C8H18
+ 25/2 O2 = 8 CO2 + 9 H2O
Balanced equation tells what reacts and what is produced but
not how you go from here to there.
Rxn Mechanism or Path is a theory of how rxn
happens.
O3 ® O2
+ O Step
1
O + O3 ® 2 O2 Step
2
2 O3 + O ® 3 O2
+O Net Rxn
Each step in a rxn mechanism is an elementary step
and the overall rxn is the sum of the elementary steps. An elementary step is thought to occur as
written and can not be broken down further.
O is an intermediate, which is a short-lived species
produced in one step and consumed in another.
Unimolecular – an elementary step involving
the rearrangement of a single molecule or ion producing products or
intermediates. The species rearranging
gets the needed energy from collisions.
A doubling of the concentration will lead to a doubling of the rate of
reaction, so a unimolecular rxn will show 1st order behavior.
Bimolecular – The collision and combination of two reactants to give products is a bimolecular reaction. The reactants can be identical or different
I + I ® I2 Rate
= k [I] [I]
O + O3
® 2 O2 Rate = k [O] [O3]
Ternmolecular Rxn – The simultaneous collision of
three particles is very unlikely.
Before, for an overall rxn, we could not predict the order
of rxn wrt the reactants. However, we
can predict the rate law for an elementary step.
In a multistep mechanism, often one step is much slower than
the others. The overall rate of rxn
cannot exceed the rate of the slowest step.
This step is called the rate determining step.
Mechanism with a slow initial step.
consider
C (CH3) I + OH- ® C (CH3)OH + I Exp Rate Law Rate = k [C (CH3) I]
A proposed mechanism has two steps with the first step being
a slow step
C (CH3) I ® C (CH3)+
+ I- R1 = k1 [C (CH3) I]
C (CH3)+ + OH- ® C (CH3)OH R2 = k2 [C (CH3)+]+ [OH-]
C (CH3) I + OH- ® C (CH3)OH + I
The first step is very slow. The intermediate produced reacts very fast and does not limit the speed of the reaction. Thus doubling the [OH- ] has very
little effect on the rxn rate. So the
overall rate law just depends on the [C (CH3) I].
2 NO + Br2 ® 2 NOBr Rate = k [NO]2 [Br2]
The proposed mechanism that does not involve a termolecular
mechanism is
1) NO + Br2
= NOBr2 (fast,
equilibrium)
2) NOBr2 + NO ® 2 NOBr (slow)
Step 2 is the RDS.
So we predict a rate law of
Rate = k2 [NOBr2] [NO]
Since it is hard to measure the intermediate [ ] (since
there is little of it) we would like to exclude it from our equation. We will try and eliminate it. The intermediate can either decompose to
form the starting reactants or it can quickly react to form products. So the forward and backward Step 1 are in
equilibrium
k1 [NO] [Br2] = k-1 [NOBr2]
Solving for [ NOBr2]
= k1 [NO] [Br2] / k-1
This can be substituted back into the rate law for step 2
Rate = k2
k1 [NO] [Br2] [NO] / k-1 = k [NO]2
[Br2]
Which fits the observed rate law.
Note, a mechanism must predict the experimentally observed
rate law and must sum to give the overall rxn.
Just because a mechanism fits the data, doesn’t mean it is the truth,
but it is consistent with the data.
A substance that changes the speed of a rxn without undergoing a permanent chemical change in itself. It provides a different mechanism and often speeds up a rxn by lowering the activation energy.
CFCl3 ® (UV
light) CFCl2 + Cl
Cl + O3 ® ClO + O2 Ea
= 2.1 kJ/mol
ClO + O ® Cl + O2
Uncatalyzed ozone decomposition
O3 (hu) ® O2 + O Ea
= 14.0 kJ/mol
This reaction has a lower Ea and the Cl is not
consumed and so it greatly speeds up the decomposition of ozone.
Compare the ratio of the catalyzed to the uncatalyzed rate
constants for the ozone decomposition at 25°C (assume
A is the same).
kcat
= A e-E(cat)/RT
kun
= A e-E(un) /RT Solve
for k(cat) / k(uncat)
2CO + 2 NO ®
(Pt/Rh) N2 + 2CO2
Note the catalysts lowers the Ea for the forward
and backward rxn, so it speeds up the rxn both directions.
Heterogeneous Catalysis involve different phase,
for example gas and liquid reactants might react faster on a solid surface.
A Homogeneous catalysis is in the same phase as the
reactants and products.
Inhibitor Something that slows a rxn. down.