Chemical Kinetics CHAPTER 14 Part B Chemistry: The Molecular Nature of Matter, 6 th edition By...

Chemical Kinetics

CHAPTER 14

Part B

Chemistry: The Molecular Nature of Matter, 6th editionBy Jesperson, Brady, & Hyslop

2

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

3



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

4Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Integrated Rate Laws



5

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


6

Integrated Rate Laws First Order Integrated Rate Law

• Corresponding to reactions – A products

• Integrating we get

• Rearranging gives

• Equation of line y = mx + b


7

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


8

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


9

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


10

Integrated Rate Laws Graphically determining Order

Make two plots:

1. ln [A] vs. time

2. 1/[A] vs. timeo 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]


11

Integrated Rate Laws Graphically determining Order


12

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 + Cl2Integrated Rate Laws


13

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

45

0 200 400 600 800 1000 1200

time (min)

1/[

SO

2C

l 2]

(L

/mo

l)

Reaction is 1st order in SO2Cl2

Example: SO2Cl2 SO2 + Cl2Integrated Rate Laws


14

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


15

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/[

HI]

(L

/mo

l)

Reaction is second order in HI.


16

GroupProblem

A plot for a zeroth order reaction is shown. What is the proper label for the y-axis in the plot ?

A. Concentration

B. ln of Concentration

C. 1/Concentration

D. 1/ ln Concentration

0 200 400 600 800 1000 1200time (min)

Zeroth Order Plot


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


18

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


19

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


20

Half Life (t1/2)Integrated Rate Laws


21

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?


22

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?


23

GroupProblem

The half-life of I-132 is 2.295 h. What percentage remains after 24 hours?


24

Half Life (t1/2): Carbon-14 DatingIntegrated Rate Laws


25

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


26

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?


Collision Theory



28

Collision Theory

Reaction Rates

Collision Theory

As the concentration of reactants increaseoThe 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


29

Collision Theory

Reaction Rates

Rate of reaction proportional to number of effective collisions/sec among reactant molecules

Effective collision o One that gives rise to product

e.g. At room temperature and pressureo H2 and I2 molecules undergoing 1010 collisions/sec

o Yet reaction takes a long timeo Not all collisions lead to reaction

Only very small percentage of all collisions lead to net change


30

Collision Theory

Molecular Orientation


31

Collision Theory

Temperature

As T increaseso More molecules have Ea

o So more molecules undergo reaction


32

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


Transition State Theory



34

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


35

Transition State

Molecular Basis of Transition State Theory

Reaction Coordinate (progress of reaction)

Po

ten

tial

En

erg

y

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


36

Transition State

Exothermic Reactions

Reaction Coordinate (progress of reaction)

Po

ten

tial

En

erg

y Exothermic reaction• Products lower PE than reactants

Exothermic Reaction H = –

Products


37

Transition State

Exothermic Reactions

o Hreaction < 0 (negative)

o Decrease in PE of systemo 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


38

Transition State

Endothermic Reactions

Endothermic Reaction H = +

Hreaction = Hproducts – Hreactants


39

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


40

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

C

Transition State


41

GroupProblem

Draw the transition state complex, or the activated complex for the following reaction:

CH3CH2O- + H3O+ CH3CH2OH + H2O


Activation Energies



43

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/mol

oT = Temperature in K


44

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


45

Ea Arrhenius Equation: Graphing Example

Given the following data, predict k at 75 ˚C using the 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


46

Ea Arrhenius Equation: Graphing Example

-7.02

-7.01

-7.00

-6.99

f(x) = − 36.0249500146524 x − 6.90800232199209R² = 0.999727511024561

1/T (K–1)

ln k


47

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:


48

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


49

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?


50

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 times

B. Rate increases approximately 5000 times

C. Rate does not increase

D. Rate increases approximately 3 times

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