Answer to Question #166340 in General Chemistry for Piyush Mishra

Question

For absorption of certain gas from an air steam by water the value of KG was found as 2x10-6 

kmol/m2-s-kPa. The absorption takes place at 298 K and at 1 atm pressure at a particular 

location of the column the gas phase concentration is 5 mol % and liquid phase concentration 

is 0.2 mol %.Only 10 %of the total resistance lies in the liquid phase.The solution obeys 

Henry’s law and the value of m=1.5 at 298 K and 1atm. calculate the individual mass transfer 

coefficient, flux and interfacial concentration.

Answer 1

Ky = KGP = (2×10-6)(101.3) = 2.026*10-4 kmol/m2

⋅s

Since 1/Ky =1/ky +m/kx for gas phase resistance that accounts for 85% of the total resistance:

Ky/0.85 =2.026*10-4 kmol/m2⋅s/0.85=2.38*10-4 kmol/m2⋅s

kx = mKy/0.15 = 1.5*2.026*10-4 kmol/m2⋅s/0.15=2.026*10-3 kmol/m2⋅s

mxA=1.5*2*10-3=3*10-3

NA = Ky(yA − yA*) = 2.026*10-4 kmol/m2

⋅s(0.05 − 3*10-3)=9.5222e-6

Na=9.5222e-6/2.38*10-4=0.04

Yai=0.04/1.5=0.02667