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Course Syllabus Chemistry I/II Ions and Common Substances Nomenclature Competency Sheet Examination I Objectives: Thermochemistry Activation Energy Problem Set Periodic Table

Chemistry 202Edward A. Mottel

Department of Chemistry

Rose-Hulman Institute of Technology

04/18/23

About Me

Edward A. Mottel Department of Chemistry F-111 x8315

04/18/23

Goals of the Course

Thermochemistry Equilibria Acid-Base Reactions Electrochemistry

04/18/23

Requirements

• Laboratory• Chem 201/202 Laboratory Manual

“An Introduction to Experimental Chemistry” (Bookstore)

• Safety Goggles (Voucher from Bookstore)• Quadrille ruled bound laboratory notebook• Pen• Proper Attire

04/18/23

Course Syllabus

• Course grade depends on• Three Exams• Final Exam• Competency tests and quizzes• Problem Sets / Assignments• Laboratory Notebook & Reports

Activation Energy

Edward A. Mottel

Department of Chemistry

Rose-Hulman Institute of Technology

04/18/23

Activation Energy

Reading assignment:• Chang: Chapter 15.4

This lecture continues a discussion of the factors that affect the rate of a chemical reaction.

The topic of activation energy is introduced along with the Arrhenius equation.

Reaction Rate

- dPA

dt= k [A]

n

Factors that affect the rate of reaction

concentrationor pressure

catalyst

particle sizetemperature

Reaction Coordinate DiagramCollision Theory

Ene

rgy

Reactants Products

Ea

H

For a chemical reactionto occur, the reactantsmust collide successfullywith enough energyto overcome theactivation energy (Ea).

If the products aremore stable than the

reactants, H < 0, the reaction is exothermic.

04/18/23

Ene

rgy

Reactants Products

Reaction Coordinate DiagramCatalyst

Why does therate of reaction

change when a catalystis added?

Why does H for thereaction stay the same?

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The Rate Constant

k = A e-Ea/RT

- dPN2O5

dt = k (PN2O5)n

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Arrhenius EquationFactors Affecting the Rate of Reaction

k = A e-Ea/RT

Rate constant

Geometric factor Activation energy

Temperaturedependence

(kelvin)

Rewrite the Arrhenius equation in straight line form.

The Arrhenius equation allows the rate of a reactionto be calculated at different temperatures.

Inverse Temperature (K-1)

ln (

Rat

e C

on

stan

t)

-Ea/R

Arrhenius EquationLinear Form

Arrhenius Equation

The rate constant for a chemical reaction is4.0 x 10-4 s-1 at 280. K

and2.0 x 10-3 s-1 at 300. K.

What is the activation energy of this reaction?

What rate constant would be expected at 320. K?

k Temp(K) ln kInverse Temp K

0.00040 280 -7.82 0.003570.0020 300 -6.21 0.00333

-6760 16.318

lnk320 -4.806k320 0.0082

Activation Energy Determination

y = -6759.6x + 16.318

R2 = 1

-8.00

-7.00

-6.00

0.00320 0.00340 0.00360

Inverse Temp (1/K)

ln k

Boltzmann Distribution

energy

num

ber

of m

olec

ules

T2 > T1

average

The rate of many organic chemical reactions doublewith every 10 degree change in temperature.

Reaction Rates vary with Temperature

If you run a fever of a 103 °F,how much “faster” is your body running,

than at a normal temperature?

The rate of many organic chemical reactions doubleswith every 10 degree Celcius change in temperature

around room temperature.

k Temp(K) ln kInverse Temp K

0.00040 280 -7.82 0.003570.0020 300 -6.21 0.00333

-6760 16.318

lnk320 -4.806k320 0.0082

Activation Energy Determination

y = -6759.6x + 16.318

R2 = 1

-8.00

-7.00

-6.00

0.00320 0.00340 0.00360

Inverse Temp (1/K)

ln k

k Temp(K) ln kInverse Temp K

0.00040 280 -7.82 0.003570.0020 300 -6.21 0.00333

-6760 16.318

lnk320 -4.806k320 0.0082

Activation Energy Determination

y = -6759.6x + 16.318

R2 = 1

-8.00

-7.00

-6.00

0.00320 0.00340 0.00360

Inverse Temp (1/K)

ln k

Temp(oC)Time(s) Temp(K) InvTempK k lnk9.5 106 282.6 0.003539 59.4 4.08

14.4 73 287.5 0.003478 86.3 4.4619.9 45 293 0.003413 140.0 4.94

25.1 30 298.2 0.003353 210.0 5.35 -6644 27.593130.2 22 303.3 0.003297 286.4 5.66 173.51 0.59291

0.998 0.033361466.1 3Formaldehye-Bisulfite Activation

Energy Determination

y = -6643.7x + 27.593

R2 = 0.998

4.00

4.50

5.00

5.50

6.00

0.0032 0.0033 0.0034 0.0035 0.0036

Inverse Temp (1/K)

ln k

k Temp(K) ln kInverse Temp K

1.00000 293 0.00 0.00341 98.6 37 3102.0000 303 0.69 0.00330 103 39.4 312.4

-6154 21

lnk98.6 1.15 3.164lnk103 1.31 3.695

1.168

Body Temperature Kinetics

y = -6153.7x + 21.002

R2 = 1

-0.20

0.80

0.00320 0.00340 0.00360

Inverse Temp (1/K)

ln k