What is Henry’s law??

pi= p*iL Xil

pi / Xil = p*iL

for non ideal solutions (low solubility)

pi= i Xil p*i

dividing bylor the molar volume of the mixture (sometimes called Vmix)

X/Vl = Concentration= Ci

pi / l = i Cip*i

pi / Ci = ip*il = const???= KiHl ;

If air – water K iH

Is the product of i,p*i Vmix a constant??

Is it different for different compounds??

Does it vary with temperature???

Does it change with concentration?

Does it change with salt or ionic content?

How do we measure it?

Chatper 6 Henry’s Law

traditionally

(dimensionless Henry’s law const.)

As Henry’s law values increase there is a tendency for higher gas phase concentrations over water i.e. partitioning is toward air

for high vapor pressure compounds the fugacity in the gas phase is high

fi = i Xifi* pure liquid

(fi*pure liquid = p*i pure liquid)

High activity(i) coefs. favor partitioning to the gas phase i.e. Lower KiH and lower ‘s favor the liquid phase. Polar compounds?

Wash out ratios or W and how fast does the atmosphere clean up during a rain

Usually defined as the conc. in rain/conc. In air

W = Ciw/Cia = 1/Kiaw

W x Cia = conc in the rain, Ciw , with unitsof moles/ ccwater

or Ciw in units of moles i /cc = moles i/g H20

The rain has an intensity, I with units of

grams of rain sec-1 cm-2

so now

I x Ciw = g rain sec-1 cm-2 x moles i/g H20

Since W = Ciw/Cia = 1/Kiaw

I x 1/Kiaw x Cia = moles of i from the atmosphere hitting the surface of the earth in the rain per sec-1 cm-2

And this is a flux

We will learn in the old book Chapter 10,

Flux / (conc x depth ) = 1st order rate constant in

C = Co e-kt

So if you know the rain intensity, Kiaw and the height of the atmosphere, you can estimate how fast the atmosphere will “clean” up with a given rain intensity???

______

Flux = I x 1/Kiaw x Cia = moles of i from the atmosphere hitting the surface of the earth per sec-1 cm-2

If the mixing height of the atmosphere is 300 m

and we have a rain that gives an 1” of water in 2 hours

I = 2.5g cm-2 /(2x60x60 sec) = 3.47x10-4 g cm-2 sec-1

Kiaw phenol = 2x10-5

krate constat = I x 1/Kiaw x Cia / (Cia x30,000 cm)
in units of 1/sec

in units of 1/sec = 0.00059 sec-1

C/Co = e-kt ; t = 2 hours = 2x60x60 sec

C/Co = 0.0145 or 98.5% of the phenol will be cleaned out in the air in the rain

Concentration effects on KiH

Ciw = Xi / VwVw = molar vol. H2O

Under dilute conditions KiH is directly proportional to the:

  • activity coef.
  • saturated vapor pressure
  • molar volume of water

What is the effect of concentration on KiH?

P*iW

water

organic

at saturation the vapor pressure pi =p*iw

pi = iXi p*i pure liquid

The question becomes how does KiHsat differ from KiH ?

If the activity coef. changes with increases in concentration of Ciw then KHsat will change?

Why?

The old book suggests from benzene partitioning data, that little difference may exist between KiHsat and KiH

for benzene K’iaw = (Cair/Ciw) a difference of <4% was observed between saturated and dilute water solutions….

This means that KiH can sometimes be approx. from KiHsat and estimated from

Example

If the Ciwsat for chlorobenzene = 4.3x10-3 mol/L at 25oC

and p*iL = 1.6x10-2 atm what is the KiH

A simple way of changing iw into iwsat (this does not always work)

for infinitely dilute solutions

Comparison of iw and iwsat

iw -logCiwsat Ciw satiwsat
(Tab 5.2) (p618) mol/L1/(CsatVmix)
(old book)

benzene 24001.640.02292425
toluene120002.250.00569879
chlorobenz190002.350.0044712437
hexCl-benz9.8E+85.562.75E-62.0E+7
octanol370002.350.0044718656

Why are iwvalues sometimes greater than iwsat?

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Effect of Temperature

by analogy

so substituting

excess heat ofsolution

1

vapHi- HEiW = awHiHHenry

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What are the effects of salts?

(Setschenow, 1889)

let’s say we want to calculate the equilibrium distribution of anthracene in sea water,

ie KiH w,salt

if we transform Setschenow’s equation


the Henry’s law for salt water is

for anthracene Kis= 0.3, assume [salt] = 0.5 M

and KiH = 0.078 atm L mol-1

so KIH,w,salt= 0.078x10(0.3)x(0.5)=0.11 atm L mol-1

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Estimating Henry’s Law values

Hine and Mookerjee (1975)

Log Kiaw =nj x functional groupi

OH

for phenol

there are

(older data)(new data)

6 aromatic carbons at: -.33/carbon-0.264

5 aromatic C-H groups, at: 0.21/group+0.154

and one C-O group at: 0.74-0.596 (C-OH)

and one OH group at: -3.21-3.232

(old data)

log K’H= 6x(-.33)+5(.21)+0.74+(-3.21) = -3.40

(New data) log Kiaw= -4.64

K’H = 0.0004 ; new book Kiaw= 0.000023

from p*iL / Csatw= 0.00041

Example Problem: Consider a well sealed flask with 100 ml of H2O and 900 ml air. At equilibrium estimate the amount of chlorobenzene in the air and in the water if the sum (total) in both phases is 10 g.

fw = the fraction in the water phase

fw = chlorobenzene mass in water/total mass

Using the Hine and Mookerjee

Cl

K’H = Kiaw= 0.1622

fw = 1/{(1+0.1662)900/100}=0.41

the concentration in the aqueous phase Cw is

Cw = fw Mtot /Vw

Cw = 0.41x10g /0.1L = 41 g/Lwater

Ca= 0.59x10g /0.9 L = 6.6g/Lair

Experimental Measurements

1.

air

toluene

McAuliffe (1971)

fract in H2O =

Vwv = vol of water

for dilute systems

Kiaw= Cia/Ciw = Dg,w( a gas/water part. coef.)

fract in H2O=

each time we take a step

Cia,n= (fact in H2O)n Ciw,o Kiaw

taking the logs of both sides and substituting for

fract in H2O and remembering that Kiaw=Dgw

+log (Ciw oDgw)


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Using fugacities to model environmental systems(Donald Mackay ES&T, 1979)

Consider the phase equilibrium of five environmental compartments. Is it possible to tell where an environmental pollutant will concentrate?

A

BCE

D

where A= air, B= lake, C= Soil, D= Sediment, E= biota and suspended solids

When a system is at equilibrium the escaping tendencies in each phase are equal

fA = fB = fC = fD = fE
For Example: oxygen in water at 0.3 mol/m3 and in air at 8mol/m3 exert the same escaping tendency of 0.2 atm and are thus in equilibrium with the same fugacity.

1. Fugacitys are linearly related to conc.

oxygen in water at 0.03 mol/m3 exerts

a fugacity of one tenth the fugacity of 0.3mol/m3.

fA = fB = fC = fD = fE
Fugacities can be translated into concentrations

fi Zi = C

where Z is called a fugacity capacity value
------

3. the mass Mtotal = Ci Vi = fi Zi Vi

if the system is at equilibrium

Mtotal = fi ZiVi

Mi = fi Zi VI

4. Calculating Z values

Zifi = Ci; Zi= C/f

In air f is equal to the partial pressure,pi

piV = nRT, pi = Cair RT, so Ziair = 1/RT

at 298K , RT= 0.082 liter atm K-1 mol-1x298K

RT= 0.025 m3 atm mol-1
------

In water pi = KiH Ciw and Ciw = Z fiw

pi = KiH Ziwfiw Ziw = pi /{fw KH}= 1/KH

We will use a representative value of
KiH= 1x10-4 m3 atm mol-1
On soils, sediment, and suspended solids

Cwi + S ----> Cis

; Cis = KisxCiwxS

Cis =Zi spx fis andCiw = pi /KiH

Zi sp = KiwS x 1/KiH x pi x S/fis = Ki sp x S/ KiH

For suspended solids at 1,000 mg/m3 and
a Ki sp of 10-4 m3/mg, Zsp= 103 mol atm-1 m-3

For sediment and soils at 2x109 mg/m3 and
a Kiws of 5x10-5 m3/mg, Zs,s= 109 mol atm-1 m-3

For Aquatic Biota

ZB = B  y  Kiow/KiH

where B is the volB/vol H20= 5x10-6 m3/m3,

y=octanol fract. of B = 0.2, Kiow=105; ZB=104

------

Let’s look at the Equilibrium Distribution of a toxic compound with an atmospheric concentration of 4 x 10-10 mol/m3.(fi x Zi = C and Mi = fi Zi Vi)

ZVolfiM%g/m3.
(m3)(atm)(moles)

air40101010-1140.35
water10410610-1110-10.0110-5
s solids10310610-1110-20.0010.01
Sed10910410-111029.10.05
Soil10910510-1110390.50.5
Aq biota 10410610-1110-10.010.2

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