AP Unit 8: Kinetics

reaction ratesrate laws

reaction mechanismscatalysts

introduction in thermodynamics, we looked at a

reaction such as: 2 C8H18 + 25 O2 16 CO2 + 18 H2O

we calculated how much energy would be released IF the reaction happened

but we never asked how fast (or how completely) the reaction would happen. with the above reaction, for instance,

does a puddle of gasoline usually react with atmospheric oxygen quickly?NO!!

the key event in reactions Collisions!!!

Reacting molecules must collide Molecules must be properly aligned Molecules must meet with enough energy

to break the existing bonds Reaction rate depends upon:

how frequently reactant molecules collide what fraction of the collisions are

effective (i.e. have proper energy and alignment)

frequency of effective collisions is increased by

increasing temperature larger fraction of molecules have sufficient KE (to

provide activation E) at higher T increasing solution concentration

molecules collide more often at higher C increasing gas pressure

higher P (or smaller V) is gas equivalent of increased solution concentration

increasing surface area of solid (A:V ratio) dividing solid into smaller pieces increases fraction

of surface molecules, allowing them to be struck more often

adding a catalyst

reaction rates Is “rate” the same thing as “time”?

Does “6 hours” tell you how fast a car moves?

No!! RATE involves “something” time

for a car: miles/hour, meters/sec, etc for a reaction:

grams/second moles/hour etc....

reaction rates which reaction is faster?

5000 molecules 3000 molecules

which reaction is faster? 50 seconds 20 seconds

which reaction is faster? 5000 molecules in 50 seconds 3000 molecules in 20 seconds

can’t be answered!

can’t be answered!

can’t be answered!

can’t be answered!

= 100 molec/s

= 150 molec/sfaster→faster→

NH4+ + NO2

– N2 (g) + 2 H2O (l)

of N2 production

7 0.500 0.250 ???

rate = k [NH4+]x [NO2

–]y

3 A + 2 B C + 4 D

–15.0×10-4

+8.0×10-7

2 NO (g) + 2 H2 (g) N2 (g) + 2 H2O (g)

of N2 production

4 0.33 0.50 ???

rate = k [NO]x [H2]y

change in concentration(and reaction rate) over time

reaction rate often is a function of the concentration of reactant(s)

rate therefore decreases as reactants are consumed

change in concentration(and reaction rate) over time

just how fast the rate decreases (if it does) is not always obvious by simple inspection of graph or data

change in concentration(and reaction rate) over time

we will need a reliable strategy to identify the degree of rate change over time

integrated rate laws zero order:

first order:

second order:

additional written work was done in class to make this complete

Arrhenius equation the experimental rate constant, k,

has three underlying contributors A

Ea

T

additional written work was done in class to make this complete

activation energy

reactants

products

time

pote

nti

al

en

erg

y H

EA

* now we will examine

the energy “hump” energy must be added to

break reactants’ bonds this added energy is

called the activation energy, EA

EA provided by the kinetic energy of colliding molecules

* represents the activated complex

previously we examined H

*the activated complex

also called the transition state is a weird intermediate “molecule”

not a normal reactant or product molecule very unstable has high energy (reactant PE + collision KE) very short-lived probably has an “illegal” Lewis structure

exists (very briefly) after reactant molecules collide and before they separate into product molecules

catalysts catalysts decrease the

activation energy with lower activation energy,

more collisions succeed collisions succeed more

often even without T increase

reaction rate thus increases catalysts cause NO change

in a reaction’s ΔH!

reactants

products

time

pote

nti

al

en

erg

y

withcatalyst

withoutcatalyst

catalyst calculations

reactants

products

time

pote

nti

al

en

erg

y

withcatalyst

withoutcatalyst

75–

260–

200–

300–

catalyst decreases the activation energy: from

by

to

100 kJ

40 kJ

60 kJ

ΔH remains unchanged at

–125 kJ

reaction mechanism is a model of what happens to

atoms and electrons (bonds) step-by-step as reactant molecules collide an activated complex forms product molecules are released from

the activated complex example: CH4 + 2 O2 CO2 + 2

H2O

reaction mechanism

reaction mechanism

reaction mechanism

reaction mechanism

reaction mechanism(as reactions rather than animations)

step 1: CH4 + O2 “CH4O2”step 2: “CH4O2” + O2 CO2 + 2 H2O

net:CH4 + O2 + “CH4O2” + O2 “CH4O2” + CO2 + 2 H2O

CH4 + 2 O2 CO2 + 2 H2O

relationship between reaction mechanism and rate law the stoichiometric coefficients in

the rate-limiting elementary step of the reaction mechanism exactly match the respective exponents in the rate law

relationship between reaction mechanism and rate law

step 1: CH4 + O2 “CH4O2”step 2:“CH4O2” + O2 CO2 + 2 H2Ooption 1: step 1 is slow

slow

fast

rate = k1[CH4]1[O2]1

option 2: step 2 is slow

fast

slow

rate = k2[CH4O2]1[O2]1

But CH4O2 is not a valid reactant molecule.Since step 1 is fast, equilibrium is established:

rate1,for = rate1,rev

k1[CH4][O2] = k -1[CH4O2]

[CH4O2] =k1[CH4][O2]

k -1

rate=k2 [O2]k1[CH4][O2]

k -1

=kx[CH4][O2]2

option 3:some other mechanism

O2+O2...

rate = k [O2]2

step 1:

catalysts catalysts cancel in overall reaction

stoichiometry catalyst ultimately is not consumed in a

reaction it enters as a reactant it is a product in a later step it is thus recycled in the next reaction cycle

one catalyst atom/molecule therefore can catalyze GAZILLIONS of cycles of the reaction since the catalyst is regenerated during each reaction cycle, ready to be used again

catalyst example ozone is destroyed in the upper

atmosphere MUCH faster because Cl atoms (from CFCs) catalyze the reaction

Cl + O3 ClO + O2 (step 1)

ClO + O Cl + O2 (step 2)

Cl+O3+ClO+O ClO+O2+Cl+O2O3 + O 2

O2

catalysts vs. intermediates both catalysts and intermediates cancel

in net reaction intermediate (ClO): generated in an

earlier step, consumed in a later step catalyst (Cl): consumed in an earlier

step, released (regenerated) in a later step

homework example

forward EA = 25 kJ H=–80 kJ

reactants

products

time

pote

nti

al en

erg

y

(kJ)

H=–80kJ

EA=25 kJ

reactants

products

timep

ote

nti

al en

erg

y

(kJ)

H

EA

reverse EA,rev = H=

105 kJ+80 kJ

100–

125–

20–

100–

125–

20–

homework example

forward EA = H=

reactants

products

time

pote

nti

al en

erg

y

(kJ)

H=

EA=

reverse EA,rev = Hrev=

30–

195–

150–

homework example

forward EA = H=

reactants

products

time

pote

nti

al en

erg

y

(kJ)

H

EA

reverse EA,rev = Hrev=

70–

250–

220–

reverse reactions: easy or difficult?

forward exothermic EA relatively small

reactants

products

time

pote

nti

al

en

erg

y H

EA

reactants

products

time

pote

nti

al

en

erg

y H

EA

reverse endothermic EA quite large

but vice versa for endothermic forward reaction:forward reaction more difficult; reverse reaction easier

easier more difficult

reverse reactions theoretically, any reaction can be

reversed, i.e. the products can be turned back into the reactants

practically speaking, however, this is very difficult, especially for exothermic reactions, because the reverse reaction is endothermic

(which is not spontaneous) the activation energy is very large for

the reverse reaction