ChE 452 Lecture 03 Variations In Rate With Temperature 1.

ChE 452 Lecture 03

Variations In Rate With Temperature

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Objectivge For Today

Review Arrhenius’ law Develop some rules of thumb that

allow one to estimate activation barriers from very little data.

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Variations in Rate With Temperature Huge

Rate variations with temperature are much larger than variations in rate with concentration Factor of two in concentration gives

factor of 2-4 in rate Factor of 2 in temperature gives 107

variation in rate

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Homogeneous Reactions

Rate increases with increasing temperature

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3.1 3.2 3.3 3.4 3.5 3.61

10

100

1000 10 °C20 °C30 °C40 °C

Gro

wth

Rate

, #/h

r

1000/T, K-1

Figure 2.10 The rate of E. Coli growth as a function of temperature adapted from

Bailey and Ollis [1977].

Heterogeneous Reactions

Rate has a maximum at intermediate temperatures

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Rat

e, M

olec

ules

/cm

-se

c2

Temperature, KTemperature, K

1E+13

1E+12

1E+11

600400 800600 800400

P =2.E-7 torrCO

A

B

C

2P =2.5E-8 torrO

DEF

Figure 2.18 The rate of the reaction CO + 2 O2 CO2 on Rh(111). Data of Schwartz, Schmidt and

Fisher[1986].

Models For Variations In Rate With Temperature

Key processes• Molecules get hot• Cross a barrier

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Reaction Cordinate

En

erg

y

ReactantsProducts

Barrier

A‡

Figure 7.5 Polanyi’s picture ofexcited molecules.

Models For Variations In Rate With Temperature

Key processes• Molecules get hot• Cross a barrier

Models• Perrin’s Model: energy transfer

dominates• Arrhenius’ Model: barrier dominates

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Perrin’s Model

Assume energy transfer dominates.

k = kT Tn

k = rate constant

kT = preexponential

n = constant between 1 and 4

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(2.24)

Arrhenius’ Model

Assume barrier dominates

k = koexp(-Ea/kBT)

ko = preexponential

Ea = activation barrier, kJ/molecule

kB = Boltzman’s constant, 1.381x10-23

J/K

T = temperature (K)9

(2.26)

Arrhenius’ Model In Kcal/mole

Assume barrier dominates

k = koexp(-Ea/RT)

ko = preexponential

Ea = activation barrier, kcal/molecule

R = Gas law constant, 1.98x10-3

kcal/mole/K

T = temperature (K)10

Real Data Somewhere In Between

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0.5 1 1.5 2 2.5 3 3.5 40.1

1

10

100

1000 3004005006001000 800 700

620 torr

100 torr

20 torr

1000/T, K

Rat

eTemperature,K

-1

Figure 2.6 The rate of the reaction CH + N2 HCN + N as a function of the temperature. Data of Becker, Gelger and Wresen[1996].

Best Fit Of Real Data Uses A Combined Expression

Tk/E-m0

m1BAe(T)k=k

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Arrhenius’ effect much larger than Harcourt and Essen

(2.28)

Rates Double Or Triple When The Temperature

Rises By 10 K

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Reaction Temperature range, C

Rate Change with a 10-K Temperature

Change

3.6-30.4 2.03

23.5-43.6 2.87

24.5-43.6 2.68

0-61 3.0

CH COOCH CH H O

CH COOH CH CH OH3 2 3 2

H

3 3 2

CH

C = CH3

2

CH Cl NaOH

H NaCl H O2

2 2

CH CH CH Cl NaOH

CH CH CH NaCl3 2 2

3 2

HPO3 H O H PO2 3 4

Table 2.6 The variation in rate of a series of reactions with a 10-K change in temperature

Also Works For Biological Processes

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Table 2.7 The variation in the respiration rate of plants with a 10 change in temperature. Data of Clausen[1890].

Wheat 2.47

Lilac 2.48

Lupine 2.463.1 3.2 3.3 3.4 3.5 3.61

10

100

1000 10 °C20 °C30 °C40 °C

Gro

wth

Rate

, #/h

r

1000/T, K-1

Figure 2.10 The rate of E. Coli growth as a function of temperature adapted from

Bailey and Ollis [1977].

Crickets Chirping Example

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3.3 3.35 3.4 3.45 3.5 3.553000

4000

5000

6000

7000

8000

9000

10000

1000/T, KC

hirp

Rat

e, C

hirp

s/hr

10 °C15 °C20 °C25 °C

3.3 3.35 3.4 3.45 3.5 3.551000

10000

2000

3000

4000

5000

6000

7000

80009000

1000/T, K

Chi

rp R

ate,

Chi

rps/

hr

10 °C15 °C20 °C25 °C

-1-1

Figure 2.11 The rate that crickets chirp as a function of temperature. Data for field

crickets (Gryllys pennsylvanicus)

Examples From You Own Life

Why does bread taste better warm? Why does beer taste better cold? Why do you refrigerate, freeze food? Why do you maintain a body

temperature? Why do amphibians stop moving when

it is cold Why a fever?

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Next Key Implications Of Arrhenius’ Law

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(2.32)

(2.33)Figure 2.8 A plot of ½ vs. EA at 100, 200, 300,

400, and 500 K.

0 10 20 30 40 501E-6

1E-2

1E+2

1E+6

1E+10

1E+14 1,000,000 yrs

100 yrs

1 yr

1 day1 hr1 min

1 sec

Ea, Kcal/mole

Hal

f Life

, sec

onds

500K

400K300K200K100K

utemina T)Kkcal15/1(E (2.31)

ondseca T]K)molkcal06.0[(E

autemin Ekcal

molK15T

Example Question

Assume that you are measuring the kinetics of sponification of ethyl acetate in the unit ops lab, and you find that you get 50% conversion in 20 minutes at 25 C. What is the activation energy of the reaction?

How long should it take to get to 50% conversion at 35 C.

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Example

Assume that a reaction has an activation barrier of 35 kcal/mole. Approximately what temperature do we want to run it?

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Summary: Cont

Arrhenius’ law and Perrin’s model Arrhenius: barrier dominates Perrin: energy transfer dominates

Truth in between Biological processes follow the same

rate laws as chemical processes Leads to simple ways to estimate

activation barriers

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