws
IntegratedRate Laws
First Order Integrated Rate Law
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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•Corresponding to reactions
–A products
•Integrating we get
•Rearranging gives
•Equation of line y = mx + b
IntegratedRate Laws
IntegratedRate Laws
First Order Integrated Rate Law
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Yields straight line
oIndicative of first orderkinetics
oSlope = –k
oIntercept = ln [A]0
oIf we don't knowalready
Slope = –k
IntegratedRate Laws
IntegratedRate Laws
2nd Order Integrated Rate Law
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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•Corresponding to special second order reaction
–2B products
•Integrating we get
•Rearranging gives
•Equation of line y = mx + b
IntegratedRate Laws
IntegratedRate Laws
2nd Order Integrated Rate Law
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Yields straight line
oIndicative of 2nd orderkinetics
oSlope = +k
oIntercept = 1/[B]0
Slope
= +k
IntegratedRate Laws
IntegratedRate Laws
Graphically determining Order
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Make two plots:
1. ln [A] vs. time
2. 1/[A] vs. time
oIf ln [A] is linear and 1/[A] is curved, thenreaction is 1st order in [A]
oIf 1/[A] plot is linear and ln [A] is curved, thenreaction is 2nd order in [A]
oIf both plots give horizontal lines, then 0th orderin [A]
IntegratedRate Laws
IntegratedRate Laws
Graphically determining Order
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Time, min
[SO2Cl2], M
ln[SO2Cl2]
1/[SO2Cl2] (L/mol)
0
0.1000
-2.3026
10.000
100
0.0876
-2.4350
11.416
200
0.0768
-2.5666
13.021
300
0.0673
-2.6986
14.859
400
0.0590
-2.8302
16.949
500
0.0517
-2.9623
19.342
600
0.0453
-3.0944
22.075
700
0.0397
-3.2264
25.189
800
0.0348
-3.3581
28.736
900
0.0305
-3.4900
32.787
1000
0.0267
-3.6231
37.453
1100
0.0234
-3.7550
42.735
Example: SO2Cl2 SO2 + Cl2
IntegratedRate Laws
IntegratedRate Laws
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Reaction is 1st order in SO2Cl2
Example: SO2Cl2 SO2 + Cl2
IntegratedRate Laws
IntegratedRate Laws
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Example: HI(g) H2(g) + I2(g)
IntegratedRate Laws
IntegratedRate Laws
Time
(s)
[HI]
(mol/L)
ln[HI]
1/[HI]
(L/mol)
0
0.1000
-2.3026
10.000
50
0.0716
-2.6367
13.9665
100
0.0558
-2.8860
17.9211
150
0.0457
-3.0857
21.8818
200
0.0387
-3.2519
25.840
250
0.0336
-3.3932
29.7619
300
0.0296
-3.5200
33.7838
350
0.0265
-3.6306
37.7358
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Example: HI(g) H2(g) + I2(g)
IntegratedRate Laws
IntegratedRate Laws
Reaction is second order in HI.
Group
Problem
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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A plot for a zeroth order reaction is shown. What isthe proper label for the y-axis in the plot ?
A. Concentration
B. ln of Concentration
C. 1/Concentration
D. 1/ ln Concentration
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Half Life (t1/2) for first order reactions
IntegratedRate Laws
IntegratedRate Laws
Half-life = t½ We often use the half life to describe how fast areaction takes place
First Order Reactions
oSet
oSubstituting into
oGives
oCanceling gives ln 2 = kt½
oRearranging gives
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Half Life (t1/2) for First Order Reactions
IntegratedRate Laws
IntegratedRate Laws
Observe:
1. t½ is independent of [A]o
oFor given reaction (and T)
oTakes same time for concentration to fall from
o2 M to 1 M as from
o5.0 10–3 M to 2.5 10–3 M
2. k1 has units (time)–1, so t½ has units (time)
ot½ called half-life
oTime for ½ of sample to decay
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Half Life (t1/2)
IntegratedRate Laws
IntegratedRate Laws
Does this mean that all of sample is gone intwo half-lives (2 × t½)?
No!
oIn 1st t½, it goes to ½[A]o
oIn 2ndt½, it goes to ½(½[A]o) = ¼[A]o
oIn 3rd t½, it goes to ½(¼[A]o) = ⅛[A]o
oIn nth t½, it goes to [A]o/2n
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Half Life (t1/2)
IntegratedRate Laws
IntegratedRate Laws
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Half Life (t1/2): First Order Example
IntegratedRate Laws
IntegratedRate Laws
131I is used as a metabolic tracer inhospitals. It has a half-life, t½ = 8.07 days.How long before the activity falls to 1% ofthe initial value?
Group
Problem
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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The radioactive decay of a new atom occurs so that after 21days, the original amount is reduced to 33%. What is therate constant for the reaction in s–1?
Group
Problem
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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The half-life of I-132 is 2.295 h. Whatpercentage remains after 24 hours?
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Half Life (t1/2): Carbon-14 Dating
IntegratedRate Laws
IntegratedRate Laws
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Half Life (t1/2): Second Order Reactions
IntegratedRate Laws
IntegratedRate Laws
How long before [A] = ½[A]o?
ot½, depends on [A]o
ot½, not useful quantity for a second order reaction
Group
Problem
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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The rate constant for the second order reaction 2A → B is 5.3× 10–5 M–1 s–1. What is the original amount present if, after 2hours, there is 0.35 M available?
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Collision Theory
Collision Theory
CHAPTER 14 Chemical Kinetics
CHAPTER 14 Chemical Kinetics
CollisionTheory
CollisionTheory
Reaction Rates
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Collision Theory
As the concentration of reactants increase
oThe number of collisions increases
o Reaction rate increases
As temperature increases
oMolecular speed increases
oHigher proportion of collisions withenough force (energy)
oThere are more collisions per second
oReaction rate increases
CollisionTheory
CollisionTheory
Reaction Rates
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Rate of reaction proportional to number of effectivecollisions/sec among reactant molecules
Effective collision
oOne that gives rise to product
e.g. At room temperature and pressure
oH2 and I2 molecules undergoing 1010 collisions/sec
oYet reaction takes a long time
oNot all collisions lead to reaction
Only very small percentage of all collisions lead tonet change
CollisionTheory
CollisionTheory
Molecular Orientation
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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CollisionTheory
CollisionTheory
Temperature
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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As T increases
oMore molecules have Ea
oSo more molecules undergo reaction
CollisionTheory
CollisionTheory
Activation Energy (Ea)
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Molecules must possess certain amount of kinetic energy(KE) in order to react
Activation Energy, Ea = Minimum KE needed forreaction to occur
oGet energy from collision with other molecules
oIf molecules move too slowly, too little KE, they justbounce off each other
oWithout this minimum amount, reaction will not occureven when correctly oriented
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Transition StateTheory
Transition StateTheory
CHAPTER 14 Chemical Kinetics
CHAPTER 14 Chemical Kinetics
TransitionState
TransitionState
Molecular Basis of Transition State Theory
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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KE decreasing as PE increases
KE
KE
PE
KE
KE
Is the combined KE ofboth moleculesenough to overcomeActivation Energy
TransitionState
TransitionState
Molecular Basis of Transition State Theory
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Reaction Coordinate (progress of reaction)
Potential Energy
Activation energy (Ea)
= hill or barrier
between reactants
and products
Heat of reaction (H)= difference in PE between
products and reactants
Hreaction = Hproducts – Hreactants
Products
TransitionState
TransitionState
Exothermic Reactions
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Reaction Coordinate (progress of reaction)
Potential Energy
Exothermic reaction
• Products lower PE than reactants
Exothermic
Reaction
H = –
Products
TransitionState
TransitionState
Exothermic Reactions
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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oHreaction < 0 (negative)
oDecrease in PE of system
oAppears as increase in KE
oSo the temperature of the system increases
oReaction gives off heat
oCan’t say anything about Ea from size of H
oEa could be high and reaction slow even ifHrxn large and negative
oEa could be low and reaction rapid
TransitionState
TransitionState
Endothermic Reactions
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Endothermic
Reaction
H = +
Hreaction = Hproducts – Hreactants
TransitionState
TransitionState
Endothermic Reactions
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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oHreaction > 0 (positive)
oIncrease in PE
oAppears as decrease in KE
oSo temperature of the system decreases
oHave to add E to get reaction to go
oEa Hrxn as Ea includes Hrxn
oIf Hrxn large and positive
oEa must be high
oReaction very slow
TransitionState
TransitionState
Activated Complex
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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oArrangement of atoms at top of activation barrier
oBrief moment during successful collision when
obond to be broken is partially broken and
obond to be formed is partially formed
Example
Transition State
Group
Problem
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Draw the transition state complex, or the activatedcomplex for the following reaction:
CH3CH2O-+H3O+CH3CH2OH +H2O
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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ActivationEnergies
ActivationEnergies
CHAPTER 14 Chemical Kinetics
CHAPTER 14 Chemical Kinetics
Ea
Ea
Arrhenius Equation
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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The rate constant is dependent on Temperature, whichallows us to calculate Activation Energy, Ea
Arrhenius Equation:
Equation expressing temperature-dependence of k
oA = Frequency factor has same units as k
oR = gas constant in energy units
= 8.314 J mol–1 K–1
oEa = Activation Energy—has units of J/mol
oT = Temperature in K
Ea
Ea
Calculating Activation Energy
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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•Method 1. Graphically
•Take natural logarithm of both sides
•Rearranging
•Equation for a line
•y = b + mx
Arrhenius Plot
•Plot ln k (y axis) vs. 1/T (x axis) yield astraight line
•Slope = -Ea/R
•Intercept = A
Ea
Ea
Arrhenius Equation: Graphing Example
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Given the following data,predict k at 75 ˚C usingthe graphical approach
k (M/s)
T, ˚C
T, K
0.000886
25
298
0.000894
50
348
0.000908
100
398
0.000918
150
448
?
75
348
ln (k) = –36.025/T – 6.908
ln (k) = –36.025/(348) – 6.908 = – 7.011
Ea
Ea
Arrhenius Equation: Graphing Example
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Ea
Ea
Arrhenius Equation
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Sometimes a graph is not needed
oOnly have two k s at two Ts
Here use van't Hoff Equation derived fromArrhenius equation:
Ea
Ea
Arrhenius Equation: Ex Vant Hoff Equation
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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CH4 + 2 S2 CS2 + 2 H2S
k (L/mol s)
T (˚C)
T (K)
1.1 = k1
550
823 = T1
6.4 = k2
625
898 = T2
Group
Problem
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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Given that k at 25 ˚C is 4.61 × 10–1 M/s and that at 50 ˚C it is4.64 × 10–1 M/s, what is the activation energy for the reaction?
Group
Problem
Jesperson, Brady, Hyslop. Chemistry: TheMolecular Nature of Matter, 6E
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A reaction has an activation energy of 40 kJ/mol. What happensto the rate if you increase the temperature from 70˚C to 80 ˚C?
A. Rate increases approximately 1.5 times
B. Rate increases approximately 5000 times
C. Rate does not increase
D. Rate increases approximately 3 times