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Chemical Kinetics CHAPTER 14 Part B Chemistry: The Molecular Nature of Matter, 6 th edition
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Transcript of Chemical Kinetics CHAPTER 14 Part B Chemistry: The Molecular Nature of Matter, 6 th edition
Chemical Kinetics
CHAPTER 14
Part B
Chemistry: The Molecular Nature of Matter, 6th editionBy Jesperson, Brady, & Hyslop
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CHAPTER 14 Chemical Kinetics
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Learning Objectives: Factors Affecting Reaction Rate:
o Concentrationo Stateo Surface Areao Temperatureo Catalyst
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
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CHAPTER 14 Chemical Kinetics
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Lecture Road Map:
① Factors that affect reaction rates
② Measuring rates of reactions
③ Rate Laws
④ Collision Theory
⑤ Transition State Theory & Activation Energies
⑥ Mechanisms
⑦ Catalysts
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Integrated Rate Laws
CHAPTER 14 Chemical Kinetics
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Integrated Rate Laws Concentration & Time
Rate law tells us how speed of reaction varies with concentrations.
Sometimes want to know o Concentrations of reactants and products at
given time during reactiono How long for the concentration of reactants to
drop below some minimum optimal value
Need dependence of rate on time
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Integrated Rate Laws First Order Integrated Rate Law
• Corresponding to reactions – A products
• Integrating we get
• Rearranging gives
• Equation of line y = mx + b
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Integrated Rate Laws First Order Integrated Rate Law
Yields straight lineo Indicative of first order
kineticso Slope = –ko Intercept = ln [A]0
o If we don't know already
0]ln[]ln[ AktA t Slope = –k
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Integrated Rate Laws 2nd Order Integrated Rate Law
• Corresponding to special second order reaction – 2B products
• Integrating we get
• Rearranging gives
• Equation of line y = mx + b
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Integrated Rate Laws 2nd Order Integrated Rate Law
Yields straight lineo Indicative of 2nd order
kineticso Slope = +ko Intercept = 1/[B]0
0][1
][1
Bkt
B t
Slope= +k
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Integrated Rate Laws Graphically determining Order
Make two plots:1. ln [A] vs. time2. 1/[A] vs. time
o If ln [A] is linear and 1/[A] is curved, then reaction is 1st order in [A]
o If 1/[A] plot is linear and ln [A] is curved, then reaction is 2nd order in [A]
o If both plots give horizontal lines, then 0th order in [A]
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Integrated Rate Laws Graphically determining Order
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Time, min [SO2Cl2], M ln[SO2Cl2] 1/[SO2Cl2] (L/mol)
0 0.1000 -2.3026 10.000100 0.0876 -2.4350 11.416200 0.0768 -2.5666 13.021300 0.0673 -2.6986 14.859400 0.0590 -2.8302 16.949500 0.0517 -2.9623 19.342600 0.0453 -3.0944 22.075700 0.0397 -3.2264 25.189800 0.0348 -3.3581 28.736900 0.0305 -3.4900 32.787
1000 0.0267 -3.6231 37.4531100 0.0234 -3.7550 42.735
Example: SO2Cl2 SO2 + Cl2Integrated Rate Laws
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First Order Plot for SO2Cl2 Decomposition
-3.8
-3.6
-3.4
-3.2
-3.0
-2.8
-2.6
-2.4
-2.2
0 200 400 600 800 1000 1200time (min)
ln[S
O2C
l 2]
Second order plot for SO2Cl2
Decomposition
10
15
20
25
30
35
40
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0 200 400 600 800 1000 1200time (min)
1/[S
O2C
l 2]
(L/m
ol)
Reaction is 1st order in SO2Cl2
Example: SO2Cl2 SO2 + Cl2Integrated Rate Laws
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Example: HI(g) H2(g) + I2(g)Integrated Rate 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
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Example: HI(g) H2(g) + I2(g)Integrated Rate Laws
First Order Plot for HI Decomposition at 508 oC
-3.8
-3.6
-3.4
-3.2
-3.0
-2.8
-2.6
-2.4
-2.2
0 50 100 150 200 250 300 350time (s)
ln[H
I]
Second order plot for HI Decomposition at 508 oC
10
15
20
25
30
35
40
0 50 100 150 200 250 300 350time (s)
1/[H
I] (L
/mol
)
Reaction is second order in HI.
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GroupProblem
A plot for a zeroth order reaction is shown. What is the proper label for the y-axis in the plot ?A. ConcentrationB. ln of ConcentrationC. 1/ConcentrationD. 1/ ln Concentration
0 200 400 600 800 1000 1200time (min)
Zeroth Order Plot
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Half Life (t1/2) for first order reactionsIntegrated Rate Laws
Half-life = t½ We often use the half life to describe how fast a reaction takes place First Order Reactions
o Set
o Substituting into
o Gives
o Canceling gives ln 2 = kt½
o Rearranging gives
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Half Life (t1/2) for First Order ReactionsIntegrated Rate Laws
Observe: 1. t½ is independent of [A]o
o For given reaction (and T)o Takes 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)o t½ called half-life
oTime for ½ of sample to decay
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Half Life (t1/2)Integrated Rate Laws
Does this mean that all of sample is gone in two half-lives (2 × t½)?No!
o In 1st t½, it goes to ½[A]o o In 2nd t½, it goes to ½(½[A]o) = ¼[A]o o In 3rd t½, it goes to ½(¼[A]o) = ⅛[A]o
o In nth t½, it goes to [A]o/2n
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Half Life (t1/2)Integrated Rate Laws
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Half Life (t1/2): First Order ExampleIntegrated Rate Laws
131I is used as a metabolic tracer in hospitals. It has a half-life, t½ = 8.07 days. How long before the activity falls to 1% of the initial value?
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GroupProblem
The radioactive decay of a new atom occurs so that after 21 days, the original amount is reduced to 33%. What is the rate constant for the reaction in s–1?
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GroupProblem
The half-life of I-132 is 2.295 h. What percentage remains after 24 hours?
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Half Life (t1/2): Carbon-14 DatingIntegrated Rate Laws
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Half Life (t1/2): Second Order ReactionsIntegrated Rate Laws
How long before [A] = ½[A]o?
o t½, depends on [A]o
o t½, not useful quantity for a second order reaction
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GroupProblem
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 2 hours, there is 0.35 M available?
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Collision Theory
CHAPTER 14 Chemical Kinetics
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Collision Theory
Reaction Rates
Collision TheoryAs the concentration of reactants increase
oThe number of collisions increaseso Reaction rate increases
As temperature increasesoMolecular speed increasesoHigher proportion of collisions with
enough force (energy) oThere are more collisions per secondoReaction rate increases
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Collision Theory
Reaction Rates
Rate of reaction proportional to number of effective collisions/sec among reactant moleculesEffective collision
o One that gives rise to producte.g. At room temperature and pressure
o H2 and I2 molecules undergoing 1010 collisions/seco Yet reaction takes a long timeo Not all collisions lead to reaction
Only very small percentage of all collisions lead to net change
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Collision Theory
Molecular Orientation
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Collision Theory
Temperature
As T increaseso More molecules have Eao So more molecules undergo reaction
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Collision Theory
Activation Energy (Ea)
Molecules must possess certain amount of kinetic energy (KE) in order to react
Activation Energy, Ea = Minimum KE needed for reaction to occur
o Get energy from collision with other moleculeso If molecules move too slowly, too little KE, they just
bounce off each othero Without this minimum amount, reaction will not occur
even when correctly oriented
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Transition State Theory
CHAPTER 14 Chemical Kinetics
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Transition State
Molecular Basis of Transition State Theory
KE decreasing as PE increases
KE KE
PE
KE KE
Is the combined KE of both molecules
enough to overcome Activation Energy
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Transition State
Molecular Basis of Transition State Theory
Reaction Coordinate (progress of reaction)
Pote
ntia
l Ene
rgy
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
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Transition State
Exothermic Reactions
Reaction Coordinate (progress of reaction)
Pote
ntia
l Ene
rgy Exothermic reaction• Products lower PE than reactants
Exothermic Reaction H = –
Products
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Transition State
Exothermic Reactions
o Hreaction < 0 (negative)o Decrease in PE of system
o Appears as increase in KEo So the temperature of the system increases
o Reaction gives off heat o Can’t say anything about Ea from size of H
o Ea could be high and reaction slow even if Hrxn large and negative
o Ea could be low and reaction rapid
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Transition State
Endothermic Reactions
Endothermic Reaction H = +
Hreaction = Hproducts – Hreactants
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Transition State
Endothermic Reactions
o Hreaction > 0 (positive)o Increase in PE
o Appears as decrease in KEo So temperature of the system decreases
o Have to add E to get reaction to goo Ea Hrxn as Ea includes Hrxn o If Hrxn large and positive
o Ea must be high o Reaction very slow
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Transition State
Activated Complex
o Arrangement of atoms at top of activation barrier o Brief moment during successful collision when
o bond to be broken is partially broken and o bond to be formed is partially formed
Example
N CH3C C NH3CH3CN
CTransition State
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GroupProblem
Draw the transition state complex, or the activated complex for the following reaction:
CH3CH2O- + H3O+ CH3CH2OH + H2O
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Activation Energies
CHAPTER 14 Chemical Kinetics
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Ea Arrhenius Equation
The rate constant is dependent on Temperature, which allows us to calculate Activation Energy, Ea
Arrhenius Equation: Equation expressing temperature-dependence of k
oA = Frequency factor has same units as koR = gas constant in energy units
= 8.314 J mol–1 K–1
oEa = Activation Energy—has units of J/moloT = Temperature in K
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Ea Calculating Activation Energy
• Method 1. Graphically• Take natural logarithm of both sides
• Rearranging
• Equation for a line • y = b + mxArrhenius Plot• Plot ln k (y axis) vs. 1/T (x axis) yield a
straight line• Slope = -Ea/R• Intercept = A
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Ea Arrhenius Equation: Graphing Example
Given the following data, predict k at 75 ˚C using the graphical approach
k (M/s) T, ˚C T, K0.00088
6 25 2980.00089
4 50 3480.00090
8 100 3980.00091
8 150 448? 75 348
ln (k) = –36.025/T – 6.908
ln (k) = –36.025/(348) – 6.908 = – 7.011
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Ea Arrhenius Equation: Graphing Example
-7.02
-7.01
-7.00
-6.99f(x) = − 36.0249500146524 x − 6.90800232199209R² = 0.999727511024561
1/T (K–1)
ln k
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Ea Arrhenius Equation
Sometimes a graph is not neededo Only have two k s at two Ts
Here use van't Hoff Equation derived from Arrhenius equation:
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Ea Arrhenius Equation: Ex Vant Hoff Equation
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
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GroupProblem
Given that k at 25 ˚C is 4.61 × 10–1 M/s and that at 50 ˚C it is 4.64 × 10–1 M/s, what is the activation energy for the reaction?
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GroupProblem
A reaction has an activation energy of 40 kJ/mol. What happens to the rate if you increase the temperature from 70˚C to 80 ˚C?A. Rate increases approximately 1.5 timesB. Rate increases approximately 5000 timesC. Rate does not increaseD. Rate increases approximately 3 times