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CE 541
Physical Chemistry
What is Physical Chemistry
Is the science that deals with laws that is
related (or that govern) chemical phenomena
such as:
 Gas
laws
 Oxidation-reduction reactions
 Equilibrium relationship
Thermodynamics
Is the study of energy
changes
accompanying
physical
or
chemical
processes
Thermodynamics
(a). Heat and Work
Heat
Is a form of energy passing from one body to another as a result of
temperature difference
Heat Units
Calorie [heat required to raise the temperature of one gram of
water one degree Celsius,  C]
British thermal unit (Btu) [heat required to raise the temperature
of one pound of water by one degree Fahrenheit,  F]
Btu = 252 Calories = 1,054 Joules (J)
Specific heat of a substance [heat required to raise
the temperature of 1 gram of the substance by
one degree Celsius,  C]
Where
q
C
MT
C = specific heat
 q = heat added in Calories or Joules
 M = weight of the substance, grams
 T = raise in temperature,  C

For water at 15  C
C = 1.000 Cal or 4.184 J / gram-degree C
Heat of fusion : heat required to melt a substance
at its normal melting temperature.
Heat of vaporization : heat required to evaporate
a substance at its normal temperature of boiling
Work
Measured in:
Force  Distance
Work (dw) is equivalent to:
Pressure  Volume Change
Then,
(dw) = P  dV
Work Units



foot – pounds (ft-lb)
Joules ( 1 Cal = 4.184 J)
Btu (1 Btu = 778 ft-lb)
Work and Heat are forms of Energy. Therefore,
Work = Heat
Thermodynamics
(b). Energy
Conservation-of-energy Law
"Any heat or work which flow into or out of the
system must result in a change in the total energy
stored in the system"
E = q - 
E = change in energy
 q = heat flowing into the system
  = work done by the system





q is positive (+ve) if the system absorbs the
heat
q is negative (-ve) if the system gives off the
heat
 is +ve if the system does the work
 is –ve if the surroundings do the work on
the system
If the chemical system does not expand or
contract (volume is constant), then:
E = qv

qv = heat absorbed in a constant-volume system
In Environmental Engineering Applications, most
of the systems are open, so they operate under
constant pressure rather than constant volume
Thermodynamics
(c). Enthalpy (H)
H = E + PV
Where
H = enthalpy
 E = internal energy of the system
 P = pressure on the system
 V = volume of the system

Enthalpy
A thermodynamic function of a system, equivalent
to the sum of the internal energy of the system
plus the product of its volume multiplied by the
pressure exerted on it by its surroundings.
At constant pressure system,
Heat absorbed by the system = qp
Work done by the system can be obtained by
integrating
dw = P dv   = P (V2 – V1)
then, change in internal energy is:
E = E2- E1 = qp -  = qp – [P (V2 – V1)]
Or
(E2 + PV2) – (E1 + PV1) = qp
(E2 + PV2) is the final enthalpy
(E1 + PV1) is the initial enthalpy
So,
H2 – H1 = qp  H = qp


(T and P are constant)
+ve heat means endothermic reaction (absorbs heat)
-ve heat means exothermic reaction (evolves heat)
Change of enthalpy or heat of a given reaction can
be found in Tables such as (Table 3-1)
To calculate heat of a reaction:
Write a balanced equation
 Find standard enthalpy of reactants
 Find standard enthalpy of products

Then,
Heat = (products) – (reactants)
Example (Problem 3.1)
Determine the heat of combustion of ethane gas.
1
CH 3 CH 3 ( g )  3 O2 ( g )  2CO2 ( g )  3H 2 O( g )
2
from.Table.3  1

H 298
..of ..CH 3CH ( g )  84.68

H 298
..of ..O2 ( g )  0

H 298
..of ..CO2 ( g )  393.5

H 298
..of ..H 2 O( g )  241.8

then,.H 298
. for.combustion  2(393.5)  3(241.8)  (84.68)  1428.kJ / mol
The enthalpy of a chemical element (at its standard state)
at 25 C and 1 atm is zero. For example, at standard
states, O2 is gas, Mercury is liquid, Sulfur is crystal.
Study Examples A and B on page 54
Thermodynamics
(d) Entropy
Is based on the second law of thermodynamics, which states "All
systems tend to approach a state of equilibrium"
In chemistry, we are interested in entropy to check the position of
the equilibrium of a chemical process.
Where



dqrev
dS 
T
S = entropy of the system
T = absolute temperature
qrev = amount of heat that the system absorbs if a chemical change
is brought about in an infinitely slow reversible manner
S  S 2  S1  
2
1



dqrev
T
+ve S indicates that change can occur
spontaneously
-ve S indicates that change tends to occur in
reverse direction
Zero S indicates that system is at equilibrium
Entropy
For a closed thermodynamic system,
entropy is a quantitative measure of the
amount of thermal energy not available
to do work.
Thermodynamics
(e). Free Energy
In environmental engineering processes, both entropy and
energy are needed in order to determine which processes
will occur spontaneously.
G = H – TS
Where




G = free energy
H = enthalpy (J)
T = absolute temperature ( K) [ K =  C + 273]
S = entropy (J /  K)
At constant temperature and pressure:
G = H - TS
Since
H = E + PV
Then
H = H2 – H1 = (E2 + P2V2) – (E1 + P2V1)
At constant P
H = E + PV
Since
E = q - 
Then
H = q -  + PV
From
S  S 2  S1  
2
1
dqrev
T
At constant T
TS = qrev
If the system change is very slow, then energy loss
is MINIMUM
q = qrev
 = max
In this case,
G  q rev   max  PV  q rev
 G   max  PV
PV represents the work that is wasted during the
expansion of the system. Therefore:-G is the
difference between the maximum work and the
wasted work, which can be described as the useful
work available from the system change. So:
 G   useful .......(T .and .P.are.cons tan t )
If a system changes from a to b, then:
-ve G means that the system or process can
proceed
 +ve G means that the system or process can
proceed in the reverse direction (b to a)
 Zero G means that the system or process is at
equilibrium and can not proceed in either
direction.

At standard state of elements and at 25 C and 1
atm, the free energy ( G ) is zero.
For values of G
, see Table 3-1.

298

298
Consider the following reaction:
aA + bB  cC + dD
Taking into consideration the concentration of
reactants and products:
G  G

298
c
d
[C ] [ D]
 RT ln
a
b
[ A] [ B]
Where
G = reaction free-energy change (J)


G

298 = standard free-energy change (J)
 R = universal gas constant = 8.314 J / K-mol =
1.99 cal / K-mol
 T = absolute temperature in Kelvin ( K)
 [ ] = activities of A, B, C, and D
c
d
[
C
]
[
D
]
 and
 .reaction.quotient ,.Q
a
b

[ A] [ B]
At equilibrium;
G = zero
So,
 [C ]c [ D] d
G   RT ln 
a
b
[
A
]
[
B
]


[C ]c [ D]d
K
a
b
[ A] [ B]
At equilibrium



 equilibrium
K = equilibrium constant
Therefore,
G   RT ln K

Comparison between Q and K
Q < K means the reaction proceeds from left to
right
 Q > K means the reaction proceeds from right
to left
 Q = K means the reaction is at equilibrium

Study Examples A, B, C, and D page 58-59.
Thermodynamics
(f). Temperature Dependence of K
From relationship between
G and K
 G and H

d ln K H 

dT
RT 2
In environmental engineering practices, the
temperature range is limited and, therefore, H is
constant. So,
KT 2
H 
ln

KT1
R
Study Example page 60
 T1  T2 


 T1T2 
Osmosis
Flow direction from dilute solution to concentrated
solution is more rapidly than the other direction
(concentrated  diluted)
In order to oppose that flow, pressure to the salt solution can be
applied to produce equilibrium. That pressure is called osmotic
pressure ()

A
RT P

ln
VA
PA





 = osmotic pressure, atm
R = 0.0882 l-atm / mol-K
T = absolute temperature, K
VA = volume per mole of solvent = 0.018 liter ( for water)
PA and PA = vapor pressure of solvent in the dilute and
concentrated solutions, respectively
For dilute solutions, the reduction in vapor
pressure of a solvent is directly proportional to
the concentration of particles in solution. So,
  cRT

c = molar concentration of particles
In environmental engineering Reverse Osmosis
is used to demineralized brackish waters
Dialysis and Electro-Dialysis
Dialysis is a phenomena that is related to the
principle of OSMOSIS
Main Membrane Processes


Are used to separate substances (solutes) from
a solution (solvent)
The main membrane processes are
Dialysis
 Electro-dialysis
 Reverse osmosis


Driving forces that cause mass transfer of
solutes are:
Difference in concentration (dialysis)
 Difference in electric potential (electro-dialysis)
 Difference in pressure (reverse osmosis)

Dialysis

Consists of :
Separating solutes of different ionic or molecular size
 Solution
 Selectively permeable membrane


The driving force is the difference in the solute
concentration across the membrane
Batch Dialysis Cell

Solution to be dialyzed is separated from solvent by a
semi-permeable membrane



Small ions and molecules pass from solution to solvent
Large ions and molecules do not pass due to relative size of
membrane pore
The mass transfer of solute through the membrane is
given by
M  KAC




M = mass transferred per unit time (gram/hour)
K = mass transfer coefficient [gram/(hr-cm2)(gram/cm3)]
A = membrane area (cm2)
C = difference in concentration of solute passing through the
membrane (gram/cm3)
Applications of Dialysis

Sodium hydroxide was recovered from textile
wastewater at:
Flowrate = 420 – 475 gal/day
 Recovery of 87.3 to 94.6%


Dialysis is limited to small flows due to small
mass transfer coefficient (K)
Electro-Dialysis


The driving force is an electromotive force
If electromotive force is applied across the
permeable membrane:
An increased rate of ion transfer will occur
 This results in decrease in the salt concentration of the
treated solution


The process demineralizes
Brackish water and seawater to produce fresh water
 Tertiary effuents

How it Works?

When direct current is applied to electrodes:
All cations (+vely charged) migrate towards cathode
 All anions (-vely charged) migrate towards anode
 Cations can pass through the cation-permeable
membrane (C) but can not pass through (A)
 Anions can pass through the anions-permeable
membrane (A) but can not pass through (C)



Alternate compartments are formed
Ionic concentration in compartments is less
than or greater than that in the feed solution
The Membrane

Membranes used in electro-dialysis are:
Porous
 Sheet-like
 Its structural matrix is made of synthetic ion exchange
resin

Current Requirement

Can be calculated from Faraday’s laws of electrolysis:

One Faraday (F) of electricity (96,500 ampere-seconds or coulombs) cause one
gram equivalent weight of a substance to migrate from one electrode to another
FQNEr
I
Ec






I = current in amperes
F = Faraday’s constant (96,500 ampere-seconds per gram equivalent weight
removed)
Q = solution flowrate (liters/second)
N = normality of the solution (gram eq weight per liter)
Er = electrolyte removal as a fraction
Ec = current efficiency as a fraction
Current Requirement

If the number of cells in a stack = n, then
FQNEr
I
nEc




Electro-dialysis stack usually have 100 to 250 cells
(200 to 500 membranes)
Ec for a electro-dialysis stack and feed water must be
determined experimentally
Ec is 0.90 or more
Er is usually 0.25 to 0.50
Cell Capacity

The capacity of the cell to pass an electric
current depends on:
Current density [ = current / membrane area (ma/cm2)]
 Normality of the feed (number of gram equivalent
weight per liter of solution)
 Current density / normality ratio
 This ratio may vary from 400 to 700

Power Requirement


The resistance (R) of an electro-dialysis stack
treating a particular feed must be determined
experimentally
If resistance (R) and current (I) are known:
Required Voltage, E = RI
 Required Power, P = RI2


R = ohms; I = amperes; E = volts; and P = watts
Applications

Electrical energy requirement is directly
proportional to the amount of salt removed

So, electrical cost is governed by



Dissolved salt content of the feed water
The desired dissolved solids content of the product water
Energy consumption increases with deposition
of scale upon the membrane

Consequently, electro-dialysis is not used to deionize
seawater
Applications

Electro-dialysis is used in demineralization of brackish
water



Brackish water having TDS concentration of 500 mg/l can be demineralized using electro-dialysis to produce a product water of 500
mg/l TDS
Membrane replacement and power costs are about 40% of total cost
Electro-dialysis have been used to de-mineralize
secondary effluents





Scale formation
Organic fouling
25 to 50% TDS can be removed in single pass
Coagulation, settling, filtration and activated carbon adsorption can
used as pre-treatment processes to reduce organic fouling OR by
cleaning the membrane using an enzyme detergent solution
Scale formation can be reduced by adding small amount of acid to
the feed
Electro-Dialysis Installations
Reverse Osmosis
Diffusion is the movement of molecules from a region of
higher concentration to a region of lower concentration.
Osmosis is a special case of diffusion in which the
molecules are water and the concentration gradient occurs
across a semipermeable membrane. The semipermeable
membrane allows the passage of water, but not ions (e.g.,
Na+, Ca2+, Cl-) or larger molecules (e.g., glucose, urea,
bacteria). Diffusion and osmosis are thermodynamically
favorable and will continue until equilibrium is reached.
Osmosis can be slowed, stopped, or even reversed if
sufficient pressure is applied to the membrane from the
'concentrated' side of the membrane.
Reverse Osmosis
Reverse osmosis occurs when the water is moved across
the membrane against the concentration gradient, from
lower concentration to higher concentration. To
illustrate, imagine a semipermeable membrane with
fresh water on one side and a concentrated aqueous
solution on the other side. If normal osmosis takes
place, the fresh water will cross the membrane to dilute
the concentrated solution. In reverse osmosis, pressure
is exerted on the side with the concentrated solution to
force the water molecules across the membrane to the
fresh water side.
Reverse Osmosis
Reverse osmosis is often used in commercial and
residential water filtration. It is also one of the
methods used to desalinate seawater. Sometimes
reverse osmosis is used to purify liquids in which
water is an undesirable impurity (e.g., ethanol).
Reverse Osmosis - Pros and Cons
The semi-permeable membrane used in reverse osmosis
contains tiny pores through which water can flow. The
small pores of this membrane are restrictive to such
organic compounds as salt and other natural minerals,
which generally have a larger molecular composition
than water. These pores are also restrictive to bacteria
and disease-causing pathogens. Thus, reverse osmosis is
incredibly effective at desalinating water and providing
mineral-free water for use in photo or print shops. It is
also effective at providing pathogen-free water. In areas
not receiving municipally treated water or at particular
risk of waterborne diseases, reverse osmosis is an ideal
process of contaminant removal.
Reverse Osmosis - Pros and Cons
The reverse osmosis process contains several downsides
which make it an inefficient and ineffective means of
purifying drinking water. The small pores in the
membrane block particles of large molecular structure
like salt, but more dangerous chemicals like pesticides,
herbicides, and chlorine are molecularly smaller than
water (Binnie et al, 2002). These chemicals can freely
pass through the porous membrane. For this reason, a
carbon filter must be used as a complimentary measure
to provide safe drinking water from the reverse osmosis
process. Such chemicals are the major contaminants of
drinking water after municipal treatment.
Reverse Osmosis - Pros and Cons
Another downside to reverse osmosis as a
method of purifying drinking water is the
removal of healthy, naturally occurring minerals
in water. The membrane of a reverse osmosis
system is impermeable to natural trace minerals.
These minerals not only provide a good taste to
water, but they also serve a vital function in the
body’s system. Water, when stripped of these
trace minerals, can actually be unhealthy for the
body.
Reverse Osmosis - Pros and Cons
Reverse osmosis also wastes a large portion of
the water that runs through its system. It
generally wastes two to three gallons of water for
every gallon of purified water it produces.
Reverse osmosis is also an incredibly slow
process when compared to other water treatment
alternatives.
Module Types



Spiral Wound
Hollow Fiber
Tubular
Membrane Installations
Costs
Principles of Solvent Extraction
Objectives:


To recover valuable constituent
For analytical purposes
Solvents are used (immiscible)






Petroleum ether
Diethyl ether
Benzene
Hexane
Dichloromethane
Others
water
Solute  (
solvent.( immiscible)
Solute will be distributed based on their solubilities in
water and solvent
Csolvent Cs

K
Cwater Cw

K = distribution coefficient
Usually a solvent is selected so that K is greater than 1
If K = 9 and volumes of water and solvent are
equal, then:
In the first extraction step 90% will be extracted
 After 3 extraction steps with fresh solvent, 99.9%
of the solute will be extracted

The mathematics are simple
In practice, it may be unjustified to use equal
volumes of water and solvent. In such cases:
Cs (W0  W1 ) / Vs
K

Cw
(W1 / Ww )
W0 = weight of solute in water (originally)
 W1 = weight of solute in water (remained after one
extraction)
 Vs = volume of solvent
 Vw = volume of water

Vw
W1  W0
KVs  Vw
In the second step of extraction
Vw
W2  W1
KVs  Vw
In terms of the original sample
Vw
Vw
W2  W0
KVs  Vw KVs  Vw
 Vw

W2  W0 

KV

V
w 
 s
2
After n extractions
 Vw

Wn  W0 

 KVs  Vw 

n
Wn = remaining weight of solute in water after n
extractions