1. reaction rates rate laws reaction mechanisms catalysts
AP Unit 8: Kinetics
reaction rates
rate laws
reaction mechanisms
catalysts

2. introduction in thermodynamics, we looked at a reaction such as:
2 C8H O2  16 CO 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!!

3. 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)

5. 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

6. 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....

7. reaction rates which reaction is faster? 5000 molecules 3000 molecules
50 seconds
20 seconds
5000 molecules in 50 seconds
3000 molecules in 20 seconds
can’t be answered!
can’t be answered!
= 100 molec/s
= 150 molec/s
faster→

10. NH4+ + NO2–  N2 (g) + 2 H2O (l) rate = k [NH4+]x [NO2–]y find x
of N2 production
???
rate = k [NH4+]x [NO2–]y
find x
find initial rates of H2O production (1-7)
find y
find k (with units)
find initial rates of NH4+ (and NO2–) consumption
find rate7

11. 3 A B  C D
–15.0×10-4
+8.0×10-7

12. 2 NO (g) + 2 H2 (g)  N2 (g) + 2 H2O (g)
of N2 production
???
rate = k [NO]x [H2]y
find x
find rates of H2O production (1-4)
find y
find k (with units)
find rates of NO (and H2) consumption (1-4)
find rate4

13. 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

14. 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

15. change in concentration (and reaction rate) over time
we will need a reliable strategy to identify the degree of rate change over time

16. integrated rate laws zero order: first order: second order:
additional written work was done in class to make this complete
integrated rate laws
zero order:
first order:
second order:

17. 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

21. activation energy * previously we examined H
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 *
reactants
products
time
potential energy EA H

23. * 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

24. 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!
without
catalyst
reactants
products
time
potential energy
with
catalyst

25. catalyst calculations
without
catalyst
catalyst decreases the activation energy:
from by to
reactants
products
time
potential energy
100 kJ
300–
260–
with
catalyst
200–
40 kJ
75–
60 kJ
ΔH remains unchanged at
–125 kJ

28. 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

29. reaction mechanism

30. reaction mechanism

31. reaction mechanism

32. reaction mechanism

33. reaction mechanism (as reactions rather than animations)
CH4 + O2  “CH4O2”
step 1:
“CH4O2” + O2  CO2 + 2 H2O
step 2:
net:
CH4 + O2 + “CH4O2” + O2  “CH4O2” + CO2 + 2 H2O
CH4 + 2 O2  CO2 + 2 H2O

34. 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

35. relationship between reaction mechanism and rate law
CH4 + O2  “CH4O2”
step 1:
slow
fast 
“CH4O2” + O2  CO2 + 2 H2O
step 2:
fast
slow
option 1: step 1 is slow
option 2: step 2 is slow
rate = k1[CH4]1[O2]1
rate = k2[CH4O2]1[O2]1
But CH4O2 is not a valid reactant molecule.
Since step 1 is fast, equilibrium is established:
option 3:
some other mechanism
rate1,for = rate1,rev
k1[CH4][O2] = k -1[CH4O2]
O2+O2...
step 1:
k1[CH4][O2]
[CH4O2] =
k -1
rate = k [O2]2
k1[CH4][O2]
rate=k [O2]
=kx[CH4][O2]2
k -1

36. 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

37. 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 + O (step 2)
Cl+O3+ClO+O  ClO+O2+Cl+O2
O3 + O  2 O2

38. 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

43. homework example forward reverse H EA EA = 25 kJ H=–80 kJ EA,rev =
reactants
products
time
potential energy (kJ)
H=–80kJ
EA=25 kJ
reactants
products
time
potential energy (kJ) H EA
125–
125–
100–
100–
20–
20–
forward
EA = 25 kJ
H=–80 kJ
reverse
EA,rev =
H=
105 kJ
+80 kJ

44. homework example forward reverse H= EA= EA = H= EA,rev = Hrev= 195–
reactants
products
time
potential energy (kJ)
H=
EA=
195–
150–
30–
forward
EA =
H=
reverse
EA,rev =
Hrev=

45. homework example forward reverse H EA EA = H= EA,rev = Hrev= 250–
reactants
products
time
potential energy (kJ) H EA
250–
220–
70–
forward
EA =
H=
reverse
EA,rev =
Hrev=

46. reverse reactions: easy or difficult?
easier
more difficult
reactants
products
time
potential energy H EA
reactants
products
time
potential energy H EA
forward
exothermic
EA relatively small
reverse
endothermic
EA quite large
but vice versa for endothermic forward reaction:
forward reaction more difficult; reverse reaction easier

47. 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