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AP Unit 8: Kinetics
reaction ratesrate laws
reaction mechanismscatalysts
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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!!
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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)
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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
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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....
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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→
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NH4+ + NO2
– N2 (g) + 2 H2O (l)
of N2 production
7 0.500 0.250 ???
rate = k [NH4+]x [NO2
–]y
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3 A + 2 B C + 4 D
–15.0×10-4
+8.0×10-7
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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
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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
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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
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change in concentration(and reaction rate) over time
we will need a reliable strategy to identify the degree of rate change over time
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integrated rate laws zero order:
first order:
second order:
additional written work was done in class to make this complete
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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
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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
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*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
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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
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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
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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
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reaction mechanism
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reaction mechanism
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reaction mechanism
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reaction mechanism
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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
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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
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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:
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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
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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
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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
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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–
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homework example
forward EA = H=
reactants
products
time
pote
nti
al en
erg
y
(kJ)
H=
EA=
reverse EA,rev = Hrev=
30–
195–
150–
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homework example
forward EA = H=
reactants
products
time
pote
nti
al en
erg
y
(kJ)
H
EA
reverse EA,rev = Hrev=
70–
250–
220–
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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
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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