Question

In a wetted-wall tower, an air-H2S mixture is flowing by a film of water which is flowing as a thin film down a vertical plate. The H2S is being absorbed from the air to the water at a total pressure of 1.50 atm abs and 30^oC. The value of kc of 9.567×10^-4 m/s has been predicted for the gas-phase mass-transfer coefficient. At a given point the mole fraction of H2S in the liquid at the liquid-gas interface is 2.0×10^-5 and pA of H2S in the gas is 0.05 atm. The Henry’s law equilibrium relation is pA(atm) = 609xA (mole fraction in liquid). Calculate the rate of absorption of H2S.

(a) 1.480×10^-3 kmol/m^2.s

(b) 1.486×10^-3 kmol/m^2.s

(c) 1.485×10^-3 kmol/m^2.s

(d) 1.487×10^-3 kmol/m^2.s

This question was posed to me in unit test.

My question is from Convective Mass Transfer in portion Mass Transfer of Bioprocess Engineering

Answer 1

Right answer is (b) 1.486×10^-3 kmol/m^2.s

Explanation: The rate of absorption of H2S per unit area of the thin film is given by:

The partial pressure of H2S in the gas phase at the interface is determined from Henry’s law and the mole fraction of H2S in the liquid at the liquid-gas interface.

pAi = 609xAi = 609×2.0×10^-5 = 1.218×10^-2 atm

The mole fraction of H2S in the gas phase at the interface is then