CE 520 Environmental Engineering Chemistry
Lecture 3 Dr. S.K. Ong
KINETICS

Thermodynamic calculations allow us to determine the direction of the reaction and provide information whether a reaction will occur or not but do not tell us how fast the reaction will occur

In a rigorous manner, thermodynamic calculations thus equilibrium state truly pertain to a "closed" system (i.e., systems with a finite mass which do not exchange materials with their surroundings).

Many reactions are thermodynamically favorable but reactions may not occur in geological time periods of the natural system or within the hydraulic retention time of the engineered systems such as a treatment plant.
Furthermore, most natural and engineered systems are open systems where there are inflows and outflows of masses with their surroundings.

Kinetics or reaction rates are used in open systems such as the design of water and wastewater systems.
CHARACTERIZATION OF REACTION KINETICS
Consider a simple reaction
Plot:
If the rate of change of species A with time (r
A
) is proportional to [A] then: where k
1
is the reaction rate constant (units = 1/time)and the negative sign accounts for decreasing concentrations of
A with time. Note the difference between r
A
(reaction rate) and k
1
(reaction rate constant). Similarly for the rate of change of species B is given by:
(Note dB/dt and dA/dt changes with time, not for k
1
)
For a simple reversible reaction:
The rate of change of A can be written as: similarly,

At equilibrium, the forward rate of reaction is equal to the reverse rate of reaction, i.e., r
A
= r
B
= 0; therefore 0 = - k
1
[A] + k
2
[B] or
General Reaction k
1 a A + b B ====> c C + d D
The rate of disappearance of A is equal to
The order of the reaction or molecularity of the reaction is given by (a + b)
Other examples: A ==> B + C
M + N ===> P
R + 2 S ==> T + U the stoichiometric equation.
Example: R + 2S ==> T + U unimolecular or first order bimolecular or second order termolecular or third order
However, it is possible for the reaction to proceed as whereby m

a and n

b. In this case, the reaction written as a A + b B ===> c C + d D does not proceed as written but most probably occur with one or more intermediate steps. For example:
A + A ==> 2A* + B ===> B A 2 ====> C + D

The various sequences of reactions or intermediate steps are termed as elementary or simple reactions. Note that a chemical reaction is simple or elementary when the reaction rate equation representing the reaction looks like the same as the stoichiometric equation of the simple reaction.

Note that m and n may be positive or negative whole numbers or fractions.

If the order of the reaction is greater than 2 there is a good chance that there are intermediate steps.

The rate of disappearance/appearance of any species is related stoichoimetrically as shown below. The relative number of moles of species reacting and the relative number of moles of products being formed should conform to dT
 k dt dS
 
2 dt
[ R ][ S ]
2 k [ R ]{ S ]
2 since

1
2 dS
 dT dt dt

Special Cases
Zero Order – reaction independent of concentration, example A == > B dA
  k dt
A

A o
 kt
A
 dA
A o
  t
 o k
First Order dA
  k dt
[ A ] dt
A

A o dA
A
  t
 o k ln


[ A
[ A o
]
]


  kt or ln dt therefore
A useful equation:


[ A o
[ A
] o

] x


  kt where x = amount reacted at time t
Example: All radioactive decays are first order and irreversible; 14 C ==> 13 C ==> 12 C. If the half-life of 14 C to
12 C is 5,760 years, calculate the rate constant for the radioactive reaction. Using: ln


A
A o


  kt where A/Ao = ½ and T = 5,760 years k = - ln (0.5)/5,760 = t
1 / 2
 ln 2

0 .
69 k k
The term relaxation time, characteristic time, life time (

= 1/k) which is the time for reaction concentration to reach
1/e or 0.368 of its initial value.
With the exception of radioactive decay, not many reactions of environmental interest are truly first order.
However, many reactions under certain reaction conditions appear to be first order. Example: k
A + B == > C + D
Then d [ A dt
]
  k [ A ][ B ] integrated form
1
[ B o
]

[ A o
] ln
[ A o
][ B ]
[ B o
][ A ]
 kt
Note that if [B o
] remains fairly constant, i.e.,
[B o
] >> [A o
] and [B]

[B o
] the equation can be simplified to
[B o
] is very large in comparison to [A], then
1
[ B o
] ln
[ A o
]
[ A ]
 kt or ln
[
[
A
A o
]
]
 k [ B o
] t
 k
' t where k’ = k[B o
] have a pseudo-first order reaction.

Second Order dA
  k dt
[ A ]
2
A

A o dA
[ A ]
2
  t
 o k dt
Half life for 2 nd order t
1 / 2

1 k [ A o
]
Summary of Special Cases
Reactions Rate Equation Generalized Equation Units for k
___________________________________________________________________________
Zero Order dA/dt = - k
(independent of concentrations)
First Order dA/dt = - k [A] concentration/time
1/time
Second Order dA/dt = - k [A] 2 1/(time • concentration)
____________________________________________________________________________
Example: Water Chemistry Section 2.9 Problem 5 A ===> products
Evaluate whether reaction is 1st or 2nd order. Calculate the rate constant, note that A o
= 1.0.
Time (sec)
0
A (mM/liter)
1.00 ln (A/A o
) 1/A
11
20
48
105
0.5
0.25
0.1
0.05
Plot ln (A/A o
) vs t and 1/A vs t, check for linearity

Other models
There are many different kinetic models that may be modeled by including various environmental conditions:
Examples :
(i) Production inhibition model S == > P where [P] is the concentration of the product
[S] is the concentration of the reactant
 max
= maximum reaction rate constant
K m
= is constant and K p
= inhibition constant
Modified Gompertz equation is a three parameter model where: M = cumulative mass of A produced at incubation time t,
P = mass of A production potential (mg)
Rm = maximum A production rate (mg/day), and

= lag phase (days)
Consecutive Reactions
Example: Nitrification, i.e., ammonia oxidation by nitrifying bacteria in wastewater treatment plants
Dechlorination of chlorinated solvents
Rate equations are as follows:
Solution : Initial conditions [A] = [A o
], [B] = [B o
], [C] = [C o
]
[ A ]

[ A o
] e
 k
1 t
[ B ]
 k
1
[ A o k
2
 k
1
]
 e
 k
1 t  e
 k
2 t


[ B o
] e
 k
2 t
Normally, one reaction will be slow, thus controlling the overall reaction rate. This step is called the rate-limiting or rate determining step. For example the conversion of nitrite to nitrate is rapid in comparison to the conversion of ammonia to nitrite. Therefore conversion of ammonia to nitrite is the rate limiting step.

Plot:
Transition State Theory

Most widely used theoretical framework for elementary solution reactions. Known as the transition state theory or the activated complex theory or absolute reaction rate theory.

Molecules come together to form a molecule of transition state which is stable for a short time but eventually if sufficient energy is available will reduce to the products of the reactions. The energy need is called the activation energy, E a
(see plot). eg.,
In the presence of a catalyst, the activation energy needed is reduced allowing reaction to proceed easily.
Examples of a catalyst in wastewater systems are ferrous salts to promote formation of OH radicals in oxidation.
Disproportionation of monochloramine in the presence of H
3
PO
4
and H
2
PO
4
2.
Another theory is the Encounter Theory which relies on diffusion controlled rates for the reaction to occur.
Temperature Dependence of Reaction Rate Constants
Use Arrhenius equation
(note equation similar to variation of equilibrium constant with temperature)
Integrating we have: where ln a is the Arrhenius constant or frequency factor plot ln k vs 1/T, then slope of line is equal to -Ea/R or two point equation
For small temperature fluctuations in natural water systems (1 atm and 4 o to 25 o C), we can write the equation as

where e

E a
RT
2
T
1 is approximately constant =

assuming that R T
2
T
1
is fairly constant. then k
2
 k
1
 ( T
2

T
1
) k
2
 k
20
 ( T
2

20 )
Comparison of Equilibrium and Kinetics
Closed System

The relationship between equilibrium (thermodynamics) constants and kinetic rate constants for simple systems can be illustrated as follows. For a single chemical reaction and closed system:
A < == > B
We have K = [B]/[A]
The assumptions are:
- the system is closed, no exchange of materials from the surroundings and the system
- requires fixed temperature and pressure
- homogeneous distribution of A and B throughout the system
We have also shown earlier that in a closed system, the forward and reverse rate constants are related to the equilibrium constant as follows:
[ B ]
[ A ]
 k
1 k
2
 forward reverse rate rate cons cons tan tan t t

K
In a closed system, the time-invariant condition is the equilibrium condition.
Continuous Flow Open System

to illustrate the relationship between equilibrium constants and rate constants for open systems, assume the above simple reaction and a reactor as shown. Mass balances for input and output of material fluxes are given by:
Mass balances for A for B
The time-invariant condition requires in an open system, this condition is called steady state condition

0

Q [ A o
]

Q [ A ]
 k
1
[ A ] V
 k
2
[ B ] V
0
 
Q [ B ]
 k
1
[ B ] V

= hydraulic residence time,
If

is large and k
1
is large (the reaction is fast), the first term on the right tends to zero, therefore an open system at steady state can be approximately equal to equilibrium condition. If not, kinetics have to be considered.