• Look at 3 levels of chemical change:
– Phenomenological or observational
• Measurement of reaction rates and interpretation of data in terms of rate laws based on mass action
– Mechanistic
• Elucidation of reaction mechanisms = the
‘elementary’ steps describing parts of a reaction sequence (or pathway)
– Statistical Mechanical
• Concerned with the details of mechanisms  energetics of molecular approach, transition states, and bond breaking/formation

• Elementary reactions are those that represent the EXACT reaction, there are NO steps between product and reactant in between what is represented
• Overall Reactions represent the beginning and final product, but do NOT include one or more steps in between.
FeS
2
+ 7/2 O
2
+ H
2
O  Fe 2+ + 2 SO
4
2+ 2 H
2 NaAlSi
3
O
8
+ 9 H
2
O + 2 H +  Al
2
Si
2
O
5
(OH)
4
+ 2 Na + + 4 H
4
SiO
4
+

• In it’s most general representation, we can discuss a reaction rate as a function of the extent of reaction:
Rate = d ξ/Vdt where ξ (small ‘chi’) is the extent of rxn, V is the volume of the system and t is time
Normalized to concentration and stoichiometry: rate = dn i
/v i
Vdt = d[C i
]/v i dt where n is # moles, v is stoichiometric coefficient, and C is molar concentration of species i

• For any reaction: X  Y + Z
• We can write the general rate law:
 d ( X
Rate = change in concentration of X with time, t dt
)
 k ( X ) n
Rate Constant
Order of reaction
Concentration of X

• ONLY for elementary reactions is reaction order tied to the reaction
• The molecularity of an elementary reaction is determined by the number of reacting species: mostly uni- or bi-molecular rxns
• Overall reactions need not have integral reaction orders – fractional components are common, even zero is possible

0
Reaction order
1
2
Rate Law d [ A ]
  k dt
Integrated
Rate Law Units for k
A=A
0
-kt mol/cm 3 s d [ A ]
  kA dt ln A=lnA
0
-kt s -1 cm 3 /mol s d [ A ]
  kA
2 dt
1
A

1
A
0
 kt

• First step in evaluating rate data is to graphically interpret the order of rxn
• Zeroth order: rate does not change with lower concentration
• First, second orders:
Rate changes as a function of concentration

d [ A ]
  k dt
• Rate independent of the reactant or product concentrations
• Dissolution of quartz is an example:
SiO
2(qtz)
+ 2 H
2
O  H
4
SiO
4(aq) log k
-
(s -1 ) = 0.707 – 2598/T

d [ A ]
  kA dt
• Rate is dependent on concentration of a reactant or product
– Pyrite oxidation, sulfate reduction are examples

d [ A ]
  kA dt
[ A ] t
[ A ]
0
 e
(
 kt ) log[ A ] t
 log[ A ]
0
 log( e
 kt
)
[ A ] ln
[ A ]
0 t
  kt log[ A ] t
 
0 .
434 kt
 log[ A ]
0
Find order from log[A] t vs t plot 
Slope=-0.434k
 k = -(1/0.434)(slope) = -2.3(slope) k is in units of: time -1

st
• Time required for one-half of the initial reactant to react t
1 2
 ln 2 k

1 ln k
[ A ]
0
0 .
5 [ A ]
0

• Rate is dependent on two reactants or products (bimolecular for elementary rxn):
• Fe 2+ oxidation is an example:
Fe 2+ + ¼ O
2
+ H +  Fe 3+ + ½ H
2
O d [ Fe
2

] dt
  k [ Fe
2

] P
O
2

0
Reaction order
1
2
Rate Law d [ A ]
  k dt
Integrated
Rate Law Units for k
A=A
0
-kt mol/cm 3 s d [ A ]
  kA dt ln A=lnA
0
-kt s -1 cm 3 /mol s d [ A ]
  kA
2 dt
1
A

1
A
0
 kt

nd
• For a bimolecular reaction: A+B  products dx
 k
2
[ A ][ B ]
 k
2
([ A ]
0
 x )([ B ]
0
 x ) dt
[ A ]
0
1

[ B ]
0
[ B ]
0
([ A ] ln
[ A ]
0
([ B ]
0
0

 x ) x )

[ A ]
0
1

[ B ]
0
[ B ]
0
[ A ] ln
[ A ]
0
[ B ]
 k
2 t
[ A ] log
[ B ] t t

0 .
43 k
2
([ A ]
0

[ B ]
0
) t
 log
[
[
B ]
0
A ]
0
[A]
0 and [B]
0 are constant, so a plot of log [A]/[B] vs t yields a straight line where slope = k
2
= k
2
([A]
0
-[B]
0
)/2.3 (when A ≠B)
(when A=B) or

nd
• For a bimolecular reaction: A+B  products dx
 k
2
[ A ][ B ]
 k
2
([ A ]
0
 x )([ B ]
0
 x ) dt
If [A]
0 or [B] reduces to:
0 are held constant, the equation above dx
 k
2
[ A ][ B ]
 k
2
([ A ]
0
 x )([ B ]
0

0 ) dt
SO – as A changes B does not, reducing to a constant in the reaction: plots as a first-order reaction

nd
• Half-lives tougher to quantify if A≠B for 2 nd order reaction kinetics – but if A=B:
1 t
1 2
 k
2
[ A ]
0
If one reactant (B) is kept constant (pseudo-1 st order rxns): t

1 2 ln k
2
2
[ A ]
0

rd
• Ternary molecular reactions are more rare, but catalytic reactions do need a 3 rd component… dx
 k
3
[ A ][ B ][ C ]
 k
2
([ A ]
0
 x )([ B ]
0
 x )([ C ]
0
 x ) dt

• NOT possible for elementary reactions
• Common for overall processes – independent of any quantity measured
 d [ A ] dt
 k
[A]
0
-[A]=kt

• For an overall reaction, one or a few (for more complex overall reactions) elementary reactions can be rate limiting
Reaction of A to P
 rate determined by slowest reaction in between
If more than 1 reaction possible at any intermediate point, the faster of those 2 determines the pathway

• For the example below, let’s determine the order of reaction A + B  C
• Next, let’s solve the appropriate rate law for k
Run #
1
2
3
Initial [A] ([A]
0
) Initial [B] ([B]
0
) Initial Rate (v
0
)
1.00 M 1.00 M 1.25 x 10 -2 M/s
1.00 M 2.00 M 2.5 x 10 -2 M/s
2.00 M 2.00 M 2.5 x 10 -2 M/s

• For an overall reaction, one or a few (for more complex overall reactions) elementary reactions will be rate limiting
Reaction of A to P  rate determined by slowest reaction in between
If more than 1 reaction possible at any intermediate point, the faster of those 2 determines the pathway

A
• Energy required for two atoms or molecules to react

• The activation energy corresponds to the energy of a complex intermediate between product and reactant, an activated complex
A + B ↔ C ± ↔ AB
It can be derived that
E
A
= RT +
D
H
C ±

• collision theory is based on kinetic theory and supposes that particles must collide with both the correct orientation and with sufficient kinetic energy if the reactants are to be converted into products.
• The minimum kinetic energy required in a collision by reactant molecules to form product is called the activation energy , E a.
• The proportion of reactant molecules that collide with a kinetic energy that is at least equal to the activation energy increases rapidly as the temperature increases.

• Svante Arrhenius, in 1889, defined the relationship between the rate constant, k, the activation energy, E
A
, and temperature in
Kelvins: k

Ae

E
A
RT or: log k
 log A

E
A
2 .
303 R 


1
T 


Where A is a constant called the frequency factor, and e
–EA/RT is the Boltzmann factor, fraction of atoms that aquire the energy to clear the activation energy

log k
 
E
A
2 .
303 R 


1
T 


 log A y = mx + b
Plot values of k at different temperatures: log k vs 1/T  slope is E
A
/2.303R to get activation energy, E
A

• E
A can be used as a general indicator of a reaction mechanism or process (rate-limiting)
Reaction / Process
Ion exchange
Biochemical reactions
E
A range
>20
5-20
Mineral dissolution / precipitation 8-36
Mineral dissolution via surface rxn 10-20
Physical adsorption
Aqueous diffusion
2-6
<5
Solid-state diffusion in minerals
Isotopic exchange in solution
20-120
18-48