Chapter 6 Henry S Law

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 by V l or the molar volume of the mixture (sometimes called Vmix)

X/Vl = Concentration= Ci

_ * pi / V l = i Ci p i * _ pi / Ci = i p i V l = 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? 1 Chatper 6 Henry’s Law

pi KiH  Ciw traditionally

atmi 1 KiH  1  atmlitersmole molesliters w

Cia Kiaw  (dimensionless Henry’s law const.) Ciw

Cia Cia KiH Cia KiH KiH Kiaw     p n / V RT RT pi i KiH

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?

3 4 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 units of moles/ cc water

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

-1 -2 I x Ciw = g rain sec cm 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

5 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

-5 Kiaw phenol = 2x10 krate constat = I x 1/Kiaw x Cia / (Cia x30,000 cm) in units of 1/sec

6 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

7 Concentration effects on KiH

Ciw = Xi / Vw Vw = molar vol. H2O

* pi  iw iw pl * KiH     iw pl Vw Ciw iw /Vw

Under dilute conditions KiH is directly proportional to the:

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

8 What is the effect of concentration on KiH? * P iW water

organic

* at saturation the vapor pressure pi = p iw

* pi = i Xi p i pure liquid

sat sat 1 pi 1 X iw  sat *  sat  iw pi  iw

psat sat i sat * K   pl Vw iH Csat w iw

sat The question becomes how does KiH differ from KiH ?

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

The old book suggests from benzene partitioning data, sat that little difference may exist between KiH 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 psat sat sat i KiH and estimated from K  iH Csat iw

Example

sat -3 o If the Ciw for chlorobenzene = 4.3x10 mol/L at 25 C

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

psat 2 sat iL 1.6x10 atm KiH K    3.6atm L / mol iH Csat 4.3x103 mol / L iw

10 K 3.6atm L mol 1 K  iH   0.15 iaw RT 0.082atm L mol 1K 1x298K

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

log sat log   i i sat 2 (1 xi )

psat K sat  iw sat p* V iH sat w iL w Ciw for infinitely dilute solutions

pi iw iw p *iL * KiH    iw piLVw Ciw iw / Vw

11  sat Comparison of iw and iw

 sat sat sat iw -logCiw Ciw iw

(Tab 5.2) (p618) mol/L 1/(CsatVmix) (old book) benzene 2400 1.64 0.0229 2425 toluene 12000 2.25 0.0056 9879 chlorobenz 19000 2.35 0.00447 12437 hexCl-benz 9.8E+8 5.56 2.75E-6 2.0E+7 octanol 37000 2.35 0.00447 18656

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

12 Effect of Temperature

* vapHi 1 ln piL    const R T by analogy E H 1 ln x sat   iw  const iw R T

sat sat xiw Ciw  so substituting Vmix excess heat of solution E H 1 lnCsat   iw  const w R T (Vmix )

* PIl KiH  sat CIw

 H  HE ln K sat   VAP i iw  const iH RT H

13 E vapHi- H iW = awHiHHenry

14 15 What are the effects of salts?

(Setschenow, 1889)

 sat  Ciw s log sat K i [ salt ]tot Ciw ,salt  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

sat sat s logC iw ,salt logCiw Ki [ salt ]tot

s sat sat Ki [ salt ]tot C iw ,salt Ciw 10

the Henry’s law for salt water is * * p s iL piL sat Ki [ salt ] KiH ,w ,salt  sat  s KiH 10 C sat Ki [ salt ] iw ,salt Ciw 10 s for anthracene Ki = 0.3, assume [salt] = 0.5 M and KiH = 0.078 atm L mol-1 (0.3)x(0.5) -1 so KIH,w,salt= 0.078x10 =0.11 atm L mol

16 17 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

18 (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

sat from p*iL / C w= 0.00041

19 20 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

CwVw 1 1 fw    CwVw  CaVa CaVa Va 1 1KH CwVw Vw

Using the Hine and Mookerjee

Cl K’H = Kiaw= 0.1622

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

21 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

22 Experimental Measurements 1. air toluene McAuliffe (1971)

CiwVwv fract in H2O = CiaVg  CiwVwv

Vwv = vol of water for dilute systems

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

23 Vwv fract in H2O= KiawVg Vwv each time we take a step

n Cia,n= (fact in H2O) Ciw,o Kiaw taking the logs of both sides and substituting for fract in H2O and remembering that Kiaw=Dgw

V wv +log (C D logCia,n  n log iw o gw) Kiaw Vg Vwv

24 25 26 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

B C E 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

27 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  Zi Vi

Mi = fi Zi VI

4. Calculating Z values

28 Zi fi = 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 Ziw fiw Ziw = pi /{fw KH}= 1/KH

We will use a representative value of -4 3 -1 KiH= 1x10 m atm mol

29 On soils, sediment, and suspended solids

Cwi + S ----> Cis

CiS KiwS  ; Cis = KisxCiwxS CiW xS

Cis =Zi sp x fis and Ciw = 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 -4 3 3 -1 -3 a Ki sp of 10 m /mg, Zsp= 10 mol atm m

For sediment and soils at 2x109 mg/m3 and -5 3 9 -1 -3 a Kiws of 5x10 m /mg, Zs,s= 10 mol atm m

For Aquatic Biota

ZB = B  y  Kiow/KiH

-6 3 3 where B is the volB/vol H20= 5x10 m /m ,

5 4 y=octanol fract. of B = 0.2, Kiow=10 ; ZB=10

------

30 Let’s look at the Equilibrium Distribution of a toxic compound -10 3 with an atmospheric concentration of 4 x 10 mol/m .(fi x Zi = C and Mi = fi Zi Vi)

Z Vol fi M % g/m3. (m3) (atm) (moles) air 40 1010 10-11 4 0.35 water 104 106 10-11 10-1 0.01 10-5 s solids 103 106 10-11 10-2 0.001 0.01 Sed 109 104 10-11 102 9.1 0.05 Soil 109 105 10-11 103 90.5 0.5 Aq biota 104 106 10-11 10-1 0.01 0.2