Part B: Diffusion of Trichloroacetic Acid Through a Stagnant Methanol Film: 23-Chem-A3 Heat and Mass Transfer, May 2018, Q4

Question 4 — Part B: Diffusion of Trichloroacetic Acid Through a Stagnant Methanol Film (25 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Component A = trichloroacetic acid ($M_A=163.4$ g/mol) diffuses through stagnant methanol B ($M_B=32.04$ g/mol) across a film $z=2$ mm thick. $D_{AB}=1.862\times10^{-9}$ m²/s. Boundary compositions and densities:

Face	wt% acid	$\rho$ (kg/m³)	$x_A$ (mole frac.)
1 (high)	6 %	1012	0.01236
2 (low)	2 %	1003	0.003986

Find. The molar (and mass) flux $N_A$ of the acid across the film.

Figure 4 — Diffusion of A through a stationary film of B: the acid mole fraction falls almost linearly across the 2-mm methanol layer. Because B does not diffuse, the log-mean $x_{B}$ enters the flux (Stefan diffusion).

Approach. Convert each wt% to a mole fraction and a molar concentration $c=\rho/M_{avg}$, average $c$ across the film, form the log-mean inert mole fraction $x_{B,lm}$, then apply the stagnant-film (Stefan) diffusion equation for A through non-diffusing B.

Mole fractions. On a 100-g basis at face 1: $n_A=6/163.4=0.0367$, $n_B=94/32.04=2.934$ mol, so $x_{A1}=0.0367/2.971=0.01236$ and $x_{B1}=0.98764$. Likewise face 2 gives $x_{A2}=0.003986$, $x_{B2}=0.99601$.
Molar concentration each face. $M_{avg}=100/(n_A+n_B)$, and $c=\rho/M_{avg}$: $$c_1=\frac{1012}{33.66\times10^{-3}}=3.01\times10^{4},\qquad c_2=\frac{1003}{32.56\times10^{-3}}=3.08\times10^{4}\ \text{mol/m}^3.$$ The film-average total concentration is $c=\tfrac12(c_1+c_2)=\boxed{3.04\times10^{4}\ \text{mol/m}^3}.$
Log-mean inert mole fraction. Since methanol is non-diffusing, $$x_{B,lm}=\frac{x_{B2}-x_{B1}}{\ln(x_{B2}/x_{B1})}=\frac{0.99601-0.98764}{\ln(0.99601/0.98764)}=0.9918.$$
Stefan flux of the acid. For A diffusing through stagnant B, $$N_A=\frac{D_{AB}\,c}{z\,x_{B,lm}}\,(x_{A1}-x_{A2})=\frac{(1.862\times10^{-9})(3.04\times10^{4})}{(0.002)(0.9918)}(0.01236-0.003986),$$ $$\boxed{N_A\approx 2.39\times10^{-4}\ \text{mol/m}^2\text{s}.}$$
Mass rate. Multiplying by $M_A$, $$n_A=N_A M_A=(2.39\times10^{-4})(0.1634)=\boxed{3.9\times10^{-5}\ \text{kg/m}^2\text{s}.}$$ This is the diffusion rate per unit film area; a real device rate would multiply by the interfacial area.

Quantity	Result
Film-average concentration $c$	$3.04\times10^{4}$ mol/m³
Log-mean inert fraction $x_{B,lm}$	0.992
Molar flux $N_A$	≈ $2.39\times10^{-4}$ mol/m²s
Mass flux	≈ $3.9\times10^{-5}$ kg/m²s

Concept & Theory

Topic. Steady one-dimensional molecular diffusion of A through a stagnant (non-diffusing) film of B — Stefan diffusion.

Key relations. $N_A=\dfrac{D_{AB}\,c}{z\,x_{B,lm}}(x_{A1}-x_{A2})$ with $x_{B,lm}=(x_{B2}-x_{B1})/\ln(x_{B2}/x_{B1})$; $c=\rho/M_{avg}$; equivalently $N_A=\dfrac{D_{AB}\,c}{z}\ln\!\dfrac{x_{B2}}{x_{B1}}$.

Why this works. Because B is stationary, its diffusive flux must be cancelled by a bulk (convective) flow, which augments A’s transport — the “drift” captured by dividing by $x_{B,lm}$ (always $<1$, so the flux is larger than pure Fick). With the acid dilute, $x_{B,lm}\approx1$ and the correction is only ≈ 0.8 %, but it is the physically correct form.

Common pitfall. Using an arithmetic-mean instead of log-mean $x_B$, or omitting the $1/x_{B,lm}$ factor entirely (equimolar counter-diffusion) — wrong here because only A crosses the film. Also, forgetting to convert wt% to mole fraction before applying the diffusion law.

Source. Treybal, Mass-Transfer Operations, Ch. 2 (molecular diffusion, diffusion of A through stagnant B); Coulson & Richardson Vol. 1, Ch. 10.

Question 4 of 6: Part B: Diffusion of Trichloroacetic Acid Through a Stagnant Methanol Film

Question 4 — Part B: Diffusion of Trichloroacetic Acid Through a Stagnant Methanol Film (25 points)