Question

. Estimate the Henry’s Law constant in

(i) in atm-m³/mol and

(ii) dimensionless

at 25^oC for the following VOCs:

a) Benzene: Solubility = 1,780 mg/L; Vapor Pressure = 0.125 atm

b) Chloroform: Solubility = 8,000 mg/L; Vapor Pressure = 0.32 atm

c) Trchloroethylene: Solubility = 1,000 mg/L; Vapor Pressure = 0.08 atm

Answer 1

Henrys law can be written as C= K*Pgas

Pgas= partial pressure of gas, C= solubilty of gas in L

given solubility of Benzene= 1780 mg/L= (1780/1000)g/L= 1.780 g/L

molar mass of Benzene= 78

hence solubility of Benzene = (1.780/78) gmoles/L =0.02282gmole/L

vapor pressure= 0.125atm

hence Henrys law constant, H = 0.125/0.02282 atm/M=5.478 atm/M

H =H'RT, H'= dimensioless henrys law constant

R= 0.0821 L.atm/mole.K and T= 25+273= 298K

H'=5.478/(0.0821*298)= 0.224

concentration of Benzene = 0.02282 mole/L= 0.02282*10^-3 mole/m3

H= 0.125/(0.02282*10^-3) atmm3/mole = 5477.651 atm/m3mole

2. For chlooform = 8000 mg/L= 8g/L, molar mass of chloroform = 120, molar concentration = 8/120 gmole/L =0.067 mole/L

henry;s kaw constant = 0.32/0.067= 4.8 atm/M, dimensionless Henrys law constant= 4.8/(0.0821*298)= 0.196

in terna of atmm3/molre = 4.8*1000=4800 atmm3/mole

3. trichloroethylene= 1000 mg/L= 1 g/L, molar mass= 131 g/mole, mole = 1/131= 0.0076 moles/L

Henrys law constant = 0.08/0.0076= 10.52 atm/M, in terms of dimensionless = 10.52/(0.0821*298)= 0.43

in terms of atmm3/mole= 10.52*1000 = 10520 atmm3/mole