Transcript M5_Chapter4
CHAPTER 4
ENVIRONMENTAL FATE
Introduction
This chapter serves as a basis to
identify the hazards associated with different substances
used and produced in the chemical process, including raw materials, products and or byproducts.
It would then be possible to do follow up with an
exposure assessment
and a
dose-response assessment
which are needed to perform
risk characterization
Substance Classification Tree
What Substances?
Physical + Chemical Properties Estimating Exposure And Environmental Effects Classifying the Substances based on risk Old Analyses Performing P2 on the substances...
Chemical Properties Used to Perform Environmental Risk Screenings
Environmental Process Relevant Properties Dispersion and Fate Volatility, density, melting point, water solubility, effectiveness of waste, water treatment.
Persistence in the Environment Atmospheric oxidation rate, aqueous hydrolysis rate, photolysis rate, rate of microbial degradation, and adsorption.
Continued on the following slide
Chemical Properties Used to Perform Environmental Risk Screenings Environmental Process Uptake by Organisms Relevant Properties Volatility, Lipophilicity, Molecular Size, Degradation Rate in Organism.
Human Uptake Toxicity and other Health Effects Transport Across Dermal Layers, Transport Rates Across Lung Membrane, Degradation Rates within the Human Body.
Dose-Response Relationships .
Boiling Point
• Distinguishes
gas and liquid partitioning
• Using the substance’s structure, it can be estimated by :
T b = 198.2 + Σ n i g i
(4.1) Where:
T b
: normal boiling point (at 1 atm) (K)
n i
: number of groups of type
i
in the molecule,
g i
: contribution of each functional group to the boiling point Corrected using :
T b (corrected) = T b – 94.84 + 0.5577*T b + 0.0007705*(T b ) 2 T b (corrected) = T b + 282.7 – 0.5209*T b
(T b 700K) (4.2) (T b > 700K) (4.3)
Example : Boiling Point Estimation
Estimate the Normal Boiling Point for diethyl ether.
Diethyl ether has the molecular structure CH 3 -CH 2 -O-CH 2 -CH 3
Solving : Group
-O 2(-CH 3 ) 2(-CH 2 )
g i contribution
25.16
2(21.98) 2(24.22) The actual boiling point for diethyl ether is 307.65 K
Example : Boiling Point Estimation (Continued) a) Using equation 4.1 : T b (K)= 198.2 + Σ n i g i
T b (K)= 198.2 + 2(21.98) + 2(24.22) + 25.16
T b = 315.76 b ) Using equation 4.2 : T b (corrected) = T b – 94.84 + 0.5577*T b - 0.0007705*(T b ) 2
T b (corr) = 315.76 – 94.84 + 0.5577(315.76) - 0.0007705(315.76) 2
T b (corrected)
=
320.2 K
Melting Point
• Distinguishes
solid and liquid partitioning
.
• Can be estimated using the substance’s boiling point : (4.4)
T m (K) = 0.5839 * T b (K)
Where :
T m
: Melting Point in Kelvins.
T b
: Boiling Point in Kelvins.
Example : Melting Point Estimation
Estimate the Melting Point for diethyl ether.
Solving :
Using equation 4.4 to calculate the T m :
T m (K) = 0.5839 * T b (K)
T m (K) = 0.5839 * 307.65 K
T m
=
179.634 K
Vapor Pressure
• Higher Vapor Pressure = Higher Air Concentrations • Can be estimated using the following equations : ln P vp = A + B/(T - C) (4.5) Where : T = T b at 1 atm ln(1 atm) = 0 = A + B/(T b – C) (4.6) ln P vp (atm) ={[A(T b – C) 2 ] / [0.97*R*T b ]}*{1/(T b – C)-1/(T – C)} (4.7) the parameters A and C can be estimated using : C = -18 + 0.19 T b A = K F *(8.75+ R ln T b ) (4.7a) (4.7b)
Vapor Pressure (continued)
For solids : ln P = -(4.4 + lnT b ) * {1.803*[(T b /T)- 1)] - [0.803*ln (T b /T)]} - 6.8(T m /T-1) (4.8) Where :
P vp
: vaporization pressure (atm).
T A
: absolute temperature and
T b
and
C
are empirical constants.
is the boiling point at 1 atm.
B
: a parameter related to the heat of vaporization.
K F R
: a correction factor.
: gas constant ; 1.987 L-atm K -1 mol -1
T m
: melting point (K).
Example : Vapor Pressure Estimation
Estimate the Vapor Pressure for diethyl ether Using the predicted value of 315.76 K:
C
=
-18 + 0.19T
b
= -18 + 0.19(320.2) =
41.9944
A
=
K f (8.75 + R ln T b )
= 1.06 [8.75 + 1.987 ln(320.2)] =
21.3962
(4.7.a) (4.7.b)
ln P vp
=
{[A(T b – C) 2 ] / [0.97*R*T b ]}*{1/(T b – C) - 1/(T – C)}
(4.7)
=
{[21.39(315.76-41.99) 2 ] / [0.97(1.987)(315.76)]}*{1/(273.76) – 1/(256)} Ln P vp = -0.6677;
P vp = 0.5128 atm
= 389.79 mm Hg.
Repeating the calculation for the experimental boiling point leads to a vapor pressure estimated of
P vp = 0.6974 atm
= 530.06 mm Hg.
Octanol-Water Partition Coefficient
• Describes
partition
between an
aqueous phase
and
it’s suspended organic phases
.
• Can be estimated using the substance’s structure :
log K ow = 0.229 + Σ n i f i
(4.9)
log K ow (corrected) = 0.229 + Σ n i f i + Σ n j c j
(4.10) Where:
K ow n i
: Octanol-Water Partition Coefficient. : number of groups
i
in the compound.
f i n j
: factor associated with the group
i
: number of groups
j
in the compound that have correction factors.
c j
: correction factor for each group
j
Example : Octanol-Water Partition Coefficient Estimation
Estimate the Octanol-Water Partition Coefficient for diethyl ether.
Solving : Group
-O 2(-CH 3 ) 2(-CH 2 )
f i contribution
-1.2566
2(0.5473) 2(0.4911) Using equation 4.9 :
log K ow = 0.229 + Σ n i f i
log K ow
=
0.229 + 2(0.5473) + 2(0.4911) + (1.2566)
log K ow =
1.0492 ≈
1.05
therefore
K ow = 11.2
Bioconcentration Factor
• Describes
partitioning
between
aqueous
and
lipid phases in living organisms.
• Higher bioconcentration factors = higher quantity of bioaccumulation in living organisms • Can be calculated using :
log BCF = 0.79*(log K ow ) – 0.40
log BCF = 0.77*(log K ow ) – 0.70 + Σ j j
Where :
BCF
: Bioconcentration Factor.
K ow j j
: octanol-water partition coefficient.
: correction factor for each group.
(4.11) (4.12)
Example : Bioconcentration Factor (BCF) Estimation
Estimate the Bioconcentration Factor for diethyl ether.
Solving :
Using equation 4.9 we obtain log K ow :
log K ow = 0.229 + Σ n i f i log K ow =
1.0492 ≈
1.05
Using equation 4.11 we can calculate
BCF
:
log BCF = 0.79*(log K ow ) – 0.40
log BCF = 0.79* (1.05) – 0.40 log BCF = 0.4295
therfore
BCF = 2.6884
Water Solubility
• Used to assess concentrations in water • Can be calculated using :
Log S = 0.342 – 1.0374 logK ow – 0.0108 (T m –25) + Σh j Log S = 0.796 –0.854 logK ow – 0.00728 (MW) + Σh j Log S = 0.693 – 0.96 los K ow
–
0.0092 (T m –25) – 0.00314 (MW) + Σh j
Where :
S
: water solubility (mol/L).
K ow T m
: octanol-water partition coefficient.
: melting point (ªC).
MW
:s the molecular weight of the substance.
h j
is the correction factor for each functional group
j
.
(4.13) (4.14) (4.15)
Example : Water Solubility Estimation
Estimate the Water Solubility for diethyl ether.
Solving :
Equation 4.9 gives the
log K ow ≈ 1.05
Using equation 4.14 we can calculate the
S
:
Log S = 0.796 –0.854 logKow – 0.00728 (MW) + Σhj
Log S = 0.796 – 0.854(1.05)
–
0.00728(74.12) + 0.0
Log S = -0.6403
Therfore :
S = 0.2289 mol/L. = 16.966 g/L = 16,966.068 mg/ L
Henry’s Law Constant
• Describes the
affinity for air over water
.
• Can be determined using :
-log H = log (air-water partition coeff) = Σ n i h i + Σ n j c j
Where :
H
: dimensionless Henry’s Law Constant.
n i h i
: number of bonds of type coefficient.
i
in the compound.
: bond contribution to the air-water partition
n j c j
: number of groups of type
j
in the molecule.
: correction factor for each group.
( 4.19)
Example : Henry’s Law Constant Estimation
Estimate the Henry’s Law Constant for diethyl ether.
H H H H H-C-C-O-C-C-H H H H H Expressed as a collection of bonds, diethyl ether consists of 10 C-H, 2 C-C bonds, and 2 C-O bonds. The uncorrected value of log (air to water partition constant) is given by : -log H = log (air-water partition coefficient) = 10(-0.1197) + 2(0.1163) + 2(1.0855) = 1.2066
log H -1
=
1.2066
Soil Sorption Coefficient
• Used to describe the Soil-Water Partitioning.
• Can be estimated by :
log K oc = 0.544 (log K ow ) +1.377
log K oc = -0.55 (log S) + 3.64
log K oc = 0.53 1 χ + 0.62 + Σ n j P j
(4.16) (4.17) (4.18) Where :
K oc :
Soil Sorption Coefficient (μg/g of organic carbon (to μg/mL of liquid)).
K ow
: Octanol-Water Partition Coefficient.
S
: Water Solubility.
1 χ :
first order Molecular Connectivity Index (from literature-appendix ).
n j P j
: number of groups of type
j
in the compound.
: correction factor for each group
j
.
Molecular Connectivity Index Calculations
The
first step
in calculating 1 χ is to
draw the bond structure of the molecule
. For example, isopentane would be drawn as: CH 3 H 3 C-CH-CH 2 -CH 3 The
second step
is to
count the number of carbon atoms to which each carbon is attached
. Each C-C bond is given a value of 1 and δ i , is the parameter that defines the quantity of carbon atoms connected to a carbon atom
i
. The diagram below gives the δ i , values for the different carbon atoms.
(
1
) CH 3 H 3 C-CH-CH 2 -CH 3 (
1
) (
3
) (
2
) (
1
)
Molecular Connectivity Index Calculations (continued) The third step is to identify the “connectedness” of the carbons connected by the bond (δ i , δ j ). For isopentane, these pairs are: (1,3) CH 3 (2,1) H 3 C-CH-CH 2 -CH 3 (1,3) (3,2) The value of 1 χ can then be calculated using the equation : 1 χ = Σ(δ i * δ j ) -0.5
For isopentane, (4.19) 1 χ = (1/√3) + (1/√3) + (1/√6) + (1/√2) =
2.68
Example : Soil Sorption Coefficient Estimation
Estimate the Soil Sorption Coefficient for diethyl ether.
Solution :
The molecular structure for diethyl ether is : CH 3 -CH 2 -O-CH 2 -CH 3 Using previously calculated values for log K ow (estimated at 1.0492) and log S (estimated at -0.6384) we can estimate the soil sorption coefficients using equations 4.16 and 4.17 : log K oc = 0.544 (log K ow ) + 1.377 =
1.9482
log K oc = -0.55 (log S) + 3.64 =
3.99
Example : Soil Sorption Coefficient Estimation
Using the molecular connectivity we can also estimate the soil sorption coefficient : First the molecular connectivity index is calculated using eq. 4.19 : CH 3 -CH 2 -O-CH 2 -CH 3 2(C-C), 2(C-O), 2(1, 2) , 2(2, 2) (molecular structure) (connection pairs) therefore : 1 χ = 2(1/√2) + 2(1/√4) =
2.414
Using equation 4.18 to calculate the soil sorption coefficient :
log K oc
log K oc
= 0.53 1 χ + 0.62 + Σ n j P j
= 0.53 1 χ + 0.62 + Σ n j P j = 0.53(2.414) + 0.62 + (-1.264)
log K oc
=
0.63542
therefore : K oc
=
4.32
Where to look up this information...
http://www.chem.duke.edu/~chemlib/properties.html
http://www.library.vanderbilt.edu/science/property.htm
http://www.library.yale.edu/science/help/chemphys.html
What do the different Properties mean?
Adapted from the Green Engineering Textbook
Estimating Environmental Persistence and Ecosystem Risks
To be discussed : – Atmospheric Lifetimes – Aquatic Lifetimes – Overall Biodegradability – Ecosystems
Estimating Atmospheric Lifetimes
• • One way to estimate the atmospheric lifetime of a compound is to
analyze the rate of oxidation
of the substance, specifically the hydroxyl radical reaction rate.
Group contributions
is again one of the approaches that can be taken to estimate this property.
• Using examples, we will show how to estimate reaction rates and half lives while using the appropriate correction factors.
Example : Atmospheric Lifetime Estimation
Dimethylsulfide (DMS, CH 3 SCH 3 ) produced by phytoplankton degredation is thought to be the major source of the sulfate and methanesulfonate aerosol found in the marine boundary layer. The primary objective of this research effort is to determine the detailed mechanism of, and final product yields from, the OH initiated gas phase oxidation of DMS.
At the low NOx levels that are characteristic of the remote marine boundary layer, reaction with OH is the initial step in DMS oxidation.
OH + CH 3 SCH 3 ⇒ Products (1)
The OH initiated oxidation of DMS proceeds via a complex, two channel, mechanism involving abstraction (1a) and reversible addition (1b, -1b). This can be described by the reaction sequence: CH CH CH 3 3 3 SCH SCH 3 3 + OH ⇒ CH 3 SCH 2 + OH + M ⇔ CH 3 + H 2 O (1a) S(OH)CH 3 + M (1b, -1b) S(OH)CH 3 + O 2 ⇒ Products (3) Because of this complex mechanism the effective rate coefficients for reaction (1) and its deuterated analog, reaction (2) depend on the partial pressure of O 2 at any total pressure. OH + CD 3 SCD 3 ⇒ Products (2) The two channel reaction mechanism implies that in the absence of O 2 measure k 1a , the abstraction rate. As we add O 2 we the effective rate increases until we measure a limiting rate (k 1a + k 1b ).
Estimating Aquatic Lifetimes
• One way to estimate the aquatic lifetime of a compound is to
analyze the rate of hydrolysis
of the substance.
• The rate of hydrolysis can be estimated by : log (hydrolysis rate) = log (hydrolysis rate of a reference compound) + Constant * σ Therefore
log (hydrolysis rate) = A + Bσ
(4.20) Where :
A
is rxn and compound class specific(depends on the reference rxn chosen)
B σ
is rxn and compound class specific (depends on type of rxn considered) is a structural parameter commonly used in linear free energy relationship.
Estimating Overall Biodegradability
• It is difficult to do an overall biodegradability analysis.
• It can be estimated using :
I = 3.199 + a 1 f 1 + a 2 f 2 + a 3 f 3 +... + a n f n + a m MW
Where :
a n
is the contribution of the functional group (see table ).
f n
is the number of different functional group.
MW
is the molecular weight.
I
is an indicator of aerobic biodegradation rate.
• Different Values (of I) represent different life times : (4.21) I value Expected degradation rate 5 Hours 4 Days 3 Weeks 2 Months 1 Years
Example : Overall Biodegradability Estimation
Estimate the Biodegradation Index for diethyl ether.
Solution :
Molecular weight of diethyl ether : MW = 74.12 g/mol Using equation 4.21, the index can be calculated :
I = 3.199 + a 1 f 1 I + a 2 f 2 + a 3 f 3 + ... + a n f n + a
= 3.199 + (- 0.0087) - 0.00221(74.12) =
m MW 3.0267
Therefor a lifetime of WEEKS
Estimating Ecosystem Risks
Compare the Fish, Guppy and Daphnids mortalities for an acrylate with log K ow = 1.22 (e.g. ethyl acrylate).
Guppies
log (1/LC 50 ) = 0.871 log K ow – 4.87
log (1/LC 50 ) = 0.871(1.22) – 4.87 = -3.80738
LC 50
=
6417.74 µmol/L
.
(4.22)
Daphnids
log LC 50
log LC 50
LC 50 = 0.00886 – 0.51136 log K
= 0.00886 – 0.51136(1.22) = -0.6149992
=
0.242 millimoles/L
=
ow 242 µmol/L.
(4.23)
Estimating Ecosystem Risks Continued
Fish
log LC 50 = -1.46 – 0.18 log K ow
log LC 50
LC 50
= = -1.46 – 0.18(1.22) = -1.6796
0.021 millimoles/L
=
21 µmol/L.
(4.24) The concentrations yielding 50% mortality are:
Guppies Daphnids
(14 day): 6417.74 µmol/L.
(48 hour): 0.242 millimoles/L = 242 µmol/L.
Fish
(96 hour): 0.021 millimoles/L = 21 µmol/L.
Environmental Fate and Exposures
Example :
If chemicals are released into a river upstream of a water treament plant,
what factors need to be taken into account to estimate the potential danger to the community
. What fraction of the chemicals are: Absorbed by river sediments.
- Taken up by living organisms.
- Reacted with other compounds.
- Volatilized into the air.
- Biodegraded.
- Removed in the treatment process.
Classification of Substances Based on Risk
By examining the table XX, we can use the calculated properties to qualitatively quantify the risk associated with the different substances Three main criteria are normally considered in the classification of the substances :
persistence, bioaccumultion and toxicity
.
There do not exist a given set of regulations or guidelines on quantifying risk, but the above parameters are used in the process.
Available Ressources
EPA (persistent, bioaccumulating and toxic substances) : http://www.epa.gov/pbt/aboutpbt.htm
http://www.epa.gov/opptintr/pbt/ Pollution Prevention, Waste Minimization and PBT Chemical Reduction : http://yosemite.epa.gov/R10/OWCM.NSF/0d511e619f047e0d882565 00005bec99/6ad9c10eb8a06bc288256506007def78?opendocument
Environment canada (existing substances evaluation) : http://www.ec.gc.ca/substances/ese/eng/psap/psap_2.cfm