Question

What is the rate expression for physical absorption of a gas A to a liquid based on two film theory? (Where, kAg is the gas film mass transfer coefficient and kAl is the liquid film mass transfer coefficient)

(a) -rA = \(\frac{p_A- H_A C_A}{\frac{1}{a} + \frac{1}{k_{Al}a}} \)

(b) -rA = \(\frac{p_A- C_A}{\frac{1}{k_{Ag} a} + \frac{1}{k_{Al} a}} \)

(c) -rA = \(\frac{p_A- H_A C_A}{\frac{1}{k_{Al}a}+  \frac{1}{k_{Al}a}} \)

(d) -rA = \(\frac{H_A C_A}{\frac{1}{k_{Al} a}+  \frac{1}{k_{Al}a}} \)

This question was posed to me by my college director while I was bunking the class.

I'm obligated to ask this question of Design of Fluid Fluid Reactors topic in section Fluid-Fluid Reactions: Kinetics & Basics of Non Ideal Flow of Chemical Reaction Engineering

Answer 1

Right choice is (c) -rA = \(\frac{p_A- H_A C_A}{\frac{1}{k_{Al}a}+  \frac{1}{k_{Al}a}} \)

Explanation: \(\frac{-r_A}{k_{Ag} a}\) = pA – pAi

\(\frac{-r_A}{k_{Al} a}\) = CA – CAi. Substituting CA = \(\frac{p_A}{H_A}\), we obtain -rA = \(\frac{p_A- H_A C_A}{\frac{1}{k_{Al}a}+  \frac{1}{k_{Al}a}}. \)