1) Diffusion Through a Film: (40 points) Hydrochloric acid (HCI) diffuses throug
ID: 701210 • Letter: 1
Question
1) Diffusion Through a Film: (40 points) Hydrochloric acid (HCI) diffuses through a thin film of water (2.0 mm thick) at 290 K The concentration of HCl at point l on one boundary of the film is 8% by mass and on the other end of the boundary, at point 2 is 5% by mass. Density of the solution mixture at points 1 and 2 are 1080 kg/m3 and 1060 kg/m3 respectively. The diffusivity of HCI in water is 2.6 x 10-8 m2/s. Calculate the molar flux of HCl assuming water to be stagnant. Assume that the net molar concentration remains almost constant across the film (or take an average of the concentrations on both sides of the film).Explanation / Answer
Temperature T = 290 K
A1 = 1080 kg/m3
A2 = 1060 kg/m3
Mass fraction at boundary 1 = 0.08
Mass fraction at boundary 2 = 0.05
Molecular weight of HCl, MA = 36.5 kg/kmol
Molecular weight of water, MB = 18 kg/kmol
Diffusivity of HCl in water DAB = 2.6 x 10^-8 m2/s
Mol fraction of solution at boundary 1
XA1 = (8/36.5)/(8/36.5 + 92/18) = 0.2191/(0.2191 + 5.111)
= 0.0411
Mol fraction of solution at boundary 2
XA2 = (5/36.5)/(5/36.5 + 95/18) = 0.1369/(0.1369 + 5.277)
= 0.02528
XB1 = 1 - XA1 = 1 - 0.0411 = 0.9589
XB2 = 1 - XA2 = 1 - 0.02528 = 0.97472
Molecular weight at point 1
M1 = 100/(8/36.5 + 92/18)
= 18.76 kg/kmol
Molecular weight at point 2
M2 = 100/(5/36.5 + 95/18)
= 18.468 kg/kmol
Cav =[( A1/M1) + (A2/M2)] /2
= [(1080/18.76) + (1060/18.468)]/2
= (57.633 + 57.440)/2
= 57.482 kmol/m3
XBM = (XB2 - XB1) / ln (XB2 /XB1)
= (0.97472 - 0.9589) / ln (0.97472/0.9589)
= 0.01618/0.01675
= 0.96678
Molar flux
NA = DAB x Cav x (XA1 - XA2) / XBM * (Z2-Z1)
= 2.6 x 10^-8 m2/s x 57.482 kmol/m3 x (0.0411 - 0.02528) /(0.96678 * 0.002 m)
= 1.222 x 10^-5 kmol/m2-s
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