Question

5. The gas "A" is absorbed into a completely stationary liquid phase where the reaction A → B occurs, characterized by the first order rate constant, k1. Let the gas-liquid interface correspond to z = 0, where the solubility of A is known. The liquid extends in the positive z-direction. The rate of reaction is much greater than the rate of diffusion, thus it can be assumed that CA O as z becomes large. Assume one dimensional diffusion. AB a. Using the conditions stated in the problem, develop an expression for CA in the liquid phase. Clearly state the reasons for the simplifications you made in your analysis. b. If k1/DAB = 7.26 cm and the solubility of "A" is about 0.84 gmol/L, at what depth is CA 2% of the interfacial equilibrium concentration?

Answer 1

The derivation is shown below:

solution 2 Consider a control volume of thickness az at a distance a from the interface. A | AB Mole balance of A in the control volume, Accumulation - Input-output + generation / consumpution. - # Assumptions a one dimensional diffusion (only 2-direction) - steady state condition (accumulation - 0) given conditions ونعلمك سيك هدم حلمك 3 o = JZ A - Izraz A + Aloz A) A is the cross section area. By Ficks Law of diffusion, Jz = - DAR ACA Flux) dz

Substituting expression of I. and ra. 0 - DAB dCAL + DAB dCAL - R, (AZ) LA Divide by az O = DAB dCAL - ACAL - R, CA Id2 12+02 dz lz) DZ Taking 0 lim 12 0 DAB d²CA - R, CA d22 - d²CA - R, CA = 0 dz? DAB This ODE is of form, dy my =o. The solution of the ODE is given by dua - yln) = C, exp (um n) + Ca exp(-5mx). where c. and ca are constants. Thus, For the governing equation, m = k/ DAB. CA/2) = 6, ep 1 e 2 + z elp |- k z IDAB

Putting in boundary conditions, CA=0 as 2. X. O c +Gole) as 20 W, Sept 2) CA = CAO at 2=0 = CAO CAO - C duplo) Substituting c, and ca ICA(Z) = Caolip(-) NDAB b given R = 7.26 cm? DAB solubility of A = 0.84 gmol/L & CAO = 0.84 god/L CA = 21 of CAO - CA = 2. CAO Too Plugging in values, 2. CA - Cho. exp| -17.26. 2). 100 + - $7.26.2 = ln 10.02) 2= - Info.022 cm 57.26 2 = 1.45 cm -