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General Chemistry IIGeneral Chemistry IICHEM 152CHEM 152

Week 3

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Week 3 Reading Assignment

Chapter 13 – Sections 13.5 (temperature), 13.7 (catalysts)

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Temperature and Rate

• We have seen that, generally, as temperature increases so does the reaction rate.

• But the powers in the rate law do NOT change

• This means that k is temperature dependent.k

Temperature

We need a microscopic modelto explain this

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Three conditions must be met at the molecular level if a reaction is to occur:

The molecules must collide;

They must be positioned so that the reacting groups are together in a transition state between reactants and products;

The collision must have enough energy to form the transition state and convert it into products.

Collision Rate Model

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Molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation.

Right Orientation

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Activation Energy• There is a minimum amount of energy required

for reaction: the activation energy, Ea.

• Just as a ball cannot get over a hill if it does not have enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.

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The higher the temperature, the more molecules have energy to overcome the

activation energy barrier.

Low T

High T

Extra moleculesat high T thatexceed the Ea

Enough Energy

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Transition State

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Reaction Coordinate Diagrams

Reaction coordinate diagrams help visualize energy changes throughout a process.

ReactionCoordinate

CH3NC CH3CN

rearrangement of

methyl isonitrile.

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Reaction Coordinate Diagrams

E Reactants

E Products∆E reaction

E activated complex

Ea

Notice Ea is not related to ∆E

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Find Ea and ∆E in each case. Identify endothermic and exothermic processes.

Your Turn

UA GenChemReaction progress

Po

ten

tia

l E

ne

rgy

SOLUTION

A key reaction in the upper atmosphere is

O3(g) + O(g) 2O2(g)

The Ea(fwd) is 19 kJ, and the Hrxn (∆E) for the reaction is -392 kJ. Draw a reaction energy

diagram for this reaction, postulate a transition state, and calculate Ea(rev).

O3+O

2O2

transition state

Ea= 19kJ

Hrxn = -392kJ

Ea(rev)= (392 + 19)kJ =

411kJ

Your Turn

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The Arrhenius Equation

RTEa

Aek

ln k = ln A - Ea/RT

ln

k2

k1

=Ea

R-

1

T2

1

T1

-

where k is the rate constant at TEa is the activation energyR is the energy gas constant

= 8.3145 J/(mol K)T is the Kelvin temperature

A is the collision frequency factor

Temperature Effects

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ln k = -Ea/R (1/T) + ln A

Graphical Analysis

.

Y = m X + b

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Suppose a chemical reaction has an activation energy of 76 kJ/mol.

By what factor is the rate of reaction at 50 oC increased over its rate at 25

oC?k2=rate constant @ 50ºC k1=rate const @ 25ºC

k2/k1 = 10.7 over 10 times faster!

Typical Problems

lnk2

k1

1323.15K

1298.15K

-=-76000 J

8.3145 J/mol K

lnk2

k1

= 2.37

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The decomposition of hydrogen iodide,

Ea = ___________ kJ/mol

2HI(g) H2(g) + I2(g)

has rate constants of 9.51x10-9 L/mol*s at 500. K and 1.10x10-5 L/mol*s at 600. K.

Find Ea.

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SOLUTION

The decomposition of hydrogen iodide,

2HI(g) H2(g) + I2(g)

has rate constants of 9.51x10-9L/mol*s at 500. K and 1.10x10-5 L/mol*s at 600. K. Find Ea.

lnk2

k1=

Ea-R

1

T2

1

T1

-

ln1.10x10-5L/mol*s

9.51x10-9L/mol*s

1

600K

1

500K-Ea = -

(8.314J/mol*K)

Ea = 1.76 x 105 J/mol = 176 kJ/mol

Answer

/

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Series of plots of concentra-tion vs. time Initial

rates Reaction orders

Rate constant (k) and actual

rate law

Integrated rate law (half-life,

t1/2)

Rate constant and reaction

order

Activation energy, Ea

Plots of concentration

vs. time

Find k at varied T

Determine slope of tangent at t0 for

each plot

Compare initial rates when [A]

changes and [B] is held constant and

vice versa

Substitute initial rates, orders, and concentrations

into general rate law: rate = k [A]m[B]n

Use direct, ln or inverse plot to

find order

Rearrange to linear form and

graph

Find k at varied T

Overview

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Catalysts speed up reactions by altering the mechanism to lower the activation energy

barrier, but they are not consumed.

MnOMnO22 catalyzes decomposition of H catalyzes decomposition of H22OO22

2 H2 H22OO22 2 H 2 H22O + OO + O22

Catalysis

Uncatalyzed reactionUncatalyzed reaction

Catalyzed reactionCatalyzed reaction

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Imagine trying to get a bunch of cattle to where you want them to go without any

help…

“Cattle-ists”

Will they get there very

quickly on their own? Will they

take the shortest path do

get there? Is there a way to

help?

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What if there was someone there to herd the cattle and guide them along a more efficient

path?

“Cattle-ists”

The end result is the same – but

the reaction takes a more efficient path.

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The metal-catalyzed hydrogenation of ethyleneH2C CH2 (g) + H2 (g) H3C CH3 (g)

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Catalytic Converters

13.6

CO + Unburned Hydrocarbons + O2 CO2 + H2Ocatalytic

converter

2NO + 2NO2 2N2 + 3O2

catalyticconverter

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CatalysisCatalysis

3) Enzymes — biological catalysts3) Enzymes — biological catalysts3) Enzymes — biological catalysts3) Enzymes — biological catalysts

Enzymes are specialized organic

substances, composed of polymers of amino acids (proteins), that

act as catalysts to regulate the speed of

the many chemical reactions involved in

the metabolism of living organisms.

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Enzyme Catalysis

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Enzymes

uncatalyzed

enzymecatalyzed

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Enzymes

Saturation Effects

Denaturization

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Summary Activity:Fireflies flash at a rate that is temperature dependent.

At 29 ˚C the average firefly flashes at a rate of 3.3 flashes every 10. seconds.

At 23 ˚C the average rate is 2.7 flashes every 10. seconds.

Use the Arrhenius equation to determine the activation energy (kJ/mol) for the flashing process.

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The activation energy of the firefly flashing process

is:

Ea = ___________ kJ/mol

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Do you know the equation for the force of

gravity?

0%0%1. Yes 2. No

Rotate your group spokesperson for today. Spokesperson: go to

www.rwpoll.com – LOG IN and enter the session ID.

Then – Answer this question: