Species balances and diffusion with reaction

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Slides by Shah (n.d.): view or download the slides.

1 Errata

• Slide 4 needs corrections to the title and generation term. Use the title “Continuity equation” and the volumetric generation term

  𝑟𝐴Δ𝑥Δ𝑦Δ𝑧$$r_A\Delta x\Delta y\Delta z$$

  The generation rate 𝑟𝐴$r_A$ has no direction subscript.

• Slide 6 mixes the reference frames for relative molar flux and convective transport. Pair 𝐉𝐴$\mathbf J_A$ with molar-average velocity 𝐯∗$\mathbf v^*$.

  Moles are conserved for any reaction with equal total stoichiometric coefficients on both sides, rather than only A →B$\mathrm{A\to B}$.

• Slide 17 places a first-order rate annotation on case 3. Move that annotation to case 4. For case 3, use constant consumption

  𝑅𝐴=−𝑘0$$R_A=-k_0$$

  The resulting concentration profile is parabolic.

• Slides 19 and 22 use inconsistent coordinates in the boundary conditions. Use 𝑥$x$ throughout

  𝑐𝐴(0)=𝑐𝐴0,𝑁𝐴𝑥(𝐿)=0$$c_A(0)=c_{A0},\qquad N_{Ax}(L)=0$$

• Slide 22 gives the wrong sign for the final surface flux. Use

  𝑁𝐴𝑥(0)=𝑐𝐴0𝐷𝐴𝐵𝑚tanh⁡(𝑚𝐿)$$N_{Ax}(0)=c_{A0}D_{AB}m\tanh(mL)$$

  The coordinate 𝑥$x$ is directed into the liquid.

2 Reading list

• Bird et al. (2006), chapters 17 and 18.
• Geankoplis (2003), chapter 6.
• Welty et al. (2001), chapter 24.

Concentrations and molar rates use kmol unless the example states another consistent basis.

3 Locate reaction before choosing the equation

Given conditions	Model and checks
No reaction in the transfer region	Set the volume source to zero; any wall reaction appears in a boundary condition.
Finite-rate reaction at a wall	Match diffusive arrival to the surface reaction rate; retain both film and reaction resistances.
Reaction throughout the volume	Retain the local source term in the continuity equation; choose zero-order or first-order kinetics as stated.
Reaction changes total gas moles	Use stoichiometry to find all species fluxes before assuming equimolar counter-diffusion.
Flow and radial diffusion in a catalytic tube	Retain convection and the required inlet, wall and symmetry conditions.

4 Continuity equation

Species continuity equation

A local species balance relating accumulation, the divergence of species flux and net production by reaction.

4.1 A fixed control volume

Figure 1: A fixed rectangular control volume in Cartesian coordinates, with volume ΔV = Δx Δy Δz.

Take a stationary rectangular volume with sides Δ𝑥$\Delta x$, Δ𝑦$\Delta y$, Δ𝑧$\Delta z$. Let 𝜌𝐴$\rho_A$ be the mass of A per volume, in kg m−3$\mathrm{kg\,m^{-3}}$, and 𝑛𝐴𝑥$n_{Ax}$ the total mass flux of A in the positive 𝑥$x$ direction, in kg m−2 s−1$\mathrm{kg\,m^{-2}\,s^{-1}}$. Production 𝑟𝐴$r_A$ is mass of A produced per volume per time, in kg m−3 s−1$\mathrm{kg\,m^{-3}\,s^{-1}}$. Production has no direction subscript. The contribution of the two faces normal to 𝑥$x$ is

[𝑛𝐴𝑥(𝑥)−𝑛𝐴𝑥(𝑥+Δ𝑥)]Δ𝑦Δ𝑧$$\left[n_{Ax}(x)-n_{Ax}(x+\Delta x)\right]\Delta y\Delta z$$

The balance over all faces is

in−out+𝑟𝐴Δ𝑥Δ𝑦Δ𝑧=∂𝜌𝐴∂𝑡Δ𝑥Δ𝑦Δ𝑧$$\text{in}-\text{out}+r_A\Delta x\Delta y\Delta z =\frac{\partial\rho_A}{\partial t}\Delta x\Delta y\Delta z$$(1)

Add the analogous 𝑦$y$ and 𝑧$z$ face contributions to Eq. 1, divide by volume, and let all three lengths approach zero. The flux difference divided by distance becomes a negative derivative. Thus

∂𝜌𝐴∂𝑡=−(∂𝑛𝐴𝑥∂𝑥+∂𝑛𝐴𝑦∂𝑦+∂𝑛𝐴𝑧∂𝑧)+𝑟𝐴=−∇⋅𝐧𝐴+𝑟𝐴$$\begin{aligned} \frac{\partial\rho_A}{\partial t} &=-\left(\frac{\partial n_{Ax}}{\partial x} +\frac{\partial n_{Ay}}{\partial y} +\frac{\partial n_{Az}}{\partial z}\right)+r_A\\ &=-\nabla\cdot\mathbf n_A+r_A \end{aligned}$$(2)

Every term is mass per volume per time. The minus sign means a net outward flux decreases the mass stored. Time 𝑡$t$ is in s; coordinates 𝑥,𝑦,𝑧$x,y,z$ are in m. Total flux already includes both convection and diffusion. In the mass-average frame, 𝐧𝐴 =𝜌𝐴𝐯 +𝐣𝐴$\mathbf n_A=\rho_A\mathbf v+\mathbf j_A$, where velocity 𝐯$\mathbf v$ is in m s⁻¹ and relative mass flux 𝐣𝐴$\mathbf j_A$ in kg m−2 s−1$\mathrm{kg\,m^{-2}\,s^{-1}}$. Substitution gives

∂𝜌𝐴∂𝑡=−∇⋅(𝜌𝐴𝐯)−∇⋅𝐣𝐴+𝑟𝐴$$\frac{\partial\rho_A}{\partial t} =-\nabla\cdot(\rho_A\mathbf v)-\nabla\cdot\mathbf j_A+r_A$$(3)

Sum Eq. 3 over species. The mass-frame definition gives ∑𝐴𝐣𝐴 =0$\sum_A\mathbf j_A=0$, while ordinary chemical reaction conserves mass, ∑𝐴𝑟𝐴 =0$\sum_A r_A=0$. With 𝜌 =∑𝐴𝜌𝐴$\rho=\sum_A\rho_A$,

∂𝜌∂𝑡+∇⋅(𝜌𝐯)=0$$\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\mathbf v)=0$$(4)

For constant mixture density,

𝜌=constant⟹∇⋅𝐯=0$$\rho=\text{constant}\quad\Longrightarrow\quad\nabla\cdot\mathbf v=0$$

Constant mixture density does not require constant species density. A dilute solute may vary substantially while the solvent determines nearly all of 𝜌$\rho$. The same overall continuity equation also applies in momentum transport.

4.2 Molar units and reference frame

Write 𝑐𝐴 =𝜌𝐴/𝑀𝐴$c_A=\rho_A/M_A$, with molar mass 𝑀𝐴$M_A$ in kg kmol⁻¹ and concentration 𝑐𝐴$c_A$ in kmol m⁻³. The molar production rate 𝑅𝐴 =𝑟𝐴/𝑀𝐴$R_A=r_A/M_A$ is in kmol m−3 s−1$\mathrm{kmol\,m^{-3}\,s^{-1}}$. Total molar flux 𝐍𝐴$\mathbf N_A$ is in kmol m−2 s−1$\mathrm{kmol\,m^{-2}\,s^{-1}}$. Its exact decomposition uses the molar-average velocity 𝐯∗$\mathbf v^*$ and its matching relative flux 𝐉𝐴$\mathbf J_A$:

𝐍𝐴=𝑐𝐴𝐯∗+𝐉𝐴∂𝑐𝐴∂𝑡=−∇⋅(𝑐𝐴𝐯∗)−∇⋅𝐉𝐴+𝑅𝐴∂𝑐∂𝑡=−∇⋅(𝑐𝐯∗)+∑𝐴𝑅𝐴$$\begin{aligned} \mathbf N_A&=c_A\mathbf v^*+\mathbf J_A\\ \frac{\partial c_A}{\partial t} &=-\nabla\cdot(c_A\mathbf v^*)-\nabla\cdot\mathbf J_A+R_A\\ \frac{\partial c}{\partial t} &=-\nabla\cdot(c\mathbf v^*)+\sum_A R_A \end{aligned}$$(5)

Here 𝑐 =∑𝐴𝑐𝐴$c=\sum_A c_A$ and ∑𝐴𝐉𝐴 =0$\sum_A\mathbf J_A=0$. Reaction conserves mass but may change molecule count. For example, N2 +3H2 →2NH3$\mathrm{N_2+3H_2\to2NH_3}$ consumes four moles and produces two. Equal total stoichiometric coefficients conserve moles; an isomerisation A →B$\mathrm{A\to B}$ is one example, rather than the only possibility.

4.3 Dilute liquids and low-density gases

For a binary dilute liquid at fixed temperature and pressure, approximate 𝜌$\rho$ and 𝐷𝐴𝐵$D_{AB}$ as constants. Diffusivity 𝐷𝐴𝐵$D_{AB}$ has units m² s⁻¹. Mass fraction 𝜔𝐴 =𝜌𝐴/𝜌$\omega_A=\rho_A/\rho$ is dimensionless. Fick’s law then becomes

𝐣𝐴=−𝜌𝐷𝐴𝐵∇𝜔𝐴=−𝐷𝐴𝐵∇𝜌𝐴$$\mathbf j_A=-\rho D_{AB}\nabla\omega_A=-D_{AB}\nabla\rho_A$$(6)

Use Eq. 4 to expand the convective divergence in Eq. 3. Divide by constant 𝑀𝐴$M_A$ to obtain

∂𝜌𝐴∂𝑡+𝐯⋅∇𝜌𝐴=𝐷𝐴𝐵∇2𝜌𝐴+𝑟𝐴∂𝑐𝐴∂𝑡+𝐯⋅∇𝑐𝐴=𝐷𝐴𝐵∇2𝑐𝐴+𝑅𝐴$$\begin{aligned} \frac{\partial\rho_A}{\partial t}+\mathbf v\cdot\nabla\rho_A &=D_{AB}\nabla^2\rho_A+r_A\\ \frac{\partial c_A}{\partial t}+\mathbf v\cdot\nabla c_A &=D_{AB}\nabla^2c_A+R_A \end{aligned}$$(7)

The dilute-liquid reduction still retains any imposed solvent velocity.

For an ideal low-density gas at fixed 𝑇$T$ and 𝑃$P$, 𝑐 =𝑃/(ℛ𝑇)$c=P/(\mathcal R T)$ is constant. Gas mole fraction is 𝑦𝐴 =𝑐𝐴/𝑐$y_A=c_A/c$, following topic 1; the source uses 𝑥𝐴$x_A$ for gas and liquid alike. The binary law is

𝐉𝐴=−𝑐𝐷𝐴𝐵∇𝑦𝐴=−𝐷𝐴𝐵∇𝑐𝐴.$$\mathbf J_A=-cD_{AB}\nabla y_A=-D_{AB}\nabla c_A.$$

Expanding Eq. 5 and summing it gives

∇⋅𝐯∗=∑𝐴𝑅𝐴𝑐∂𝑐𝐴∂𝑡+𝐯∗⋅∇𝑐𝐴=𝐷𝐴𝐵∇2𝑐𝐴+𝑅𝐴−𝑦𝐴∑𝐴𝑅𝐴$$\begin{aligned} \nabla\cdot\mathbf v^*&=\frac{\sum_A R_A}{c}\\ \frac{\partial c_A}{\partial t}+\mathbf v^*\cdot\nabla c_A &=D_{AB}\nabla^2c_A+R_A-y_A\sum_A R_A \end{aligned}$$(8)

The simpler gas convection-diffusion equation applies when ∑𝐴𝑅𝐴 =0$\sum_A R_A=0$, including a nonreacting bulk gas. For a mole-changing homogeneous reaction, retain the additional term or justify its neglect by a suitable dilute approximation. All gas films below have no homogeneous reaction; their surface reaction sets a boundary condition. This distinction avoids assuming gas volume conservation during a mole-changing reaction. Pressure 𝑃$P$ is in Pa, 𝑇$T$ in K and ℛ =8314 Pa m3 kmol−1 K−1$\mathcal R=8314\ \mathrm{Pa\,m^3\,kmol^{-1}\,K^{-1}}$ when 𝑐$c$ is in kmol m⁻³.

With no flow and no reaction, the constant-property equation reduces to Fick’s second law,

∂𝑐𝐴∂𝑡=𝐷𝐴𝐵∇2𝑐𝐴$$\frac{\partial c_A}{\partial t}=D_{AB}\nabla^2c_A$$(9)

An ordinary chemical reaction conserves total mass. Must it also conserve the total number of moles?

• Yes, for every reaction.
• Yes, whenever temperature is constant.
• No, because chemical reaction can destroy total mass.
• No. A reaction can change the total number of moles while conserving mass. ✓

Explanation

A reaction conserves mass, but can change the total number of moles. For example, one mole of nitrogen and three moles of hydrogen form two moles of ammonia. The mass is unchanged even though four moles become two.

4.4 Curved coordinates

The divergence changes with geometry, while the conservation statement does not. With cylindrical coordinates 𝑟,𝜃,𝑧$r,\theta,z$, radial distance 𝑟$r$ in m and angle 𝜃$\theta$ in radians,

∂𝑐𝐴∂𝑡=−1𝑟∂(𝑟𝑁𝐴𝑟)∂𝑟−1𝑟∂𝑁𝐴𝜃∂𝜃−∂𝑁𝐴𝑧∂𝑧+𝑅𝐴$$\frac{\partial c_A}{\partial t} =-\frac1r\frac{\partial(rN_{Ar})}{\partial r} -\frac1r\frac{\partial N_{A\theta}}{\partial\theta} -\frac{\partial N_{Az}}{\partial z}+R_A$$(10)

With spherical coordinates 𝑟,𝜃,𝜙$r,\theta,\phi$, where 𝜃$\theta$ is the polar angle and 𝜙$\phi$ the azimuth,

∂𝑐𝐴∂𝑡=−1𝑟2∂(𝑟2𝑁𝐴𝑟)∂𝑟−1𝑟sin⁡𝜃∂(𝑁𝐴𝜃sin⁡𝜃)∂𝜃−1𝑟sin⁡𝜃∂𝑁𝐴𝜙∂𝜙+𝑅𝐴$$\frac{\partial c_A}{\partial t} =-\frac1{r^2}\frac{\partial(r^2N_{Ar})}{\partial r} -\frac1{r\sin\theta}\frac{\partial(N_{A\theta}\sin\theta)}{\partial\theta} -\frac1{r\sin\theta}\frac{\partial N_{A\phi}}{\partial\phi}+R_A$$(11)

Cylindrical radial area grows as 𝑟$r$; spherical radial area grows as 𝑟2$r^2$. Thus a steady radial problem with no volumetric reaction conserves 𝑟𝑁𝐴𝑟$rN_{Ar}$ in a cylinder and 𝑟2𝑁𝐴𝑟$r^2N_{Ar}$ in a sphere. The local flux need not be constant even though the total rate is.

A cylindrical wall carries a steady outward species rate with no reaction or end effects. What happens to the local radial flux when radius doubles within the wall?

• It doubles.
• It halves. ✓
• It stays the same.
• It falls to one quarter.

Explanation

The same molar rate crosses each cylindrical surface:

˙𝑛𝐴=2𝜋𝑟𝐿𝑁𝐴𝑟$$\dot n_A=2\pi rL N_{Ar}$$

Doubling radius doubles the area. Therefore,

𝑁𝐴𝑟(2𝑟)𝑁𝐴𝑟(𝑟)=12$$\frac{N_{Ar}(2r)}{N_{Ar}(r)}=\frac{1}{2}$$

The local flux halves.

5 Case 1: diffusion of A through non-diffusing B

In a steady plane gas film with variation only in 𝑧$z$, no accumulation, transverse gradients or bulk reaction remain in Eq. 5. Therefore 𝑑𝑁𝐴𝑧/𝑑𝑧 =0$dN_{Az}/dz=0$. Binary total flux combines diffusion and molar drift,

𝑁𝐴𝑧=−𝑐𝐷𝐴𝐵𝑑𝑦𝐴𝑑𝑧+𝑦𝐴(𝑁𝐴𝑧+𝑁𝐵𝑧)$$N_{Az}=-cD_{AB}\frac{dy_A}{dz}+y_A(N_{Az}+N_{Bz})$$

With 𝑁𝐵𝑧 =0$N_{Bz}=0$, rearranging gives

𝑁𝐴𝑧=−𝑐𝐷𝐴𝐵1−𝑦𝐴𝑑𝑦𝐴𝑑𝑧$$N_{Az}=-\frac{cD_{AB}}{1-y_A}\frac{dy_A}{dz}$$(12)

Integrate the constant flux between 𝑧1$z_1$ and 𝑧2$z_2$, where 𝑦𝐴 =𝑦𝐴1$y_A=y_{A1}$ and 𝑦𝐴2$y_{A2}$. Separating 𝑑𝑦𝐴/(1 −𝑦𝐴)$dy_A/(1-y_A)$ gives

𝑁𝐴𝑧=𝑐𝐷𝐴𝐵𝑧2−𝑧1ln⁡1−𝑦𝐴21−𝑦𝐴1$$N_{Az}=\frac{cD_{AB}}{z_2-z_1}\ln\frac{1-y_{A2}}{1-y_{A1}}$$(13)

This reproduces the shell-balance result in topic 1 by deleting terms from the general equation. A heterogeneous reaction still leaves 𝑅𝐴 =0$R_A=0$ in the film. The source uses the same method to extend the geometry.

Figure 2: Radial diffusion through a stationary cylindrical wall and from a subliming sphere into still air.

Cylindrical wall

A given gas A can diffuse through the solid wall of a tube in which it flows. Find the rate of gas diffusion in terms of the tube dimensions (inside wall radius 𝑅𝑖$R_i$, outside wall radius 𝑅𝑜$R_o$), the gas-solid diffusivity, and the gas concentrations at the walls.

Assumptions:

• steady state
• flux in 𝑟$r$ direction only
• no chemical reaction
• A concentration in gas is quite dilute →𝑥𝐴 ∼0$\rightarrow x_A\sim0$
• Constant 𝐷𝐴𝐵$D_{AB}$

Solution

The radial shell in Figure 3 shows why a cylindrical balance must account for the changing area.

Figure 3: A stationary cylindrical wall and an enlarged radial shell. The molar rate equals the local flux multiplied by the cylindrical area 2πrL.

Data

Quantity	Value
Inner wall radius	𝑅𝑖$R_i$
Outer wall radius	𝑅𝑜$R_o$
Axial length	𝐿$L$
Constant gas-solid diffusivity	𝐷𝐴𝐵$D_{AB}$
Dissolved concentration at the inner face	𝑐𝐴𝑖$c_{Ai}$ in kmol m−3$\mathrm{kmol\,m^{-3}}$
Dissolved concentration at the outer face	𝑐𝐴𝑜$c_{Ao}$ in kmol m−3$\mathrm{kmol\,m^{-3}}$

Assumptions

• Steady state.
• Radial diffusion only.
• No chemical reaction.
• Dilute dissolved A in a stationary solid matrix.
• Constant diffusivity.
• Negligible end effects.

Calculations

Choose positive flux radially outwards. A steady shell has no accumulation or reaction, so its inward and outward molar rates are equal. From Eq. 10, the governing radial balance is

𝑑𝑑𝑟(𝑟𝑁𝐴𝑟)=0$$\frac{d}{dr}(rN_{Ar})=0$$

Consequently, the total outward molar rate is independent of radius,

˙𝑛𝐴=2𝜋𝑟𝐿𝑁𝐴𝑟$$\dot n_A=2\pi rL N_{Ar}$$

The source denotes this rate by 𝑊𝐴$W_A$. In a stationary solid, transport of dilute dissolved A follows Fick’s law,

𝑁𝐴𝑟=−𝐷𝐴𝐵𝑑𝑐𝐴𝑑𝑟$$N_{Ar}=-D_{AB}\frac{dc_A}{dr}$$

Substitute the flux into the rate expression, separate the variables and integrate between the two wall faces,

˙𝑛𝐴∫𝑅𝑜𝑅𝑖𝑑𝑟𝑟=−2𝜋𝐿𝐷𝐴𝐵∫𝑐𝐴𝑜𝑐𝐴𝑖𝑑𝑐𝐴˙𝑛𝐴ln⁡𝑅𝑜𝑅𝑖=2𝜋𝐿𝐷𝐴𝐵(𝑐𝐴𝑖−𝑐𝐴𝑜)$$\begin{aligned} \dot n_A\int_{R_i}^{R_o}\frac{dr}{r} &=-2\pi LD_{AB}\int_{c_{Ai}}^{c_{Ao}}dc_A\\ \dot n_A\ln\frac{R_o}{R_i} &=2\pi LD_{AB}(c_{Ai}-c_{Ao}) \end{aligned}$$

Solving for the rate gives

˙𝑛𝐴=2𝜋𝐿𝐷𝐴𝐵(𝑐𝐴𝑖−𝑐𝐴𝑜)ln⁡(𝑅𝑜𝑅𝑖)$$\dot n_A=\frac{2\pi LD_{AB}(c_{Ai}-c_{Ao})}{\ln\left(\frac{R_o}{R_i}\right)}$$(14)

The concentration profile follows from the same differential equation. Since 𝑟 𝑑𝑐𝐴/𝑑𝑟$r\,dc_A/dr$ is constant, integration from 𝑅𝑖$R_i$ to an arbitrary radius 𝑟$r$ gives

𝑐𝐴(𝑟)−𝑐𝐴𝑖=𝐶1ln⁡(𝑟𝑅𝑖)$$c_A(r)-c_{Ai}=C_1\ln\left(\frac{r}{R_i}\right)$$

At 𝑟 =𝑅𝑜$r=R_o$, impose 𝑐𝐴 =𝑐𝐴𝑜$c_A=c_{Ao}$ to obtain

𝐶1=𝑐𝐴𝑜−𝑐𝐴𝑖ln⁡(𝑅𝑜𝑅𝑖).$$C_1=\frac{c_{Ao}-c_{Ai}}{\ln\left(\frac{R_o}{R_i}\right)}.$$

Therefore

𝑐𝐴−𝑐𝐴𝑖𝑐𝐴𝑜−𝑐𝐴𝑖=ln⁡(𝑟𝑅𝑖)ln⁡(𝑅𝑜𝑅𝑖)$$\frac{c_A-c_{Ai}}{c_{Ao}-c_{Ai}} =\frac{\ln\left(\frac{r}{R_i}\right)}{\ln\left(\frac{R_o}{R_i}\right)}$$(15)

The profile recovers both specified wall concentrations. When 𝑐𝐴𝑖 >𝑐𝐴𝑜$c_{Ai}>c_{Ao}$, its gradient is negative and the rate is positive outwards. The units of Eq. 14 are

m (m2 s−1)(kmol m−3)=kmol s−1.$$\mathrm m\,(\mathrm{m^2\,s^{-1}})(\mathrm{kmol\,m^{-3}})=\mathrm{kmol\,s^{-1}}.$$

The concentration-based resistance is

ln⁡(𝑅𝑜𝑅𝑖)2𝜋𝐿𝐷𝐴𝐵,$$\frac{\ln\left(\frac{R_o}{R_i}\right)}{2\pi LD_{AB}},$$

in s m−3$\mathrm{s\,m^{-3}}$. Define the log-mean area,

𝐴lm=2𝜋𝐿(𝑅𝑜−𝑅𝑖)ln⁡(𝑅𝑜𝑅𝑖)$$A_{\mathrm{lm}}=\frac{2\pi L(R_o-R_i)}{\ln\left(\frac{R_o}{R_i}\right)}$$

The rate can then be written as

˙𝑛𝐴=𝐷𝐴𝐵𝐴lm(𝑐𝐴𝑖−𝑐𝐴𝑜)𝑅𝑜−𝑅𝑖,$$\dot n_A=\frac{D_{AB}A_{\mathrm{lm}}(c_{Ai}-c_{Ao})}{R_o-R_i},$$

just as cylindrical conduction can be written using a log-mean area.

The specified concentrations must be concentrations dissolved in the solid. A bulk gas concentration requires a solubility relation before it can be used as a wall boundary condition. For example, 𝑐𝐴 =𝑆𝐴𝑝𝐴$c_A=S_Ap_A$, with 𝑆𝐴$S_A$ in kmol m−3 Pa−1$\mathrm{kmol\,m^{-3}\,Pa^{-1}}$, gives the permeability 𝐷𝐴𝐵𝑆𝐴$D_{AB}S_A$.

Answer

The outward molar rate is

˙𝑛𝐴=2𝜋𝐿𝐷𝐴𝐵(𝑐𝐴𝑖−𝑐𝐴𝑜)ln⁡(𝑅𝑜𝑅𝑖).$$\dot n_A=\frac{2\pi LD_{AB}(c_{Ai}-c_{Ao})}{\ln\left(\frac{R_o}{R_i}\right)}.$$

It is positive when the dissolved concentration is greater at the inner wall. The logarithmic concentration profile is given by Eq. 15.

Naphthalene sphere

A spherical ball of solid, nonporous naphthalene is suspended in still air. The naphthalene ball slowly sublimes, releasing naphthalene into the surrounding air. It may be assumed that the sublimation occurs as a molecular diffusion-limited process. Through consideration of the diffusion equation, estimate the time required to reduce the diameter of the ball from 2 to 0.5 cm when the surrounding air is at 347 K and 1 atm.

Data for naphthalene:

• MW = 128 g/mol
• Density = 1.145 g/cm3$^3$
• Diffusivity in air = 8.19 ×10−6 m2/s$8.19\times10^{-6}\ \mathrm{m^2/s}$
• Vapour pressure = 666 Pa at 347 K

Solution

The spherical geometry in Figure 2 applies at each instantaneous ball radius. Vapour diffuses outwards through air while the solid surface slowly recedes.

Data

Quantity	Value
Temperature	𝑇 =347 K$T=347\ \mathrm K$
Pressure	𝑃 =1 atm =101325 Pa$P=1\ \mathrm{atm}=101325\ \mathrm{Pa}$
Diffusivity	𝐷𝐴𝐵 =8.19 ×10−6 m2 s−1$D_{AB}=8.19\times10^{-6}\ \mathrm{m^2\,s^{-1}}$
Saturation pressure	𝑝sat𝐴 =666 Pa$p_A^{\mathrm{sat}}=666\ \mathrm{Pa}$
Solid density	𝜌𝑠 =1145 kg m−3$\rho_s=1145\ \mathrm{kg\,m^{-3}}$
Molar mass	𝑀𝐴 =128 kg kmol−1$M_A=128\ \mathrm{kg\,kmol^{-1}}$
Initial radius	𝑅𝑖 =0.010 m$R_i=0.010\ \mathrm m$
Final radius	𝑅𝑓 =0.0025 m$R_f=0.0025\ \mathrm m$

The supplied dimensions are diameters. The table gives the corresponding radii.

Assumptions

• Spherically symmetric diffusion.
• Pseudo-steady gas profile as the surface recedes.
• Isothermal sublimation at 347 K$347\ \mathrm K$.
• Ideal gas behaviour.
• Constant pressure, diffusivity and solid density.
• Vapour-solid equilibrium at the surface.
• Negligible naphthalene concentration far from the ball.
• Zero net air flux through the solid surface.
• Negligible natural convection and gas-phase reaction.

Calculations

From Eq. 11 at a fixed instantaneous radius, the steady radial balance gives

𝑑(𝑟2𝑁𝐴𝑟)𝑑𝑟=0.$$\frac{d(r^2N_{Ar})}{dr}=0.$$

The same total rate crosses each concentric sphere,

˙𝑛𝐴=4𝜋𝑟2𝑁𝐴𝑟$$\dot n_A=4\pi r^2N_{Ar}$$

Air cannot cross the solid surface, so 𝑁𝐵𝑟 =0$N_{Br}=0$. Writing Eq. 12 in radial form retains the resulting molar flow,

𝑁𝐴𝑟=−𝑐𝐷𝐴𝐵1−𝑦𝐴𝑑𝑦𝐴𝑑𝑟$$N_{Ar}=-\frac{cD_{AB}}{1-y_A}\frac{dy_A}{dr}$$

Substituting the rate and integrating between radii 𝑅1$R_1$ and 𝑅2$R_2$ gives

˙𝑛𝐴4𝜋𝑐𝐷𝐴𝐵∫𝑅2𝑅1𝑑𝑟𝑟2=−∫𝑦𝐴2𝑦𝐴1𝑑𝑦𝐴1−𝑦𝐴$$\frac{\dot n_A}{4\pi cD_{AB}}\int_{R_1}^{R_2}\frac{dr}{r^2} =-\int_{y_{A1}}^{y_{A2}}\frac{dy_A}{1-y_A}$$

Evaluating both integrals and solving for the rate gives

˙𝑛𝐴=4𝜋𝑐𝐷𝐴𝐵1𝑅1−1𝑅2ln⁡1−𝑦𝐴21−𝑦𝐴1$$\dot n_A=\frac{4\pi cD_{AB}}{\frac{1}{R_1}-\frac{1}{R_2}}\ln\frac{1-y_{A2}}{1-y_{A1}}$$(16)

For an effectively infinite surrounding gas, take 𝑅1 =𝑅$R_1=R$, 𝑅2 →∞$R_2\to\infty$ and 𝑦𝐴2 =0$y_{A2}=0$. Surface equilibrium and the ideal gas relation give

𝑦𝐴𝑠=𝑝sat𝐴𝑃=666101325=0.00657291$$y_{As}=\frac{p_A^{\mathrm{sat}}}{P}=\frac{666}{101325}=0.00657291$$𝑐=𝑃𝑅𝑢𝑇=1013258314(347)=0.0351218 kmol m−3$$c=\frac{P}{R_uT}=\frac{101325}{8314(347)}=0.0351218\ \mathrm{kmol\,m^{-3}}$$

Define the constant composition factor,

𝐹=ln⁡11−𝑦𝐴𝑠=0.00659461$$F=\ln\frac{1}{1-y_{As}}=0.00659461$$

The instantaneous sublimation rate is then ˙𝑛𝐴 =4𝜋𝑅𝑐𝐷𝐴𝐵𝐹$\dot n_A=4\pi RcD_{AB}F$. The sphere loses moles as its radius decreases. Differentiating its solid inventory gives

˙𝑛𝐴=−𝑑𝑑𝑡(𝜌𝑠𝑀𝐴4𝜋𝑅33)=−𝜌𝑠𝑀𝐴4𝜋𝑅2𝑑𝑅𝑑𝑡$$\dot n_A=-\frac{d}{dt}\left(\frac{\rho_s}{M_A}\frac{4\pi R^3}{3}\right) =-\frac{\rho_s}{M_A}4\pi R^2\frac{dR}{dt}$$

Equate surface loss to diffusion into the gas. After cancelling 4𝜋𝑅$4\pi R$ and separating variables,

𝑅 𝑑𝑅=−𝑀𝐴𝑐𝐷𝐴𝐵𝐹𝜌𝑠 𝑑𝑡∫𝑅𝑓𝑅𝑖𝑅 𝑑𝑅=−𝑀𝐴𝑐𝐷𝐴𝐵𝐹𝜌𝑠∫𝑡0𝑑𝑡𝑡=𝜌𝑠(𝑅2𝑖−𝑅2𝑓)2𝑀𝐴𝑐𝐷𝐴𝐵𝐹$$\begin{aligned} R\,dR&=-\frac{M_AcD_{AB}F}{\rho_s}\,dt\\ \int_{R_i}^{R_f}R\,dR&=-\frac{M_AcD_{AB}F}{\rho_s}\int_0^t dt\\ t&=\frac{\rho_s(R_i^2-R_f^2)}{2M_AcD_{AB}F} \end{aligned}$$(17)

Substitute the radii and properties on a consistent kmol basis,

𝑡=1145(0.0102−0.00252)2(128)(0.0351218)(8.19×10−6)(0.00659461)=2.210×105 s=61.40 h$$t=\frac{1145(0.010^2-0.0025^2)} {2(128)(0.0351218)(8.19\times10^{-6})(0.00659461)} =2.210\times10^5\ \mathrm s=61.40\ \mathrm h$$(18)

The gas adjustment time is

𝑅2𝑖𝐷𝐴𝐵=0.01028.19×10−6=12.2 s,$$\frac{R_i^2}{D_{AB}}=\frac{0.010^2}{8.19\times10^{-6}}=12.2\ \mathrm s,$$

much shorter than the sublimation time. This supports a pseudo-steady gas profile. Replacing 𝐹$F$ by the dilute approximation 𝑦𝐴𝑠$y_{As}$ gives 61.60 h, only 0.33 percent higher.

The integrated balance can also be expressed as a diameter-square law,

𝑑2=𝑑2𝑖−𝐾𝑡$$d^2=d_i^2-Kt$$𝐾=8𝑀𝐴𝑐𝐷𝐴𝐵𝐹𝜌𝑠$$K=\frac{8M_AcD_{AB}F}{\rho_s}$$

Here 𝐾$K$ has units m2 s−1$\mathrm{m^2\,s^{-1}}$. The composition factor must remain in 𝐾$K$ because the vapour pressure determines the driving force. A different thermal model would be needed if sublimation cooling changes the surface temperature appreciably.

Answer

The diffusion-limited model predicts 𝑡 =2.21 ×105 s$t=2.21\times10^5\ \mathrm s$, or about 61.4 h, for the diameter to decrease from 2 cm to 0.5 cm.

6 Case 2: diffusion with a surface reaction

6.1 A surface reaction and flux stoichiometry

Heterogeneous reaction

A reaction at an interface between phases, such as a gas reacting at a catalyst surface. It supplies a species-flux boundary condition.

A gas film of thickness 𝛿$\delta$ surrounds a catalyst. Coordinate 𝑧$z$ increases from the bulk-film edge at 𝑧 =0$z=0$ toward the surface at 𝑧 =𝛿$z=\delta$. Bulk A has fraction 𝑦𝐴0$y_{A0}$. At the surface, the irreversible instantaneous reaction 2A →B$\mathrm{2A\to B}$ consumes two moles of A for each mole of B returning through the film. There is no homogeneous reaction in the gas. Constant 𝑇$T$, 𝑃$P$ and 𝐷𝐴𝐵$D_{AB}$ make 𝑐$c$ and diffusivity constant; this case need not be dilute.

Figure 4: Molar fluxes through a nonreacting gas film toward a catalyst where two moles of A form one mole of B. Positive z points toward the catalyst. The outward B flux has half the magnitude of the inward A flux, leaving a net molar flux toward the surface.

The stoichiometric boundary fixes 𝑁𝐵𝑧 = −𝑁𝐴𝑧/2$N_{Bz}=-N_{Az}/2$. Steady nonreacting gas conserves both fluxes throughout the film. Substitute the ratio into the total-flux law,

𝑁𝐴𝑧=−𝑐𝐷𝐴𝐵𝑑𝑦𝐴𝑑𝑧+12𝑦𝐴𝑁𝐴𝑧𝑁𝐴𝑧=−𝑐𝐷𝐴𝐵1−𝑦𝐴2𝑑𝑦𝐴𝑑𝑧$$\begin{aligned} N_{Az}&=-cD_{AB}\frac{dy_A}{dz}+\frac12y_AN_{Az}\\ N_{Az}&=-\frac{cD_{AB}}{1-\frac{y_A}{2}}\frac{dy_A}{dz} \end{aligned}$$(19)

The net flux 𝑁𝐴𝑧 +𝑁𝐵𝑧 =𝑁𝐴𝑧/2$N_{Az}+N_{Bz}=N_{Az}/2$ points toward the catalyst. Molar drift carries A along with diffusion. More generally 𝑁𝐵 = −𝛽𝑁𝐴$N_B=-\beta N_A$ gives denominator 1 −(1 −𝛽)𝑦𝐴$1-(1-\beta)y_A$. Equimolar counter-diffusion has 𝛽 =1$\beta=1$, stagnant B has 𝛽 =0$\beta=0$, and this reaction has 𝛽 =1/2$\beta=1/2$.

At a catalyst, 2𝐴 →𝐵$2A\to B$. Positive flux points toward the catalyst. If 𝑁𝐴 >0$N_A>0$, which flux relation holds in the steady nonreacting gas film?

• 𝑁𝐵 = −𝑁𝐴/2$N_B=-N_A/2$. ✓
• 𝑁𝐵 = −𝑁𝐴$N_B=-N_A$.
• 𝑁𝐵 =0$N_B=0$.
• 𝑁𝐵 =2𝑁𝐴$N_B=2N_A$.

Explanation

For every two moles of A consumed, one mole of B leaves the catalyst. Since positive flux points toward the catalyst,

𝑁𝐵=−𝑁𝐴2$$N_B=-\frac{N_A}{2}$$

The negative sign means that B moves away from the catalyst. The net molar flux is

𝑁𝐴+𝑁𝐵=𝑁𝐴2$$N_A+N_B=\frac{N_A}{2}$$

It points toward the catalyst.

6.2 Integration and instantaneous boundary

Since 𝑑𝑁𝐴𝑧/𝑑𝑧 =0$dN_{Az}/dz=0$, substitute Eq. 19 and remove constant 𝑐𝐷𝐴𝐵$cD_{AB}$. Integrate twice,

𝑑𝑑𝑧(11−𝑦𝐴2𝑑𝑦𝐴𝑑𝑧)=011−𝑦𝐴2𝑑𝑦𝐴𝑑𝑧=𝐶1−2ln⁡(1−𝑦𝐴2)=𝐶1𝑧+𝐶2$$\begin{aligned} \frac{d}{dz}\left(\frac1{1-\frac{y_A}{2}}\frac{dy_A}{dz}\right)&=0\\ \frac1{1-\frac{y_A}{2}}\frac{dy_A}{dz}&=C_1\\ -2\ln\left(1-\frac{y_A}{2}\right)&=C_1z+C_2 \end{aligned}$$(20)

Here 𝐶1$C_1$ is in m⁻¹ and 𝐶2$C_2$ dimensionless. The factor two comes from the derivative of 1 −𝑦𝐴/2$1-y_A/2$. The source rewrites this as

1−𝑦𝐴2=𝐾𝑧1𝐾2.$$1-\frac{y_A}{2}=K_1^zK_2.$$

We use a dimensionless coordinate ratio instead of a dimensional exponent. Instantaneous reaction leaves 𝑦𝐴(𝛿) =0$y_A(\delta)=0$, while bulk supply sets 𝑦𝐴(0) =𝑦𝐴0$y_A(0)=y_{A0}$. Thus

𝐶2=−2ln⁡(1−𝑦𝐴02)$$C_2=-2\ln\left(1-\frac{y_{A0}}{2}\right)$$

and 𝐶1 = −𝐶2/𝛿$C_1=-C_2/\delta$. Substitution gives

1−𝑦𝐴(𝑧)2=(1−𝑦𝐴02)1−𝑧𝛿𝑁𝐴𝑧=2𝑐𝐷𝐴𝐵𝛿ln⁡11−𝑦𝐴02$$\begin{aligned} 1-\frac{y_A(z)}2&=\left(1-\frac{y_{A0}}{2}\right)^{1-\frac{z}{\delta}}\\ N_{Az}&=\frac{2cD_{AB}}{\delta}\ln\frac1{1-\frac{y_{A0}}{2}} \end{aligned}$$(21)

Both endpoints hold. The flux is positive toward the catalyst. At small 𝑦𝐴0$y_{A0}$ the logarithm is approximately 𝑦𝐴0/2$y_{A0}/2$, so

𝑁𝐴𝑧≈𝑐𝐷𝐴𝐵𝑦𝐴0𝛿,$$N_{Az}\approx\frac{cD_{AB}y_{A0}}{\delta},$$

the dilute Fick result. At bulk fractions 0.2 and 0.8 the exact rate exceeds that approximation by about 5.36 and 27.7 percent respectively. Complementary 𝑦𝐵 =1 −𝑦𝐴$y_B=1-y_A$ ensures binary fractions sum to one. The film is a model of a continuous boundary layer; 𝛿$\delta$ is an equivalent thickness determined by flow, rather than a sharp physical boundary. Thinning it increases rate per area. Increasing total catalyst area can also increase total rate, even when a faster intrinsic rate per area cannot.

6.3 Finite surface kinetics

If reaction is finite, the A-consumption law is

𝑁𝐴𝑧=𝑘′′1𝑐𝐴𝑠=𝑘′′1𝑐𝑦𝐴𝑠$$N_{Az}=k_1''c_{As}=k_1''cy_{As}$$

at the surface. Surface coefficient 𝑘′′1$k_1''$ is in m s⁻¹ and 𝑐𝐴𝑠$c_{As}$ is the unknown dissolved or gas surface concentration appropriate to this model. The exact film equation is

𝑁𝐴𝑧=2𝑐𝐷𝐴𝐵𝛿ln⁡1−𝑦𝐴𝑠21−𝑦𝐴02𝑦𝐴𝑠=𝑁𝐴𝑧𝑘′′1𝑐$$\begin{aligned} N_{Az}&=\frac{2cD_{AB}}{\delta} \ln\frac{1-\frac{y_{As}}{2}}{1-\frac{y_{A0}}{2}}\\ y_{As}&=\frac{N_{Az}}{k_1''c} \end{aligned}$$(22)

Solve these together; a finite surface concentration is allowed. In the dilute limit define 𝑘𝑐 =𝐷𝐴𝐵/𝛿$k_c=D_{AB}/\delta$ in m s⁻¹. Equating 𝑘𝑐(𝑐𝐴0 −𝑐𝐴𝑠)$k_c(c_{A0}-c_{As})$ to 𝑘′′1𝑐𝐴𝑠$k_1''c_{As}$ yields

𝑐𝐴𝑠=𝑘𝑐𝑐𝐴0𝑘𝑐+𝑘′′1𝑁𝐴=𝑐𝐴01𝑘𝑐+1𝑘′′1$$\begin{aligned} c_{As}&=\frac{k_c c_{A0}}{k_c+k_1''}\\ N_A&=\frac{c_{A0}}{\frac{1}{k_c}+\frac{1}{k_1''}} \end{aligned}$$(23)

Damköhler number

A dimensionless ratio of reaction rate to transport rate. For this first-order surface reaction and gas film,

Da=𝑘′′1𝑘𝑐=𝑘′′1𝛿𝐷𝐴𝐵.$$\mathrm{Da}=\frac{k_1''}{k_c}=\frac{k_1''\delta}{D_{AB}}.$$

Large Da$\mathrm{Da}$ means diffusion controls; small Da$\mathrm{Da}$ means surface reaction controls.

For this film,

Da=𝑘′′1𝛿𝐷𝐴𝐵$$\mathrm{Da}=\frac{k_1''\delta}{D_{AB}}$$

Both resistances have units s m⁻¹. Their relative sizes identify the useful limiting model:

• For Da <0.1$\mathrm{Da}<0.1$, surface reaction resistance exceeds film resistance by more than tenfold. The surface concentration is close to the bulk concentration, and faster kinetics increases the rate.
• For 0.1 ≤Da ≤10$0.1\le\mathrm{Da}\le10$, both resistances can affect the rate. Retain the film balance and the surface reaction law.
• For Da >10$\mathrm{Da}>10$, film resistance exceeds surface reaction resistance by more than tenfold. A nearly zero surface concentration is a useful limiting approximation, and improving film transport increases the rate.

The cutoffs mark resistance ratios, rather than sharp changes in behaviour. Neglecting the smaller resistance at a ratio of ten overpredicts the overall rate by ten percent in this linear model.

For 𝑘′′1 =10−3$k_1''=10^{-3}$ and 𝑘𝑐 =10−5 m s−1$k_c=10^{-5}\ \mathrm{m\,s^{-1}}$, Da =100$\mathrm{Da}=100$ and 𝑐𝐴𝑠/𝑐𝐴0 =1/101$c_{As}/c_{A0}=1/101$. Increasing transport is more effective than increasing an already fast surface reaction. If the ratio is 0.01 instead, the surface concentration is about 99 percent of the bulk value and kinetics controls. This definition is for a first-order surface coefficient in m s⁻¹; inserting the homogeneous coefficient in s−1$^{-1}$ would leave a dimensional quantity. The homogeneous comparison below uses a volumetric reaction coefficient and its own diffusion scale.

A dilute film model has surface Damköhler number Da =𝑘′′1/𝑘𝑐 =100$\mathrm{Da}=k_1^{\prime\prime}/k_c=100$. Which change most directly increases the rate when the bulk concentration is held fixed?

• Increase an already very fast surface reaction coefficient alone.
• Make the diffusion film thicker.
• Increase the film mass transfer coefficient. ✓
• Reduce the bulk concentration.

Explanation

The ratio of the surface reaction coefficient to the film coefficient is

Da=𝑘′′1𝑘𝑐=100$$\mathrm{Da}=\frac{k_1^{\prime\prime}}{k_c}=100$$

Reaction is much faster than transport through the film. The uptake is approximately

𝑁𝐴≃𝑘𝑐𝑐𝐴0$$N_A\simeq k_c c_{A0}$$

Increasing the film coefficient increases the rate. Making an already fast surface reaction faster has little effect.

7 Case 3: diffusion with a zero-order homogeneous reaction

7.1 Zero-order consumption

Homogeneous reaction

A reaction occurring within a phase volume. Its production or consumption rate appears as a volumetric source in the species balance.

Gas A dissolves into liquid B at an interface and diffuses down a pool of depth 𝐿$L$. Reaction A +B →AB$\mathrm{A+B\to AB}$ consumes A throughout the liquid. Treat A and product AB as dilute, the solvent as stationary, and 𝑇$T$ and 𝐷𝐴𝐵$D_{AB}$ as constant. Let 𝑥$x$ measure depth from the interface. The bottom is impermeable. Zero-order consumption has 𝑅𝐴 = −𝑘0$R_A=-k_0$, with positive 𝑘0$k_0$ in kmol m⁻³ s⁻¹, independent of concentration within the range where the kinetics remain valid.

Figure 5: A dissolves at the upper surface of a stationary liquid layer and reacts throughout its depth. The bottom is impermeable.

Steady, one-dimensional continuity is

−𝑑𝑁𝐴𝑥𝑑𝑥+𝑅𝐴=0.$$-\frac{dN_{Ax}}{dx}+R_A=0.$$

Dilution and absence of solvent flow give

𝑁𝐴𝑥=−𝐷𝐴𝐵𝑑𝑐𝐴𝑑𝑥.$$N_{Ax}=-D_{AB}\frac{dc_A}{dx}.$$

Substituting 𝑅𝐴 = −𝑘0$R_A=-k_0$ therefore gives

𝐷𝐴𝐵𝑑2𝑐𝐴𝑑𝑥2−𝑘0=0$$D_{AB}\frac{d^2c_A}{dx^2}-k_0=0$$(24)

Both terms are kmol m⁻³ s⁻¹. Integrate twice,

𝑑𝑐𝐴𝑑𝑥=𝑘0𝐷𝐴𝐵𝑥+𝑎𝑐𝐴=𝑘02𝐷𝐴𝐵𝑥2+𝑎𝑥+𝑏$$\begin{aligned} \frac{dc_A}{dx}&=\frac{k_0}{D_{AB}}x+a\\ c_A&=\frac{k_0}{2D_{AB}}x^2+ax+b \end{aligned}$$(25)

The interface sets 𝑐𝐴(0) =𝑐𝐴0$c_A(0)=c_{A0}$ by gas-liquid equilibrium, for example a measured Henry solubility at the given temperature. An impermeable bottom sets 𝑁𝐴𝑥(𝐿) =0$N_{Ax}(L)=0$, equivalent here to

[𝑑𝑐𝐴𝑑𝑥]𝑥=𝐿=0.$$\left[\frac{dc_A}{dx}\right]_{x=L}=0.$$

Thus 𝑎 = −𝑘0𝐿/𝐷𝐴𝐵$a=-k_0L/D_{AB}$ in kmol m⁻⁴ and 𝑏 =𝑐𝐴0$b=c_{A0}$ in kmol m⁻³. Use 𝑥$x$ for depth in both boundary conditions. The result and flux are

𝑐𝐴=𝑐𝐴0+𝑘02𝐷𝐴𝐵(𝑥2−2𝐿𝑥)𝑁𝐴𝑥(𝑥)=𝑘0(𝐿−𝑥)𝑁𝐴𝑥(0)=𝑘0𝐿$$\begin{aligned} c_A&=c_{A0}+\frac{k_0}{2D_{AB}}(x^2-2Lx)\\ N_{Ax}(x)&=k_0(L-x)\\ N_{Ax}(0)&=k_0L \end{aligned}$$(26)

Concentration decreases with depth and its gradient vanishes at the bottom. Surface influx equals the integrated consumption 𝑘0𝐿$k_0L$, independently checking the balance. The bottom concentration is

𝑐𝐴0−𝑘0𝐿22𝐷𝐴𝐵.$$c_{A0}-\frac{k_0L^2}{2D_{AB}}.$$

This full-depth solution requires

𝑘0𝐿22𝐷𝐴𝐵𝑐𝐴0≤1.$$\frac{k_0L^2}{2D_{AB}c_{A0}}\le1.$$

Equality puts zero concentration exactly at the bottom. Beyond it, negative concentrations are unphysical. A consumes completely at

𝑥∗=√2𝐷𝐴𝐵𝑐𝐴0𝑘0<𝐿;$$x^*=\sqrt{\frac{2D_{AB}c_{A0}}{k_0}}<L;$$

impose both 𝑐𝐴 =0$c_A=0$ and zero flux there and stop consumption below that depth. A zero-order law cannot continue consuming an absent reactant.

The full-depth zero-order consumption solution predicts a negative concentration near the impermeable bottom. How should this result be interpreted?

• The full-depth solution remains valid because the overall balance closes.
• The reactant has run out in part of the liquid, so the model must stop consumption there. ✓
• A first-order rate law can be substituted without changing the solution.
• The impermeable-bottom condition requires a negative bottom concentration.

Explanation

Zero-order consumption cannot continue where reactant is absent. The full-depth solution requires a nonnegative concentration throughout. Beyond its validity limit, locate a depletion depth with zero concentration and zero flux and stop consumption below it.

8 Case 4: diffusion with a first-order homogeneous reaction

8.1 First-order consumption and equation

Keep the same stationary dilute liquid, interface concentration and impermeable bottom, but use 𝑅𝐴 = −𝑘′′′1𝑐𝐴$R_A=-k_1'''c_A$. Volumetric coefficient 𝑘′′′1$k_1'''$ is in s⁻¹, distinct from surface 𝑘′′1$k_1''$ in m s⁻¹. Reaction now slows as A disappears. Substitute the rate and Fick’s law into the same steady balance,

𝐷𝐴𝐵𝑑2𝑐𝐴𝑑𝑥2−𝑘′′′1𝑐𝐴=0$$D_{AB}\frac{d^2c_A}{dx^2}-k_1'''c_A=0$$(27)

Define

𝑚=√𝑘′′′1𝐷𝐴𝐵$$m=\sqrt{\frac{k_1'''}{D_{AB}}}$$

Both terms have concentration-per-time units. Parameter 𝑚$m$ is in m⁻¹, so 1/𝑚$1/m$ is the penetration length. It follows by equating diffusion time ℓ2/𝐷𝐴𝐵$\ell^2/D_{AB}$ to reaction time 1/𝑘′′′1$1/k_1'''$. This is the fin equation with excess temperature replaced by concentration.

8.2 Solution, surface flux and limits

Effectiveness factor

The actual overall reaction rate divided by the rate if the reacting region were everywhere at the reference surface concentration under the same kinetic and temperature assumptions. It measures the effect of internal concentration gradients.

Fin efficiency

The heat transferred by a fin divided by the heat it would transfer if its entire surface were at the base temperature, with the same surface coefficient and surroundings. For a uniform straight fin with insulated tip,

𝜂𝑓=tanh⁡(𝑚𝑓𝐿)𝑚𝑓𝐿,$$\eta_f=\frac{\tanh(m_fL)}{m_fL},$$

where

𝑚𝑓=√ℎ𝑃𝑘𝑠𝐴𝑐,$$m_f=\sqrt{\frac{hP}{k_sA_c}},$$

𝑃$P$ is exposed perimeter and 𝐴𝑐$A_c$ is the conduction cross-section. It is analogous to reaction effectiveness.

Both ratios approach one for small internal gradients and fall below one as those gradients grow. The fin parameter 𝑚𝑓$m_f$ uses solid conductivity; the reaction parameter 𝑚$m$ uses species diffusivity and kinetics. Their similar equations do not make those parameters interchangeable.

Hatta number

A dimensionless measure of reaction relative to diffusion for a homogeneous reaction. These notes use the pool-depth convention

Ha=𝑚𝐿=𝐿√𝑘′′′1𝐷𝐴𝐵,$$\mathrm{Ha}=mL=L\sqrt{\frac{k_1'''}{D_{AB}}},$$

the pool depth divided by the reaction penetration length. In absorption film models the length is an equivalent film thickness instead of pool depth. Here 𝑘′′′1$k_1'''$ is a first-order volumetric coefficient in s−1$^{-1}$.

For this first-order pool model, Hatta number indicates how much of the liquid depth participates in reaction:

• For Ha <0.3$\mathrm{Ha}<0.3$, diffusion can replenish A throughout the depth faster than reaction consumes it. Concentration is nearly uniform, reaction controls, and uptake is approximately proportional to depth.
• For 0.3 ≤Ha ≤3$0.3\le\mathrm{Ha}\le3$, concentration gradients and reaction both affect uptake. Use the full reaction and diffusion solution below.
• For Ha >3$\mathrm{Ha}>3$, A is consumed within a thin region near the interface. Additional depth has little effect on uptake; both diffusivity and reaction kinetics determine the rate in that region.

These approximate limits apply to the stated depth convention and boundary conditions. The flux limits below differ from the exact result by about three percent at Ha =0.3$\mathrm{Ha}=0.3$ and half a percent at Ha =3$\mathrm{Ha}=3$.

The general solution is

𝑐𝐴=𝐶1𝑒𝑚𝑥+𝐶2𝑒−𝑚𝑥.$$c_A=C_1e^{mx}+C_2e^{-mx}.$$

Both constants have concentration units. At 𝑥 =𝐿$x=L$, zero gradient requires 𝐶2 =𝐶1𝑒2𝑚𝐿$C_2=C_1e^{2mL}$. At 𝑥 =0$x=0$, 𝐶1 +𝐶2 =𝑐𝐴0$C_1+C_2=c_{A0}$. Thus

𝐶1=𝑐𝐴01+𝑒2𝑚𝐿$$C_1=\frac{c_{A0}}{1+e^{2mL}}$$

and

𝐶2=𝑐𝐴0𝑒2𝑚𝐿1+𝑒2𝑚𝐿.$$C_2=\frac{c_{A0}e^{2mL}}{1+e^{2mL}}.$$

Equivalently,

𝑐𝐴=𝑐𝐴0[cosh⁡(𝑚𝑥)−tanh⁡(𝑚𝐿)sinh⁡(𝑚𝑥)]𝑐𝐴𝑐𝐴0=cosh⁡[𝑚(𝐿−𝑥)]cosh⁡(𝑚𝐿)𝑐𝐴(𝐿)=𝑐𝐴0cosh⁡(𝑚𝐿)$$\begin{aligned} c_A&=c_{A0}[\cosh(mx)-\tanh(mL)\sinh(mx)]\\ \frac{c_A}{c_{A0}}&=\frac{\cosh[m(L-x)]}{\cosh(mL)}\\ c_A(L)&=\frac{c_{A0}}{\cosh(mL)} \end{aligned}$$(28)

The difference-of-arguments identity converts the source’s expression to the ratio. Both boundaries hold and concentration stays positive for finite positive data. Differentiate Eq. 28 and multiply by −𝐷𝐴𝐵$-D_{AB}$,

𝑁𝐴𝑥(𝑥)=𝑐𝐴0𝐷𝐴𝐵𝑚sinh⁡[𝑚(𝐿−𝑥)]cosh⁡(𝑚𝐿)𝑁𝐴𝑥(0)=+𝑐𝐴0𝐷𝐴𝐵𝑚tanh⁡(𝑚𝐿)$$\begin{aligned} N_{Ax}(x)&=c_{A0}D_{AB}m\frac{\sinh[m(L-x)]}{\cosh(mL)}\\ N_{Ax}(0)&=+c_{A0}D_{AB}m\tanh(mL) \end{aligned}$$(29)

The gradient is negative at the surface, so influx must be positive in +𝑥$+x$. Integrating 𝑘′′′1𝑐𝐴$k_1'''c_A$ over 0 ≤𝑥 ≤𝐿$0\le x\le L$ reproduces the positive surface flux.

For Ha ≪1$\mathrm{Ha}\ll1$, tanh⁡Ha ≈Ha$\tanh\mathrm{Ha}\approx\mathrm{Ha}$, giving

𝑁𝐴𝑥(0)≈𝑐𝐴0𝑘′′′1𝐿.$$N_{Ax}(0)\approx c_{A0}k_1'''L.$$

For Ha ≫1$\mathrm{Ha}\gg1$, tanh⁡Ha ≈1$\tanh\mathrm{Ha}\approx1$, giving

𝑁𝐴𝑥(0)≈𝑐𝐴0√𝑘′′′1𝐷𝐴𝐵.$$N_{Ax}(0)\approx c_{A0}\sqrt{k_1'''D_{AB}}.$$

The effectiveness factor is

𝜂=tanh⁡(Ha)Ha,$$\eta=\frac{\tanh(\mathrm{Ha})}{\mathrm{Ha}},$$

the same function as the insulated-tip fin efficiency.

Figure 6: First-order concentration profiles at Hatta numbers 0.3, 1.5 and 5. Larger Hatta numbers confine consumption closer to the interface.

The curves follow Eq. 28.

For 𝐿 =10−3 m$L=10^{-3}\ \mathrm m$, 𝑘′′′1 =1 s−1$k_1'''=1\ \mathrm{s^{-1}}$ and 𝐷𝐴𝐵 =10−9 m2 s−1$D_{AB}=10^{-9}\ \mathrm{m^2\,s^{-1}}$, Ha =31.6$\mathrm{Ha}=31.6$. The reaction penetrates only about 32 𝜇m$32\ \mathrm{\mu m}$, so adding depth has little effect on uptake. Reducing 𝐿$L$ to 10−5 m$10^{-5}\ \mathrm m$ gives Ha =0.316$\mathrm{Ha}=0.316$, close to the uniform-concentration limit. Check whether a quoted length is pool depth or film thickness before applying either limit.

In the stated first-order pool model,

Ha=𝐿√𝑘′′′1𝐷𝐴𝐵≫1.$$\mathrm{Ha}=L\sqrt{\frac{k_1^{\prime\prime\prime}}{D_{AB}}}\gg1.$$

With fixed surface concentration, what does adding more pool depth do to surface uptake?

• It increases uptake in direct proportion to depth indefinitely.
• It makes concentration uniform throughout the pool.
• It reverses the surface flux.
• It changes uptake very little because A is consumed close to the surface. ✓

Explanation

The distance over which A penetrates before reacting is approximately

ℓ=√𝐷𝐴𝐵𝑘′′′1$$\ell=\sqrt{\frac{D_{AB}}{k_1^{\prime\prime\prime}}}$$

At large Hatta number this distance is much smaller than the pool depth. The surface flux approaches

𝑁𝐴0≃𝑐𝐴0√𝑘′′′1𝐷𝐴𝐵$$N_{A0}\simeq c_{A0}\sqrt{k_1^{\prime\prime\prime}D_{AB}}$$

Adding liquid below the thin reacting region changes the uptake very little.

9 Classroom example: catalytic tube

9.1 Species advection versus diffusion

Use Péclet number with species diffusivity to decide whether axial diffusion can be neglected. With mean speed ¯¯¯¯𝑣$\bar v$ and axial length 𝐿$L$, Pe𝑚,𝐿 =¯¯¯¯𝑣𝐿/𝐷𝐴𝐵$\mathrm{Pe}_{m,L}=\bar vL/D_{AB}$; small values require axial diffusion, while large values support omitting it away from end regions. The example below gives 107$10^7$ for ¯¯¯¯𝑣 =0.1 m s−1$\bar v=0.1\ \mathrm{m\,s^{-1}}$, 𝐿 =0.1 m$L=0.1\ \mathrm m$ and 𝐷𝐴𝐵 =10−9 m2 s−1$D_{AB}=10^{-9}\ \mathrm{m^2\,s^{-1}}$. This compares axial terms, so radial diffusion still supplies the wall. Using tube radius as the axial scale or evaluating this bulk estimate at a zero-velocity wall would answer a different question. The thermal version uses 𝛼$\alpha$ instead of 𝐷𝐴𝐵$D_{AB}$.

Catalytic tube

Consider a catalytic tubular reactor, where a dilute solution of solute A in solvent S is in fully developed, laminar flow in the region 𝑧 <0$z<0$. It encounters the catalytic wall in the region 0 ≤𝑧 ≤𝐿$0\le z\le L$, where solute A is instantaneously and irreversibly rearranged to an isomer B. Write the appropriate diffusion equation (in differential form) for this problem, along with the boundary conditions. Assume that the flow is isothermal and neglect the presence of B.

Solution

Figure 7: Axial liquid flow in a tube with a catalytic wall. For the instantaneous wall reaction, the labelled wall concentration is zero. Species A is consumed at the wall as the liquid flows downstream.

In this diagram, 𝑑 =2𝑅$d=2R$ and 𝑄 =¯¯¯¯𝑣𝜋𝑅2$Q=\bar v\pi R^2$. Its concentration labels denote the molar concentration 𝑐𝐴$c_A$ used below.

Data

• Tube radius 𝑅$R$
• Catalyst-coated length 0 ≤𝑧 ≤𝐿$0\le z\le L$
• Mean axial velocity ¯¯¯¯𝑣$\bar v$
• Uniform inlet concentration 𝑐𝐴0$c_{A0}$
• Wall reaction A converts instantaneously to its isomer B

The dilute liquid enters in fully developed laminar flow. The question asks for a model and boundary conditions, so no numerical solution is required.

Assumptions

• Steady state.
• Axisymmetric flow and concentration field.
• Constant properties.
• No radial velocity.
• No homogeneous reaction in the liquid.
• Uniform inlet concentration.
• Axial diffusion neglected only when axial convection dominates over the length of interest.

The specified isothermal flow and instantaneous wall reaction are supplied features of the model.

Calculations

Reaction occurs on the wall, so it enters through a boundary condition. It does not supply a volumetric 𝑅𝐴$R_A$ term in the liquid. The steady, axisymmetric form of Eq. 10 is

0=−1𝑟∂(𝑟𝑁𝐴𝑟)∂𝑟−∂𝑁𝐴𝑧∂𝑧$$0=-\frac1r\frac{\partial(rN_{Ar})}{\partial r}-\frac{\partial N_{Az}}{\partial z}$$

There is no radial convection. Axial convection remains present even though the solute is dilute. The two total-flux components are

𝑁𝐴𝑟=−𝐷𝐴𝐵∂𝑐𝐴∂𝑟$$N_{Ar}=-D_{AB}\frac{\partial c_A}{\partial r}$$𝑁𝐴𝑧=𝑣𝑧(𝑟)𝑐𝐴−𝐷𝐴𝐵∂𝑐𝐴∂𝑧$$N_{Az}=v_z(r)c_A-D_{AB}\frac{\partial c_A}{\partial z}$$

Substituting the fluxes into the balance gives the full convection-diffusion equation,

𝑣𝑧(𝑟)∂𝑐𝐴∂𝑧=𝐷𝐴𝐵𝑟∂∂𝑟(𝑟∂𝑐𝐴∂𝑟)+𝐷𝐴𝐵∂2𝑐𝐴∂𝑧2$$v_z(r)\frac{\partial c_A}{\partial z} =\frac{D_{AB}}r\frac{\partial}{\partial r}\left(r\frac{\partial c_A}{\partial r}\right) +D_{AB}\frac{\partial^2c_A}{\partial z^2}$$(30)

For fully developed laminar flow in a circular tube, the velocity profile is

𝑣𝑧(𝑟)=2¯¯¯¯𝑣[1−(𝑟𝑅)2]$$v_z(r)=2\bar v\left[1-\left(\frac{r}{R}\right)^2\right]$$

To judge axial diffusion, compare the convection scale ¯¯¯¯𝑣𝑐𝐴/𝐿$\bar v c_A/L$ with the diffusion scale 𝐷𝐴𝐵𝑐𝐴/𝐿2$D_{AB}c_A/L^2$. Their ratio is the axial Péclet number,

Pe𝐿=¯¯¯¯𝑣𝐿𝐷𝐴𝐵$$\mathrm{Pe}_L=\frac{\bar vL}{D_{AB}}$$

For illustration, use ¯¯¯¯𝑣 =0.1 m s−1$\bar v=0.1\ \mathrm{m\,s^{-1}}$, 𝐿 =0.1 m$L=0.1\ \mathrm m$ and 𝐷𝐴𝐵 =10−9 m2 s−1$D_{AB}=10^{-9}\ \mathrm{m^2\,s^{-1}}$. These are supplementary scale values rather than data supplied in the question. They give

Pe𝐿=0.1(0.1)10−9=107$$\mathrm{Pe}_L=\frac{0.1(0.1)}{10^{-9}}=10^7$$

Axial convection then dominates axial diffusion over this length, away from axial end regions. The resulting working equation is

𝑣𝑧(𝑟)∂𝑐𝐴∂𝑧=𝐷𝐴𝐵𝑟∂∂𝑟(𝑟∂𝑐𝐴∂𝑟)$$v_z(r)\frac{\partial c_A}{\partial z} =\frac{D_{AB}}r\frac{\partial}{\partial r}\left(r\frac{\partial c_A}{\partial r}\right)$$(31)

The reduced equation needs inlet data and two radial conditions. The inlet supplies uniform A. Instantaneous reaction holds the catalytic-wall concentration at zero. Centreline symmetry gives zero radial gradient. These conditions form the system

𝑐𝐴(𝑟,0)=𝑐𝐴0,0≤𝑟<𝑅𝑐𝐴(𝑅,𝑧)=0,0<𝑧≤𝐿[∂𝑐𝐴∂𝑟]𝑟=0=0,0<𝑧≤𝐿$$\begin{aligned} c_A(r,0)&=c_{A0}, &&0\le r<R\\ c_A(R,z)&=0, &&0<z\le L\\ \left[\frac{\partial c_A}{\partial r}\right]_{r=0}&=0, &&0<z\le L \end{aligned}$$

The inlet and wall data are interpreted as one-sided conditions at the inlet-wall corner, where the idealised concentration jumps. Every term in the governing equation has units kmol m−3 s−1$\mathrm{kmol\,m^{-3}\,s^{-1}}$.

If axial diffusion is retained in Eq. 30, the axial equation is second order. A finite-domain model then also needs an appropriate outlet condition, such as zero axial concentration gradient at a sufficiently downstream boundary. The Péclet estimate supports the reduced model over a specified scale; it is not a pointwise argument at the wall, where the velocity is zero.

For a finite first-order surface reaction, the wall condition becomes

−𝐷𝐴𝐵∂𝑐𝐴∂𝑟=𝑘′′1𝑐𝐴$$-D_{AB}\frac{\partial c_A}{\partial r}=k_1''c_A$$

at 𝑟 =𝑅$r=R$, because diffusive arrival must equal surface consumption.

This formulation is the mass-transfer Graetz problem. In the fully developed concentration region, a constant wall concentration gives Sh =𝑘𝑐𝑑𝑡/𝐷𝐴𝐵 =3.66$\mathrm{Sh}=k_cd_t/D_{AB}=3.66$, where 𝑑𝑡 =2𝑅$d_t=2R$ and 𝑘𝑐$k_c$ uses the difference between mixing-cup and wall concentrations. The entrance-region coefficient is larger. The thermal problem with the same geometry and boundary conditions gives Nu =3.66$\mathrm{Nu}=3.66$.

Answer

Use Eq. 31 with

𝑣𝑧=2¯¯¯¯𝑣[1−(𝑟𝑅)2]$$v_z=2\bar v\left[1-\left(\frac{r}{R}\right)^2\right]$$

when axial diffusion can be neglected. Apply uniform inlet concentration, zero A concentration at the catalytic wall, and zero radial concentration gradient at the centreline, as stated above.

If axial diffusion is significant, use Eq. 30 and provide an additional axial boundary condition.

10 Additional examples

Derive a film balance with finite surface reaction

A dilute gas species A crosses a plane film to a catalyst, where it undergoes an isomerisation 𝐴 ⟶𝐵$A\longrightarrow B$. The gas is at 500 K and 100 kPa. The bulk mole fraction is 0.020, film thickness is 1.5 mm, diffusivity is 6.0 ×10−5$6.0\times10^{-5}$ m² s⁻¹ and surface rate constant is 𝑘𝑠 =0.015$k_s=0.015$ m s⁻¹. The catalyst area is 0.040 m². Assume steady one-dimensional isothermal transport with constant pressure and diffusivity, no gas-phase reaction, and negligible normal bulk velocity. Let 𝑧 =0$z=0$ be the bulk edge and 𝑧 =𝐿$z=L$ the catalyst.

Derive the concentration profile from a shell balance. Find the surface mole fraction and molar removal rate. Which resistance is greater, and what happens in the limits of very slow and very fast surface reaction?

Solution

Figure 8: A diffuses from the bulk through a nonreacting film to the catalyst. A finite reaction rate leaves a nonzero surface concentration, determined by matching film transport and surface consumption.

Data

Quantity	Value
Temperature	𝑇 =500 K$T=500\ \mathrm K$
Pressure	𝑃 =100000 Pa$P=100000\ \mathrm{Pa}$
Bulk mole fraction	𝑦𝐴𝑏 =0.020$y_{Ab}=0.020$
Film thickness	𝐿 =0.0015 m$L=0.0015\ \mathrm m$
Diffusivity	𝐷𝐴𝐵 =6.0 ×10−5 m2 s−1$D_{AB}=6.0\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Surface reaction coefficient	𝑘𝑠 =0.015 m s−1$k_s=0.015\ \mathrm{m\,s^{-1}}$
Surface area	𝐴𝑠 =0.040 m2$A_s=0.040\ \mathrm{m^2}$

Assumptions

• Steady state.
• One-dimensional transport across a plane film.
• Ideal gas behaviour.
• Constant temperature, pressure and diffusivity.
• No gas-phase reaction.
• Negligible normal bulk velocity.

Calculations

For a film slice of area 𝑆$S$ and thickness Δ𝑧$\Delta z$, no accumulation or volumetric reaction occurs. The shell balance is

𝑆𝑁𝐴(𝑧)−𝑆𝑁𝐴(𝑧+Δ𝑧)=0⟹𝑑𝑁𝐴𝑑𝑧=0$$SN_A(z)-SN_A(z+\Delta z)=0 \quad\Longrightarrow\quad \frac{dN_A}{dz}=0$$

With

𝑁𝐴=−𝐷𝐴𝐵𝑑𝑐𝐴𝑑𝑧$$N_A=-D_{AB}\frac{dc_A}{dz}$$

and constant diffusivity, 𝑑2𝑐𝐴/𝑑𝑧2 =0$d^2c_A/dz^2=0$. Integrating twice and imposing 𝑐𝐴(0) =𝑐𝐴𝑏$c_A(0)=c_{Ab}$ and 𝑐𝐴(𝐿) =𝑐𝐴𝑠$c_A(L)=c_{As}$ gives

𝑐𝐴(𝑧)=𝑐𝐴𝑏+(𝑐𝐴𝑠−𝑐𝐴𝑏)𝑧𝐿,𝑁𝐴=𝐷𝐴𝐵𝐿(𝑐𝐴𝑏−𝑐𝐴𝑠)$$c_A(z)=c_{Ab}+(c_{As}-c_{Ab})\frac zL, \qquad N_A=\frac{D_{AB}}L(c_{Ab}-c_{As})$$

Reaction occurs at the boundary. It is absent from the film differential equation. At steady state delivery equals surface consumption:

𝐷𝐴𝐵𝐿(𝑐𝐴𝑏−𝑐𝐴𝑠)=𝑘𝑠𝑐𝐴𝑠,𝑐𝐴𝑠=𝑐𝐴𝑏1+𝑘𝑠𝐿𝐷𝐴𝐵$$\frac{D_{AB}}L(c_{Ab}-c_{As})=k_sc_{As}, \qquad c_{As}=\frac{c_{Ab}}{1+\frac{k_sL}{D_{AB}}}$$

Total gas concentration is uniform, so mole fractions have the same ratio:

𝑘𝑠𝐿𝐷𝐴𝐵=0.375$$\frac{k_sL}{D_{AB}}=0.375$$𝑦𝐴𝑠=0.0201.375=0.01455$$y_{As}=\frac{0.020}{1.375}=0.01455$$𝑐=1000008.314(500)=24.06 mol m−3$$c=\frac{100000}{8.314(500)}=24.06\ \mathrm{mol\,m^{-3}}$$𝑁𝐴=𝑘𝑠𝑐𝑦𝐴𝑠=0.015(24.06)(0.01455)=0.005249 mol m−2 s−1$$N_A=k_scy_{As}=0.015(24.06)(0.01455) =0.005249\ \mathrm{mol\,m^{-2}\,s^{-1}}$$˙𝑛𝐴=𝐴𝑠𝑁𝐴=0.040(0.005249)=2.099×10−4 mol s−1$$\dot n_A=A_sN_A=0.040(0.005249) =2.099\times10^{-4}\ \mathrm{mol\,s^{-1}}$$

Both concentration-based resistances have units s m⁻¹: 𝐿/𝐷𝐴𝐵 =25.0$L/D_{AB}=25.0$ and 1/𝑘𝑠 =66.7$1/k_s=66.7$. Their sum gives

𝑁𝐴=𝑐𝐴𝑏𝐿𝐷𝐴𝐵+1𝑘𝑠.$$N_A=\frac{c_{Ab}}{\frac{L}{D_{AB}}+\frac{1}{k_s}}.$$

Surface reaction supplies the larger resistance, but neither resistance is negligible.

As 𝑘𝑠 →0$k_s\to0$, 𝑐𝐴𝑠 →𝑐𝐴𝑏$c_{As}\to c_{Ab}$ and removal tends to zero. As 𝑘𝑠 →∞$k_s\to\infty$, 𝑐𝐴𝑠 →0$c_{As}\to0$ and removal approaches the diffusion-limited value 𝐷𝐴𝐵𝑐𝐴𝑏/𝐿$D_{AB}c_{Ab}/L$. The isomerisation exchanges one gas mole for one gas mole, consistent with zero net molar flow. Other stoichiometries require checking that assumption.

For a cylindrical reactor with axial flow, the earlier catalytic-tube example retains axial convection. Its finite-rate wall condition uses the same principle: arriving diffusive flux equals surface consumption. Its radial profile must be found with the axial concentration field, rather than assumed linear as in this plane, nonconvecting film.

Answer

The surface mole fraction is 𝑦𝐴𝑠 =0.0146$y_{As}=0.0146$ and the molar removal rate is ˙𝑛𝐴 =2.10 ×10−4 mol s−1$\dot n_A=2.10\times10^{-4}\ \mathrm{mol\,s^{-1}}$. Surface reaction supplies the larger resistance, 66.7 s m−1$66.7\ \mathrm{s\,m^{-1}}$, compared with 25.0 s m−1$25.0\ \mathrm{s\,m^{-1}}$ for the film. The slow-reaction limit gives zero removal; the fast-reaction limit gives the diffusion-limited flux 𝐷𝐴𝐵𝑐𝐴𝑏/𝐿$D_{AB}c_{Ab}/L$.

Use reaction stoichiometry to determine flux directions

At a catalyst, gas R reacts by 𝑅 ⟶2𝑃$R\longrightarrow2P$. Take the coordinate direction towards the catalyst and suppose 𝑁𝑅 =0.010$N_R=0.010$ mol m⁻² s⁻¹. Find 𝑁𝑃$N_P$ and the total molar flux. Is equimolar counter-diffusion appropriate? Compare with a reaction 𝑅 ⟶𝑃$R\longrightarrow P$ at the same reactant flux.

Solution

Data

• Surface reaction 𝑅 ⟶2𝑃$R\longrightarrow2P$
• Reactant flux toward the catalyst 𝑁𝑅 =0.010 mol m−2 s−1$N_R=0.010\ \mathrm{mol\,m^{-2}\,s^{-1}}$
• Comparison reaction 𝑅 ⟶𝑃$R\longrightarrow P$

Positive flux points toward the catalyst.

Assumptions

• Steady surface reaction.
• No surface accumulation.
• Only the stated reaction produces or consumes R and P.

Calculations

Products move away from the surface. Two moles of P leave for each mole of R arriving, so

𝑁𝑃=−2𝑁𝑅=−2(0.010)=−0.020 mol m−2 s−1$$N_P=-2N_R=-2(0.010)=-0.020\ \mathrm{mol\,m^{-2}\,s^{-1}}$$𝑁𝑅+𝑁𝑃=0.010−0.020=−0.010 mol m−2 s−1$$N_R+N_P=0.010-0.020=-0.010\ \mathrm{mol\,m^{-2}\,s^{-1}}$$

The net molar flow is away from the catalyst, so the equimolar model does not apply. In the total-flux expression

𝑁𝑅=−𝑐𝐷𝑅𝑃 𝑑𝑦𝑅𝑑𝑧+𝑦𝑅(𝑁𝑅+𝑁𝑃),$$N_R=-cD_{RP}\,\frac{dy_R}{dz}+y_R(N_R+N_P),$$

the bulk term is negative. It opposes R’s inward diffusive transport.

The flux relation can also be written 𝑁𝑅 = −𝑁𝑃/2$N_R=-N_P/2$. The negative sign expresses opposing directions; the factor comes from stoichiometry.

For 𝑅 ⟶𝑃$R\longrightarrow P$, 𝑁𝑃 = −𝑁𝑅$N_P=-N_R$ and the total molar flux is zero. That reaction permits the equimolar model under the other stated binary-film assumptions. Constant temperature and pressure alone do not determine the flux ratio.

Answer

For 𝑅 ⟶2𝑃$R\longrightarrow2P$, 𝑁𝑃 = −0.020 mol m−2 s−1$N_P=-0.020\ \mathrm{mol\,m^{-2}\,s^{-1}}$ and the total molar flux is −0.010 mol m−2 s−1$-0.010\ \mathrm{mol\,m^{-2}\,s^{-1}}$, directed away from the catalyst. Equimolar counter-diffusion does not apply. For 𝑅 ⟶𝑃$R\longrightarrow P$, 𝑁𝑃 = −0.010 mol m−2 s−1$N_P=-0.010\ \mathrm{mol\,m^{-2}\,s^{-1}}$ and the total molar flux is zero.

Distinguish reaction throughout a liquid from reaction at a wall

An idealised stationary water layer of depth 0.010 m absorbs oxygen at its upper face. That face is held at 𝑐𝐴0 =0.25$c_{A0}=0.25$ mol m⁻³. Oxygen consumption throughout the liquid follows 𝑅𝐴 = −𝑘′′′1𝑐𝐴$R_A=-k_1'''c_A$, with 𝑘′′′1 =0.0020$k_1'''=0.0020$ s⁻¹ and 𝐷𝐴𝐵 =2.0 ×10−9$D_{AB}=2.0\times10^{-9}$ m² s⁻¹. The bottom is impermeable. Assume a steady one-dimensional profile and constant properties.

Formulate the problem and estimate the oxygen influx. Explain how the model would change if consumption occurred only on a catalytic bottom instead.

Solution

Data

• Liquid depth 𝐿 =0.010 m$L=0.010\ \mathrm m$
• Fixed top concentration 𝑐𝐴0 =0.25 mol m−3$c_{A0}=0.25\ \mathrm{mol\,m^{-3}}$
• First-order volumetric rate constant 𝑘′′′1 =0.0020 s−1$k_1^{\prime\prime\prime}=0.0020\ \mathrm{s^{-1}}$
• Diffusivity 𝐷𝐴𝐵 =2.0 ×10−9 m2 s−1$D_{AB}=2.0\times10^{-9}\ \mathrm{m^2\,s^{-1}}$

The bottom is impermeable.

Assumptions

• Steady state.
• One-dimensional diffusion.
• Stationary liquid.
• Dilute oxygen.
• Constant properties.
• First-order homogeneous oxygen consumption.

Calculations

Measure 𝑥$x$ downwards from the upper surface. Consumption occurs throughout the liquid, so it is a homogeneous reaction and belongs in the local balance:

𝐷𝐴𝐵𝑑2𝑐𝐴𝑑𝑥2−𝑘′′′1𝑐𝐴=0,𝑐𝐴(0)=𝑐𝐴0,[𝑑𝑐𝐴𝑑𝑥]𝑥=𝐿=0$$D_{AB}\frac{d^2c_A}{dx^2}-k_1'''c_A=0, \qquad c_A(0)=c_{A0}, \qquad \left[\frac{dc_A}{dx}\right]_{x=L}=0$$

The reaction penetration length is

√𝐷𝐴𝐵𝑘′′′1=0.0010 m,$$\sqrt{\frac{D_{AB}}{k_1'''}}=0.0010\,\mathrm{m},$$

and

Ha=𝐿√𝑘′′′1𝐷𝐴𝐵=10.$$\mathrm{Ha}=L\sqrt{\frac{k_1'''}{D_{AB}}}=10.$$

From Eq. 29,

𝑁𝐴(0)=𝑐𝐴0√𝑘′′′1𝐷𝐴𝐵tanh⁡(Ha)=0.25√0.0020(2.0×10−9)tanh⁡(10)≃5.00×10−7 mol m−2 s−1$$\begin{aligned} N_A(0)&=c_{A0}\sqrt{k_1'''D_{AB}}\tanh(\mathrm{Ha})\\ &=0.25\sqrt{0.0020(2.0\times10^{-9})}\tanh(10)\\ &\simeq5.00\times10^{-7}\ \mathrm{mol\,m^{-2}\,s^{-1}} \end{aligned}$$

Most consumption takes place near the upper surface. Adding more depth has little effect in this large-Ha limit.

With reaction only at a catalytic bottom, the interior has no volumetric source:

𝐷𝐴𝐵𝑑2𝑐𝐴𝑑𝑥2=0.$$D_{AB}\frac{d^2c_A}{dx^2}=0.$$

Replace the impermeable-bottom condition by

−𝐷𝐴𝐵𝑐′𝐴(𝐿)=𝑘𝑠𝑐𝐴(𝐿)$$-D_{AB}c_A'(L)=k_sc_A(L)$$

for a first-order surface reaction. The two rate constants have different units, s⁻¹ in the liquid and m s⁻¹ at the surface.

A pond can require an unsteady absorption model or a steady reaction model. Its size alone does not decide. The stated consumption, flow, initial state and boundary conditions determine which model is appropriate.

Answer

The oxygen influx is 5.00 ×10−7 mol m−2 s−1$5.00\times10^{-7}\ \mathrm{mol\,m^{-2}\,s^{-1}}$ into the liquid. If reaction occurs only at the bottom, use 𝐷𝐴𝐵𝑐′′𝐴 =0$D_{AB}c_A''=0$ in the interior and the bottom condition

−𝐷𝐴𝐵𝑐′𝐴(𝐿)=𝑘𝑠𝑐𝐴(𝐿).$$-D_{AB}c_A'(L)=k_sc_A(L).$$

An expanding reaction mixture changes the film profile

A catalyst converts gas A instantaneously by 𝐴 ⟶3𝐵$A\longrightarrow3B$. A plane gas film of thickness 𝛿 =2.0$\delta=2.0$ mm separates the surface from a well-mixed bulk gas with 𝑦𝐴0 =0.20$y_{A0}=0.20$. The binary gas is at 500 K and 1.5 bar, with 𝐷𝐴𝐵 =2.0 ×10−5$D_{AB}=2.0\times10^{-5}$ m² s⁻¹. Assume steady, one-dimensional transport, constant temperature, pressure and diffusivity, and no reaction inside the film. The catalyst is impermeable except for the reactant and product exchange specified by the reaction.

Taking 𝑧$z$ from the bulk edge towards the catalyst, derive 𝑦𝐴(𝑧)$y_A(z)$ and the inward flux of A. Compare with equimolar counter-diffusion across the same film. Does a constant species flux imply a linear mole-fraction profile?

Solution

Figure 9: A crosses the gas film toward the catalyst. The surface reaction produces B, whose outward flux changes the mixture molar flow.

Data

Quantity	Value
Film thickness	𝛿 =0.0020 m$\delta=0.0020\ \mathrm m$
Bulk mole fraction	𝑦𝐴0 =0.20$y_{A0}=0.20$
Temperature	𝑇 =500 K$T=500\ \mathrm K$
Pressure	𝑃 =150000 Pa$P=150000\ \mathrm{Pa}$
Diffusivity	𝐷𝐴𝐵 =2.0 ×10−5 m2 s−1$D_{AB}=2.0\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Instantaneous catalyst reaction	𝐴 ⟶3𝐵$A\longrightarrow3B$

Assumptions

• Steady state.
• One-dimensional transport across a plane film.
• Binary ideal gas mixture.
• Constant temperature, pressure and diffusivity.
• No gas-phase reaction.
• Surface exchange occurs only through the stated reaction.

Calculations

A film shell balance gives 𝑑𝑁𝐴/𝑑𝑧 =0$dN_A/dz=0$ because there is no accumulation or gas-phase reaction. The boundary conditions are 𝑦𝐴(0) =𝑦𝐴0$y_A(0)=y_{A0}$ and 𝑦𝐴(𝛿) =0$y_A(\delta)=0$. Surface stoichiometry fixes 𝑁𝐵 = −3𝑁𝐴$N_B=-3N_A$, so the total flux is 𝑁𝐴 +𝑁𝐵 = −2𝑁𝐴$N_A+N_B=-2N_A$. Product expansion creates outward molar flow even though there is no externally imposed flow through the film.

Keep that flow in the total-flux equation:

𝑁𝐴=−𝑐𝐷𝐴𝐵𝑑𝑦𝐴𝑑𝑧−2𝑦𝐴𝑁𝐴𝑑𝑦𝐴1+2𝑦𝐴=−𝑁𝐴𝑐𝐷𝐴𝐵 𝑑𝑧$$\begin{gathered} N_A=-cD_{AB}\frac{dy_A}{dz}-2y_AN_A\\ \frac{dy_A}{1+2y_A}=-\frac{N_A}{cD_{AB}}\,dz \end{gathered}$$

Integrating from the bulk edge to position 𝑧$z$ gives

12ln⁡1+2𝑦𝐴(𝑧)1+2𝑦𝐴0=−𝑁𝐴𝑧𝑐𝐷𝐴𝐵$$\frac12\ln\frac{1+2y_A(z)}{1+2y_{A0}} =-\frac{N_Az}{cD_{AB}}$$

Using the zero surface concentration at 𝑧 =𝛿$z=\delta$ determines the flux. Substitution back into the integrated equation gives the profile:

𝑁𝐴=𝑐𝐷𝐴𝐵2𝛿ln⁡(1+2𝑦𝐴0)𝑦𝐴(𝑧)=12[(1+2𝑦𝐴0)1−𝑧𝛿−1]$$\begin{aligned} N_A&=\frac{cD_{AB}}{2\delta}\ln(1+2y_{A0})\\ y_A(z)&=\frac12\left[(1+2y_{A0})^{1-\frac{z}{\delta}}-1\right] \end{aligned}$$

The numerical concentration is 𝑐 =𝑃/(𝑅𝑇) =36.08$c=P/(RT)=36.08$ mol m⁻³, so

𝑁𝐴=36.08(2.0×10−5)2(0.0020)ln⁡[1+2(0.20)]=0.06071 mol m−2 s−1$$N_A=\frac{36.08(2.0\times10^{-5})}{2(0.0020)}\ln[1+2(0.20)] =0.06071\ \mathrm{mol\,m^{-2}\,s^{-1}}$$

An equimolar model would give

𝑁𝐴,𝑒𝑞=𝑐𝐷𝐴𝐵𝑦𝐴0𝛿=0.07217 mol m−2 s−1.$$N_{A,eq}=\frac{cD_{AB}y_{A0}}{\delta}=0.07217\,\mathrm{mol\,m^{-2}\,s^{-1}}.$$

The actual flux in this film model is 15.9 percent below that value. Outward bulk flow opposes the inward transport of A. This reverses the effect of the inward molar flow in the contracting 2𝐴 ⟶𝐵$2A\longrightarrow B$ case above.

At 𝑧 =𝛿/2$z=\delta/2$,

𝑦𝐴=√1.4−12=0.09161,$$y_A=\frac{\sqrt{1.4}-1}{2}=0.09161,$$

compared with 0.100 for a straight line. The species flux is constant, but its diffusive and bulk contributions vary with composition. Constant flux therefore does not require a linear profile. Both imposed endpoints are recovered exactly. As 𝑦𝐴0 →0$y_{A0}\to0$, the logarithm approaches 2𝑦𝐴0$2y_{A0}$ and the dilute equimolar estimate is recovered.

Answer

The inward flux is 𝑁𝐴 =0.0607 mol m−2 s−1$N_A=0.0607\ \mathrm{mol\,m^{-2}\,s^{-1}}$, 15.9 percent below the equimolar estimate. The profile is

𝑦𝐴(𝑧)=12[(1+2𝑦𝐴0)1−𝑧𝛿−1].$$y_A(z)=\frac{1}{2}\left[(1+2y_{A0})^{1-\frac{z}{\delta}}-1\right].$$

The profile is curved despite the constant species flux.

11 What you should now be able to calculate

Distinguish surface and volume reaction, combine finite reaction and diffusion resistance, and use stoichiometry to determine total molar motion.

Practice: Workshop 9 for surface reaction and coupled species fluxes.

12 Report an error

Email Ranjeet Utikar with the title Species balances and diffusion with reaction, the URL /mass-transport/diffusion-with-reaction/, and the equation or passage to check.

References

Bird, R. Byron, Warren E. Stewart, and Edwin N. Lightfoot. 2006. Transport Phenomena. Revised 2nd. John Wiley & Sons. https://www.wiley-vch.de/en/areas-interest/engineering/transport-phenomena-978-0-470-11539-8.

Geankoplis, Christie J. 2003. Transport Processes and Separation Process Principles. 4th ed. Prentice Hall.

Shah, Milinkumar. n.d. “CHEN3001-TP: Mass Transport 2.” Curtin University. Unpublished.

Welty, James R., Charles E. Wicks, Robert E. Wilson, and Gregory L. Rorrer. 2001. Fundamentals of Momentum, Heat, and Mass Transfer. 4th ed. John Wiley & Sons.