Gas Properties and Kinetics
Gas and Vapor
• Gas and vapors have some properties in
common.
– Both are composed of widely separated freely
moving molecules.
– Both will expand to fill a larger differently
sahped container.
– Both exert pressure in all directions. The
major difference lies in the internal eneryg of
the molecules.
Gas and Vapor
• Gas if it is far removed from the liquid
state. (Usually it means that the
temperature of the substance is above its
critical point)
• Vapor a substance in the gaseous state
that is not far from being a liquid.
Therefore a vapor can usally be adsorbed
onto surfaces or condensed into a liquid
relatively easily.
Vapor Pressure
• The pressure exerted by a pure
component vapor in equilibrium with a flat
surface of its pure component liquid at a
certain temperature
• is a measure of the escaping tendency or
volatility of the liquid (liquid’s vapor
pressure)
Vapor Pressure
• Vapor pressure increases rapidly with an
increase in the temperature
• The pressure/temperature relatin can be
given by Antoine equation:
Bi
log Pvi  Ai 
Ci  T
Pv in mmHg and T in oC (if Table 9.2 is used)
Equilibrium
• When a pure liquid is placed in contact
with air in a closed space, some of the
liquid will volatilize until vapor-liquid
equilibrium is established
Pi  yi P  Pvi
Pi partial pressure of component i, atm
yi mole fraction of component i in the gas
P= total pressure, atm
Example 10.1
• An airstream is presently at 1 atm and
160 F., and contains 40,000 ppm toluene.
To what temperature must the air be
cooled to remove two-thirds of the toluene
vapor?
Solution
• 40,000 ppm = 4% = 0.04 mole fraction
At equilibrium:
Pi  yi P  Pvi
160 F
40 ppm
y=0.04
Pi=0.04 atm
Sıcaklık
0.013 atm = 0.196 psi
13.33 ppm
y=0.013
Pi =0.013 atm
T=?
Solution, contd.
40 F
Solution, contd.
Bi
log Pvi  Ai 
Ci  T
Pv in mmHg and T in oC (if Table 9.2 is used)
• Pvi =0.013 atm = 10.13 mmHg
T
Bi
1,343.943
 Ci 
 219.377
Ai  log Pvi
6.95334 log(10.13)
T = 6.5 C
Comments on the Example
• In the design of a heat exchanger to cool
the air in the previous example, provide
enough surface area to not only cool the
air (remove sensible heat) bu also to
condense the vapors (remove latent heat)
• Usually chilled water or a refigerant might
be sufficiently cool to remove enough
vapor from the air to meet the pollution
control objective
• Even cooled to 6.5 C, the air stream still
contains 13.333 ppm toluene.
Diffusivities
• The matter will diffuse spontenously from a region of high
concentration to one of low concentration
• Fick’s first law describes diffusivity: the proportionality
constant between the rate of flux of matter and the
concentration gradient

M
dC
 D
A
dx
M=mass transfer rate, mol/s
A= area normal to the direction of diffusion, cm2
D=diffusivity, cm2/s
dC/dx=concentration gradient, mol/cm4
Since the concentration decreases in the direction of diffusion a minus sign is put
Diffusivities
• The diffusivity of a particular substance depends on the
substance itself and the medium through which it is moving
• In all the gas and vapor control processes in which the gas
flows past (and contacts) a solid or liquid, a laminar film
layer is formed where the pollutant passes from one phase
to the other is via molecular diffusion
• Diffusion is ofther the rate-limitin step in control processes
• More detailed discussion on diffusion will be seen in
absorption process
Gas Liquid and Gas-Solid Equilibria
• In absorption process, once a gas molecule has
diffused thorughthe stagnnt gas film it must be
absorbed into the liqud.
• Once absorbed, the pollutant then diffuse
through a stagnant ilquid film into the bulk liquid
• Even thoud the rate of absoprtion is very fast
and thus not a limitation, the extent of absorption
(solubility) is crucial to the overall objective of
mass transfer
Gas Liquid and Gas-Solid Equilibria
• Most air pollution control equipment works at a or
near atmospheric pressure and with relatively dilute
solutions
• For dilute solutions, the concentrations of the
pollutant in the gas and the liquid are often linearly
related (Henry’s Law)
Henry’s Law constant =
(M/atm)
[Ci ]
Hi 
pi
• Ci: concentration of pollutant i in the aqeuous phase
• http://www.mpch-mainz.mpg.de/~sander/res/henry.html
Henry’s Law Constants
• Hi has units depending on the Ci and Pi
unit.
• The equation eventually becomes
nonlinear at high Ci value and no longer
valid.
• Hi vary with T
Henry’s Law Constants
Tür
HA (298K),mol/Latm
-∆H/R, K
SO2
1.23
3120
H2O2
7.45E4
6620
HNO3
2.1E5
HNO2
49
4780
O3
1.13E-2
2300
O2
1.3E-3
NO2
1.0E-2
2500
NO
1.9E-3
1480
CO2
3.4E-2
2420
NH3
75
3400
CH3C(O)O2NO2
2.9
5910
HCl
727
2020
HCHO
6.3E3
6460
NO3
2.1E5
8700
OH.
25
5280
HO2.
2.0E3
6640
Adsorption
• Involves a gas-solid equilibrium that is quite
similar in principle to the solubility equilibrium in
liquid-gas systems
• Amount adsorbed on the solid depends on
–
–
–
–
–
Type of vapor
The partial pressure of the vapor
The type of solid (adsorbent)
The amount of sufrace area available for adsorption
Temperature
A Typical Adsorption Isotherm
Adsorption Types
• 1) PHYSICAL
– Exothermic
– Reversible (to separate VOCs from air)
• 2) CHEMICAL
– İnvolves the breaking and re-forming of bonds
– Much more energetic than physical adsorption
Chemical Reactions
• Once the pollutant is captured a chemical
reaction takes place
– In catalytic incineration of VOCs to from CO2
and H2O
– SO2 scrubbing to form CaSO4
Chemical Reactions-Kinetics
• Reaction rate: the rate of “disapperance” of
reactants or the rate of “apperance” of
products.
• Reaction rate =rP =-rR
• rP = rate of generation of product P
• rR = rate of generation of ractant R
• Is the rate constant?
• r is a function of
»
»
»
»
Reactants concentration
T
Catalyst or inhibitor existence
Light
Chemical Reactions-Kinetics
•
•
•
•
Consider R + S  P + Q
rP = kCxRCyS
k = rate constant f(T)
CR and CS= concentrations of reactants
(mol/L)
• x,y = exponents
Chemical Reactions-Kinetics
• Consider R + S  P + Q
• rP = kCxRCyS
• k can be expressed by an Arrhenius
equation such as:
K=Ae-E/RT
A: frequency factor
E: activation energy
R: universal gas constant, in energy units
T: absolute temperature
Example 10.3
Reactor Models
• Two ideal reactor models are commonly
used to descirbe real reactors where
reactions take place
• 1) CSTR (Continous stirred tank reactor
• 2) Plug flow
CSTR
• Continuous flow with inlet concentration (Cin)
through a tank in which the contents are rapidly
mixed
• Homogenous distribution of the all species
• Concentration in the tank equals outlet
concentration (Cout)
• At steady state:
• dC/dt = QinCin-QoutCout+riV=0
V: tank volume (L)
Q:volumetric flow rate (L/s)
Plug Flow Reactor
• Can be described as one dimensional flow through
a long tube.
• Velocity is constant at all radial position in the tube
• Axial dispersion is negligible
• The stady-state material balance for component i:
• dC/dt = QVCi,V-QV+DVCi,V+DV+riV=0
If the flow rate is constant and ri in not a function of
position then:
dCi/ri=(1/Q)dV
Example 10.4
• The reaction RP in Example 10.3 is to
occur isothermally at 640 K in a) CSTR
b)PFR. Calculate the required volume of
each reactor to give 99% convesion of R
to P when the volumetric flow rate is
constant at 100 L/s.
Example 10.4, Solution
• From Example 10.3 A=5.28E7 s-1 , E=21.53
cal/mol. k at 640 K:
k  5.28(10)7 e21.53 /(1.987*640)2.34
For a CSTR
• dC/dt = QinCin-QoutCout+riV=0 then
V
V
V
 Q(C R ,in  C R ,out )
rR
 QCR ,in (1  C R ,out / CR ,in )
 kCR ,out
100C R ,in (1  0.01)
2.34(0.01C R ,in )
 4230L
Example 10.4, Solution
For a PFR if Q is constant:
dCR/rR=dCR/-kCR=(1/Q)dV

C R ,out
C R ,in
dCR
1 V
  dV
 kCR Q 0
 1 C R ,out V
ln

k
C R ,in Q
1
V
ln(0.01) 
k
100
100
V
ln(0.01)  197L
2.34
Thermodynamics
• Reactions can be exothermic (release
heat) and endothermic ( heat input)
• These heat exchange changes the
temperature of the reacting mix and thus
affects the reaction rates
• T also affects the volumetric flow rates and
concentrations in mol/L unit
Thermodynamics
•
•
•
•
•
Many chemical reactions do not go to
100% completion
An equilibrium depending on the
temperature is establisehde between
reactants and products
SO2 + 1/2O2SO3
For this gas phase reaction equilibrium
constant Kp = PSO3/(PSO2*(PO2)1/2
Kp=1.53E-5exp(11750/T) (T in Kelvin)
Thermodynamics
•
•
•
•
As T increases Kp decreases rapidly
at 298 Kp = 2(10)12
at 750 Kp = 97
Note that equilibrium concentrations are not often
observed in industrial reactors.
Generally the equilibrium constants for exothermic
reactions decrease with increasing T while kinetic
constants always increase with temperature
Therefore it is possible for an optium T to exist for
a given reaction. It should be low enough to allow
the equilibrium to achieve the desired degree of
conversion yet high enough to allow the reactions
to proceed rapidly.
Gas Properties and Kinetics
• Gas kinetics is a branch of statistical
mechanics applied to gases
• Applying statistical mechanics to gain a
microscopic understanding of gas
properties such as
pressure,temperature,viscosity, and
diffusivity.
• Let’s examine gas transportation and
interaction between gases and particulates
Ideal Gas Law
• Ideal gas law is the result of two
experiments and is stated as:
• 1)For a given mass of gas held at a
constant T, P is inversely proportional to the
volume (Boyle’s law)
• 2) For a given mass of gas held at a
constant pressure, the V is directly
proportional to the T (Law of Charles and
Gay-Lussac)
Given in an Equation:
PV = nRT (for a fixed mass of gas)
Ideal Gas Law
PV = nRT
P in atm, V in L, T is in K, n is the number of
moles of gas in the gas sample, and R is
universal gas constant
R = 0.082 (Latm/molK)
Useful Links
• http://www.aim.env.uea.ac.uk/aim/aim.php