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Kinetics Part IV: Activation Energy

Jespersen Chapter 14 Sec 5 & 6

Dr. C. Yau

Fall 2014

The Collision TheoryThe rate of a reaction is proportional to the

number of effective collisions per second among the reactant molecules.

We already know concentration plays an important part in rxn rate:

Conc Freq Collision Rxn Rate

Only EFFECTIVE collisions lead to products.

Only a small fraction of collisions lead to products, based on two other factors:

1) Activation Energy

2) Molecular Orientation 2

Kinetic Energy DistributionEa = Activation energy = minimum energy needed for collision to be effective

Fig 14.11 p.666

Activation Energy is not affected by increase in temperature.

REMEMBER: This is the graph for Kinetic Energy.Don’t confuse with graph for Potential Energy.

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Collision Theory Of ReactionsFor a reaction to occur, three conditions must be met:

1. Reactant particles must collide.

2. Collision energy must be enough to break bonds/initiate.

3. Particles must be oriented so that the new bonds can form.

e.g. NO2Cl + Cl NO2 + Cl2

Eqn Summarizing 3 Factors in Collision Theory

Particulate Level:

Rxn Rate (molecules L-1 s-1) =

N x forientation x fKE

N = # collisions per second per liter of mixture

forientation = fraction of collisions with effective orientation

fKE = fraction of collisions with sufficient kinetic energy for effective collision (area under the curve with KE Ea

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Mathematically, fKE

has been found to be related to Ea and T in this equation:

a

aKE

aKE

-E

RTKE

-E ln =

RTSimplify by finding the anti ln of both sides of the equation...

-Eantiln ln = anti ln

RT

= e

f

f

f

Still remember what fKE stands for?6

Eqn Summarizing 3 Factors in Collision Theory

Macroscopic Level:

Equation has to be in terms of moles instead of molecules.

Conversion factor is

So we divide the previous equation by Avogadro’s number to get

reaction rate in units of mol L-1 s-1

7

23

1 mole of molecules

6.02x10 molecules

-1 -1

23 -1

Rxn Rate (molecules L s )

6.02x10 (molecules mol )mol

= (M/s)L s

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Temperature Effects

Changes in temperature affect the rate constant, k, according to the

Arrhenius equation:p is the steric factor

Z is the frequency of collisions.

Ea is the activation energy

R is the Ideal Gas Constant (8.314 J/(mol K)

T is the temperature (K)

We substitute A (the frequency factor) for (pZ))

E /RTaepZk

E /RTaAek This is an important equation to remember!

Graphical Determination of Ea

E /RTa

-Ea/RT

a

a

a

Ae

ln k = ln A + ln (e )

E ln k = ln A -

RTE 1

ln k = ln A - R T

E 1 ln k = - + ln A

R T

y = m x + b

k

How exactly do we determine Ea?

What do we plot on the x-axis? on the y-axis?

How do we find Ea on the graph?

You are expected to be able to derive this yourself.

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Example 14.12 p.670: Determine Ea in kJ/mol

What do we do with this data?

aE 1ln k = - + ln A

R T 10

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-5

4

-0.70slope =

5.0x10

= -1.4x10 (units?)

aE 1ln k = + ln A

R T

Then what?...

How do we find Ea?

-5

4

-0.70slope =

5.0x10

= - 1.4x10 ( ?)units

Ln

kLn k vs. 1/T

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2

Ea = slope x R

= 1.2x10 kJ/mol

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Determination of Ea from k at 2 temperatures

Ratio form: Can be used when A isn’t known.

a2

2 11

-E 1 1ln -

R T T

k

k

/RTaEAe k

You should be able to derive this equation for yourself. We did a similar derivation earlier this semester for Hvap, VP and T.

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ExampleGiven 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?

1a

-1 -1 -1

4 .64 10 M /s -E 1 1ln -

4 .61x10 M /s 8 .314 J m ol K 323K 298K

x

Ea= 188 J/mol = 2 x102 J/mol

a2

2 11

-E 1 1ln -

R T T

k

k

Can you think of a reason why the graphical method would give a more accurate value for Ea?

a2

1 2 1

-1a

-1 -1 -1

-1 -1a-1 -1

a

6 5 5

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-Ek 1 1ln =

k R T T

-E4.64x10 1 1ln =

4.61x10 8.314 J mol K 323 K 298 K

-Eln 1.00 = 0.00309 K - 0.00335 K

8.314 J mol K-E

0.005 = 8.314 J m

-1-1 -1

-1

2a

0

9

0

(0.00026 K )ol K

0.005 8.314 J molE = 188.66... 2x10 J/mol

0.00026

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Working With The Arrhenius Equation

Given the following data, predict k at 75°C using the graphical approach.

k (M/s) T °C

0.000886 25

0.000894 50

0.000908 100

0.000918 150

ANS k=9.01×10-4M/s

Trendline: y = -36.025x – 6.908

k = ? at T = 75oC

/RTaEAe kaE 1

ln k = - + ln AR T

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T deg C k 1/T (in K-1) ln k25 0.000886 0.003355705 -7.02879360750 0.000894 0.003095975 -7.019804783

150 0.000918 0.002364066 -6.993313167

100 0.000908 0.002680965 -7.004266179

0.00230.0024

0.00250.0026

0.00270.0028

0.00290.003

0.00310.0032

0.00330.0034

-7.03

-7.025

-7.02

-7.015

-7.01

-7.005

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-6.995

-6.99

f(x) = − 36.0249500146574 x − 6.90800232199207R² = 0.999727511024572

ln k vs. 1/T

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In the reaction 2N2O5(g) 4 NO2(g) + O2(g) the following temperature and rate constant information is obtained. What is the activation energy of the reaction?A. 102 kJ mol-1

B. -102 kJ mol-1

C. 1004 kJ mol-1

D. -1004 kJ mol-1

E. none of these

T (K) k (s-1)

338328318

4.87(10-3)1.50(10-3)4.98(10-4)

Practice with Example 14.12 p.672, Exer.26,27,28

If we are to determine Ea graphically, what do we graph?Slope = -1.224x104 K and y-intercept = 30.9, what are the units? What is Ea?

aE 1ln k = + ln A

R T

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Potential Energy Diagrams

                                                                                                                                                                                                                                                                         

                       

The product is said to be “thermodynamically favored” over the reactant. LEARN THIS TERMINOLOGY.

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Potential Energy Diagrams• demonstrate the energy needs and

products as a reaction proceeds• tell us whether a reaction is exothermic or

endothermic• tell us if a reaction occurs in one step or

several steps• show us which step is the slowest

Do not confuse PE diagram with KE diagram! Learn the terminology!

So, remember which is the KE diagram?

Potential Energy DiagramWhat would the potential energy diagram

look like for an endothermic reaction?

Make a sketch of a PE diagram for an endothermic reaction.

Where do we look to find the activation energy?

Where do we look to find the heat absorbed during the reaction?

What is thermodynamically favored?21

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Catalysts• speed a reaction, but

are not consumed by the reaction

• may appear in the rate law

• lower the Ea for the reaction

• may be heterogeneous or homogeneous

Ea of uncatalyzed rxn

Ea of catalyzed rxn

PE

Reaction Coordinate

PE Graph

KE Graph: Effect of Catalyst

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Fra

cti

on

of

Mo

lecu

les

Kinetic Energy

Ea without catalyst

Ea with catalyst

000000

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Catalytic Actions• may serve to weaken bonds through

induction• may serve to change polarity through

amphipathic/surfactant effects• may reduce geometric orientation effects• Heterogeneous catalyst: reactant and

product exist in different states.• Homogeneous catalyst: reactants and

catalyst exist in the same physical state

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Example of a heterogeneous catalyst

Note: Fe is never consumed.Catalysts do not have to be in large amounts.

Well-known “The Haber Process.”

Fe3H2 (g) + N2 (g) 2NH3 (g)

Fe