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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 30oC. 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.

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 30oC. 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. Correct Answer 1.486×10-3 kmol/m2.s

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

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