C1 — Diffusion with homogeneous reaction to a catalyst surface: 23-Chem-B1 Transport Phenomena, May 2014, Q5

Question 5: C1 — Diffusion with homogeneous reaction to a catalyst surface

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. Steady one-dimensional diffusion of A across a stagnant film $0\le z\le\delta$ ($z=0$ at the catalyst surface). Total molar concentration $c$ and diffusivity $D_{A\text{-mix}}$ constant; homogeneous consumption of A at the constant volumetric rate $R_A=-k_A$; the instantaneous surface reaction makes the wall a perfect sink for A, $y_A(0)=0$. At the film edge $y_A(\delta)=y_{A\delta}$.

Find. The mole-fraction profile $y_A(z)$ and the molar flux $N_A(z)$ (a “show that” derivation — no numbers).

Fig. C1: Species A diffuses across a stagnant film toward a catalyst surface. Homogeneous consumption ($-k_A$) makes the concentration profile sag below a straight line; the instantaneous surface reaction fixes $y_A=0$ at $z=0$.

Approach. Apply the steady species-A continuity equation with a constant (zero-order) homogeneous sink, integrate twice, and close with the surface-sink and film-edge conditions; then differentiate to obtain the flux.

Reduce the species-A equation. For steady, one-dimensional transport with a homogeneous reaction, the diffusion form (Table A.5) becomes $0=D_{A\text{-mix}}\dfrac{d^2c_A}{dz^2}+R_A$ with $R_A=-k_A$. Writing $c_A=c\,y_A$ (constant $c$), $$cD_{A\text{-mix}}\frac{d^2y_A}{dz^2}=k_A\quad\Longrightarrow\quad\frac{d^2y_A}{dz^2}=\frac{k_A}{cD_{A\text{-mix}}}.$$
Integrate twice. $\dfrac{dy_A}{dz}=\dfrac{k_A}{cD_{A\text{-mix}}}z+B_1$ and $y_A=\dfrac{k_A}{2cD_{A\text{-mix}}}z^2+B_1z+B_2.$
Apply the boundary conditions. The instantaneous surface reaction makes A vanish at the wall, $y_A(0)=0\Rightarrow B_2=0$; at the film edge $y_A(\delta)=y_{A\delta}$ gives $B_1=\dfrac{y_{A\delta}}{\delta}-\dfrac{k_A\delta}{2cD_{A\text{-mix}}}.$
Mole-fraction profile. Substituting the constants and grouping the reaction terms, $$y_A=\frac{z}{\delta}y_{A\delta}-\frac{k_A(\delta z-z^2)}{2cD_{A\text{-mix}}},$$ the required result: a straight diffusion line $\tfrac{z}{\delta}y_{A\delta}$ minus a parabolic sag from consumption ($\delta z-z^2\ge0$ on the film). $\boxed{\,y_A=\dfrac{z}{\delta}y_{A\delta}-\dfrac{k_A(\delta z-z^2)}{2cD_{A\text{-mix}}}\,}$
Flux of A. No bulk-flow term is needed: every reaction here ($A\!\rightarrow\!B$ in the film and at the surface, $B\!\rightarrow\!C$ in the film) is one mole in, one mole out, so $d(N_A+N_B+N_C)/dz=-k_A+k_A-k_B+k_B=0$, and at the catalyst each arriving A leaves as one B, so the net molar flux is zero at $z=0$ and hence everywhere. The flux of A is therefore purely diffusive, $N_A=-cD_{A\text{-mix}}\dfrac{dy_A}{dz}$ (the $B\!\rightarrow\!C$ step changes the B/C split but not the A balance). Differentiating the profile, $\dfrac{dy_A}{dz}=\dfrac{y_{A\delta}}{\delta}-\dfrac{k_A(\delta-2z)}{2cD_{A\text{-mix}}}$, so $$N_A=-cD_{A\text{-mix}}\frac{y_{A\delta}}{\delta}+\frac{k_A(\delta-2z)}{2}=k_A\!\left(\frac{\delta}{2}-z\right)-\frac{cD_{A\text{-mix}}}{\delta}y_{A\delta},$$ the required flux expression. $\boxed{\,N_A=k_A\!\left(\dfrac{\delta}{2}-z\right)-\dfrac{cD_{A\text{-mix}}}{\delta}y_{A\delta}\,}$

Quantity	Result
Mole-fraction profile	$y_A=\dfrac{z}{\delta}y_{A\delta}-\dfrac{k_A(\delta z-z^2)}{2cD_{A\text{-mix}}}$
Flux of A	$N_A=k_A\!\left(\dfrac{\delta}{2}-z\right)-\dfrac{cD_{A\text{-mix}}}{\delta}y_{A\delta}$
Check at $z=0,\ z=\delta$	$y_A=0$; $y_A=y_{A\delta}$

Concept & Theory

Topic. Steady film diffusion with a homogeneous zero-order reaction and a diffusion-controlled surface sink.

Key relations. $D_{A\text{-mix}}\,d^2c_A/dz^2=k_A$; Dirichlet conditions $y_A(0)=0$ (surface sink), $y_A(\delta)=y_{A\delta}$; diffusive flux $N_A=-cD_{A\text{-mix}}\,dy_A/dz$.

Why this works. A constant volumetric sink adds a constant to the second derivative of concentration, so the profile is necessarily quadratic: the homogeneous (linear) part carries the boundary mole fractions while the particular (parabolic) part carries the consumption. Because A is consumed along the path, its flux $N_A$ is not constant — it varies linearly with $z$, unlike pure film diffusion.

Common pitfall. Sign of the source ($-k_A$ consumption gives $y_A''>0$, a downward sag); and treating the flux as constant — the reaction makes $dy_A/dz$ position-dependent, so $N_A$ changes across the film.

Source. Bird, Stewart & Lightfoot, Ch. 18 (diffusion with homogeneous reaction, Table A.5); Welty et al., Ch. 26 (steady diffusion with chemical reaction).

Question 5 of 6: C1 — Diffusion with homogeneous reaction to a catalyst surface

Question 5: C1 — Diffusion with homogeneous reaction to a catalyst surface