Finding reaction velocity
Physical Chemistry
Lecture 7
Multi-step reaction mechanisms
One knows the form of the reaction rate for every
elementary reaction
Express disappearance of reactant or appearance of
product in terms of the rates of these elementary
steps
Include terms for every step that affects reactants
(or products), with stoichiometry
Example from Bodenstein’s work:
d[H2 ]
d[HBr]
 v2  v4
 v2  v3  v4
dt
dt
d[Br2 ]

 v1  v3  v5
dt

Elementary reactions
Reactions are often more complex than
indicated by the beginning and ending states
alone
Some reactions do occur in a single step -elementary reactions



Generally involve simple mono- or bimolecular
interactions
Order in elementary reactions is the stoichiometry
number
Examples:
H2 + O  H2O
H + Br  HBr
Reaction mechanism
A description of the course of a reaction in
terms of elementary reactions
Example from Bodenstein’s work:
H 2 (g) 
Br2
Br
H
H
Br
Br2 ( g )  2 HBr ( g )
 H2
 Br2
 HBr
 Br





2 Br
HBr  H
HBr  Br
 Br
H2
Br2
v1
v2
v3
v4
v5
Reaction mechanisms are not necessarily
unique
Reducing rate equations
Eliminate dependences on reactive
species (reactive intermediates such as
radicals)
Use approximations


Steady state for reactive species
Fast-equilibrium approximation when two
fast steps precede a slow one
Must be able to identify which are
reactive intermediates
Steady-state approximation
Reactive species have very low concentrations
After initiation, this concentration is presumed
to be independent of time
d [reactive species]
 0
dt
Gives relation among velocities of elementary
steps
Example from Bodenstein’s work:
d [ Br ]
 2 v1
dt
 v2
 v3
 v4
 2 v5
 0
1
Fast-equilibrium
approximation
Summary
One step containing an intermediate is ratelimiting
Prior step is reversible
(1) A  B  C
fast
( 2) C 
A 
B
fast
( 3) C  Product
slow
One presumes a quasi-equilibrium in the first
two steps to relate the concentrations
K eq 
k1 [C ]


k 2 [ A][ B]
d [ Product ]
 k3 [C ]  k3 K eq [ A][ B ]
dt
Every reaction is described by an equation
H2 ( gas) 
Br2 ( gas)  2 HBr ( gas)
“Simple” reaction sequences solved exactly
Generally, equation like above does NOT describe
reaction course
Often cannot get exact time dependence of
concentrations for reactions

Use a mechanism
 Overall reaction expressed in terms of elementary steps
 Not unique

“Solve” mechanism, using approximations if necessary
 Rate-limiting steps
 Steady-state approximation
 Fast-equilibrium approximation
Solving mechanism of
formation of HBr
From the reaction mechanism, identify
time changes of reactive species and
use the steady-state approximation
d [ Br ]
 2 v1  v2  v3  v4
dt
d[ H ]
 v2  v3  v4  0
dt
 2 v5
 0
These two equations give relations
1/ 2
v1
k 
 v5  [ Br ]   1  [ Br2 ]1/ 2
 k5 
 k k2 
 [H]   1 2 
 k5 
1/ 2
[ H2 ][ Br2 ]1/ 2
k 3[ Br2 ]  k 4 [ HBr ]
Example: H2 + Br2 continued
Identify the reaction velocity in terms of
change of a reactant or product
v  
d [ Br2 ]
  v1
dt
 v3
 v5
 v3
Substitution gives the final prediction of
the reaction mechanism
 k k 2k 2 
v   1 2 3
 k5 
1/ 2
[ H2 ][ Br2 ]3/ 2
k 3[ Br2 ]  k 4 [ HBr ]
2