CHE 318 Lecture 05

Diffusion in Liquid and Solid

Dr. Tian Tian

2026-01-14

Recap

Theories for predicting gas diffusivity \(D_{AB}\):
- Kinetic theory (inaccurate)
- Chapman-Enkog theory (high accuracy, hard to use)
- Fuller method (trade off)
Estimating \(\sum \nu_i\) using Fuller method table
Extrapolating \(D_{AB}\) at different \((T, P)\)
Solving examples

Learning Outcomes

After today’s lecture, you will be able to:

Recall differences between diffusion in gas and in liquid
Describe the limiting cases of mass transport in liquid
Solve EMCD, stagnant B and general cases for transport in liquid
Analyze analogs between mass transport solution in gas and liquid

Diffusion in Liquid

Recall our general transport phenomena equation in Lecture 1

\[ \text{[rate of transfer process]} = \dfrac{\text{[driving force]}}{\text{[resistance]}} \]

Let \(A\) being a soluble species, \(B\) being the liquid. The resistance in liquid for diffusive transport is much larger than in gases:

Molecular density of liquid (value) is much higher than gas (value) !

Intermolecular interaction in liquid is dominating (in comparison, kinetic effect in gases)
\(D_{AB}\vert_{l} \approx 10^{-5} D_{AB}\vert_{g}\)
\(D_{AB}\vert_{l} \approx 10^{-12} \sim 10^{-10}\ \text{m}^2\cdot\text{s}^-1\)

Diffusion in Liquid (2)

Unlike in gas, \(D_{AB}\vert_{l}\) is usually dependent on the molar fraction of \(A\)

In this lecture we will use the diluted limit
\(D_{AB}\vert_{l} \neq f(c_A)\)

Mass Transport in Liquid – Example 1 EMCD

The equimolar counter diffusion (EMCD) case in liquid also satisfies \(N_A + N_B = 0\)

From notes in Lecture 2, we can directly write the solution for \(N_A\):

\[ N_A = \frac{D_{AB}}{z_2 - z_1} (c_{A1} - c_{A2}) \]

Often for dilute A solution, we can rewrite the EMCD solution using average concentration.

EMCD Solution in Liquid

\[\begin{align} \boxed{ N_A = \frac{D_{AB} c_{A, v}}{z_2 - z_1} (x_{A1} - x_{A2}) } \end{align}\]

We define \(c_{av}\) as the average total concentration in A+B (not A alone!):

\[ c_{av} = \left(\frac{\rho}{M}\right) = \frac{1}{2} \left(\frac{\rho_1}{M_1} + \frac{\rho_2}{M_2}\right) \]

\(\rho_1\) and \(\rho_2\) are average density of solution at points 1 and 2
\(M_1\) and \(M_2\) are average molecular weight of the solution at points 1 and 2
What is the assumption if we can use the average \(c_{av}\) in this case?

Example 2 Stagnant \(B\) (\(N_B = 0\))

EMCD in liquid is very rare, stagnant B case more frequent
Typical setup: diffusion through liquid films, where liquid molecules cannot permeate the barrier

Recall gas-phase equation:

\[ N_A = \frac{D_{AB} \, p_T \, (p_{A1} - p_{A2})} {R T \, (z_2 - z_1) \, p_{B,m}} \]

We have similarly in liquid:

\[\begin{align} \boxed{ N_A = \frac{D_{AB}c_{A, v}}{(z_2 - z_1)} \frac{(x_{A1} - x_{A2})}{x_{B,m}} },\quad x_{B,m} = \frac{x_{B2} - x_{B1}}{\ln\!(x_{B2} / x_{B1})} \end{align}\]

Stagnant \(B\): Further Discussions

\(x_{B,m}\) is the log-mean value of \(x_B\)
In diluted A through stagnant B, we often have \(x_{B1} \approx x_{B2} \approx 1\)
A simpler approximation can be made \(x_{B,m} \approx \frac{x_{B1} + x_{B2}}{2}\)
We can even roughly approximate \(x_{B, m} = 1\)!

\[\begin{align} N_A &= \frac{D_{AB}c_{A, v}}{(z_2 - z_1)} \frac{(x_{A1} - x_{A2})}{x_{B,m}} \\ &\approx \frac{D_{AB}c_{A, v}}{(z_2 - z_1)} \frac{(x_{A1} - x_{A2})}{(x_{B1} + x_{B2}) / 2} \\ &\approx \frac{D_{AB}}{(z_2 - z_1)}(c_{A1} - c_{A2}) \end{align}\]

Diffusivity in Liquid

Experimentally measured \(D_{AB}\) in liquid:

Prediction of Diffusivity in Liquid

Molecular diffusion in liquid encounters much more collision than gas phase!
Kinetic theory is not applicable!
A few semi-empirical laws exist:

Einstein-Stokes Equation

Model molecules A as spheres through a fluid B
Drag force of A in B predicted from Stoke’s law

\[ D_{AB} = \mu_{AB} k_B T \]

where \(\mu_{AB}\) is the mobility of A in B

Einstein-Stokes Equation for \(D_{AB}\)

Stokes–Einstein correlation for diffusivity in liquids:

\[ D_{AB} = \frac{9.96 \times 10^{-16}\, T} {\eta \, V_A^{0.333}} \]

\(D_{AB}\) in \(\mathrm{m^2/s}\)
\(T\) : temperature (K)
\(\eta_B\) : viscosity of B (kg / m·s)
\(V_A\) : molar volume of solute \(A\)
- evaluated at normal boiling point
- units: m\(^3\)/kg mol

✔ Good for:
- Molecular weight \(> 1000\)
- \(V_A > 0.5\) m\(^3\)/kg mol
✘ Poor accuracy otherwise

Semi-Empirical Wilke–Chang Equation for Liquids

\[ D_{AB} = \frac{1.173 \times 10^{-16}\, (\phi M_B)^{1/2}\, T} {\eta_B \, V_A^{0.6}} \]

\(D_{AB}\) in \(\mathrm{m^2/s}\)
\(T\) : temperature (K)
\(\eta_B\) : viscosity of solvent \(B\) (kg / m·s)
\(M_B\) : molecular weight of solvent \(B\)
\(V_A\) : molar volume of solute \(A\) (m\(^3\)/kgmol)
\(\phi\) : solvent association parameter (see next slide)

Typical error:
- 10–15% for aqueous systems
- ~25% for non-aqueous solvents
Valid temperature range:
- \(278\ \text{K} < T < 313\ \text{K}\)

Wilke–Chang Equation, \(\phi\) values

Association parameter \(\phi\)

Water: \(\phi = 2.6\)
Methanol: \(\phi = 1.9\)
Ethanol: \(\phi = 1.5\)
Benzene: \(\phi = 1.0\)
Non-associating solvents: \(\phi \approx 1.0\)

Diffusion in Solids

Slowest mode of mass transfer
Vital to industry:
- Packaging
- Catalysts
- Biological processes
We will focus on two types of solids

Type 1: Homogeneous Solids

Uniform solid matrix
Diffusion follows Fick’s law
Well-defined diffusion path

Examples

\(O_2\) diffusion through plastic
\(H_2O\) diffusion through paint

Type 1: General Flux Equation for Solids

For diffusion of \(A\) in a homogeneous solid:

\[ N_A = - c_T D_{AB} \frac{d x_A}{d z} + x_A (N_A + N_B) \]

In solids:

Solid matrix is stationary
\(v_m = 0 \;\Rightarrow\; (N_A + N_B) = 0\)

\[ \boxed{ N_A = - c_T D_{AB} \frac{d x_A}{d z} } \]

Final Flux Expression (Concentration Form)

Key properties for diffusion in solids:

\(D_{AB}\) is independent of pressure
\(D_{AB} \neq D_{BA}\)

For steady-state diffusion through a slab:

\[ \boxed{ N_A = \frac{D_{AB}\,(c_{A1} - c_{A2})} {(z_2 - z_1)} } \]

Assumptions:

Slab geometry
Homogeneous solid matrix
Constant \(D_{AB}\)

How Do We Know \(c_{A}\)?

For this type of problem, we are often interested in the gas solubility in solid, where \(c_A \propto p_{A}\). When expressed using the solubility \(S\) :

\[ \boxed{ c_A = \frac{S\, p_A}{22.414} } \]

Unit:

\(c_A\): \(\text{kg mol}\cdot\text{m}^{-3}\)
\(S\): \(\text{m}^3\) (STP at 0 ℃ and 1 atm)
\(p_A\): atm

Often we also express the permeability of a gas (\(P_{M}\)) in solid using \(P_M = S D_{AB}\)

Type 2: Inhomogeneous Solids

Non-uniform structure
Diffusion through:
- Pores
- Fixed or tortuous paths in solid matrix
Requires modified Fick’s law

Examples:

Brita water filter
Porous catalysts

Flux in Porous Solids

Steady-state diffusion of \(A\) through a porous solid:

\[ N_A = \frac{\varepsilon}{\tau} \frac{D_{AB}\,(c_{A1} - c_{A2})} {(z_2 - z_1)} \]

\(\varepsilon\): open void fraction
- typically 0.1–0.9
\(\tau\): tortuosity
- typically 1.5–5 for solids

Effective Diffusivity

Porous-media effects are lumped into an effective diffusivity:

\[ \boxed{ D_{AB,\text{eff}} = \frac{\varepsilon}{\tau}\, D_{AB} } \]

Accounts for reduced area and increased path length
Used directly in Fick’s law for porous solids

Example Problems

General Steps

See handwritten notes for step-by-step solutions.

Draw the physical scheme
Identify the diffusion case
List assumptions
Write the governing flux equation
Apply boundary conditions (optional)
Solve for the molar flux

Example 1: EMCD Basics

Adapted from Geankoplis 6.2-1

Ammonia (\(A\)) is diffusing through nitrogen (\(B\)) in a straight tube of length \(L = 0.10\) m at steady state.
The system is maintained at a total pressure of \(P_T = 1.0132 \times 10^5\) Pa and a temperature of \(T = 298\) K.
The partial pressure of ammonia at \(z_1\) is \(p_{A1} = 1.013 \times 10^4\) Pa, and at \(z_2\) is \(p_{A2} = 0.507 \times 10^4\) Pa.
The binary diffusivity of ammonia in nitrogen is \(D_{AB} = 0.230 \times 10^{-4}\) m\(^2\)/s.

Determine values for fluxes of A and B.

Example 2: EMCD in Two-Bulb Setup

Two bulbs with \(V_1=V_2\) are connected by a narrow tube of length \(L=0.15\) m and diameter \(d=1\) mm. The system is at \(T=25\ ^\circ\)C and \(P=1\) atm. Species \(A\) is \(N_2\) and species \(B\) is \(H_2\), with \(D_{AB}=0.784\) cm\(^2\)/s.

At \(t=0\):

Left bulb: \(x_{N_2}=1.00\), \(x_{H_2}=0.00\)
Right bulb: \(x_{N_2}=0.00\), \(x_{H_2}=1.00\)

At time \(t=t_1\):

Left bulb: \(x_{N_2}=0.80\)
Right bulb: \(x_{N_2}=0.25\)

Determine the molar fluxes \(N_A\) and \(N_B\) at \(t=t_1\) (include direction).
Find the value for \(v_{Ad}\) (diffusive velocity)

Example 3: Diffusion Through Stagnant B with Changing Path Length

Adapted from Geankoplis 6.2-3

Water vapor diffuses through a stagnant gas in a narrow vertical tube, dry air is constantly blown at the top of tube.
At time \(t\), the liquid level is a distance \(z\) from the tube top (i.e., the diffusion path length is \(z\)).
As diffusion proceeds, the liquid level drops slowly, so \(z\) increases with time.

Derive an expression for the time \(t_F\) required for the level to drop such that the diffusion path length changes from \(z=z_0\) at \(t=0\) to \(z=z_F\) at \(t=t_F\).

Hint: use pseudo steady-state assumption

Example 4: Determine \(D_{AB}\) Through Evaporation

Adapted from Griskey 10-2

Sample setup as example 4, a vertical tube of diameter \(D=0.01128\) m contains a liquid volatile species \(A\) (chloropicrin, \(CCl_3NO_2\)) evaporating into stagnant air (\(B\)) at 1 atm. The gas-phase diffusion of \(A\) occurs through the air column above the liquid surface.

At \(t=0\), the distance from the tube top to the liquid surface is \(z_0 = 0.0388\) m, after \(t=1\) day, the distance is \(z_1 = 0.0412\) m.

Vapor pressure at the interface: \(p_{A1} = 3178.3\) Pa
Liquid density: \(\rho_A = 1650\) kg/m\(^3\)
Molecular weight: \(M_A = 164.39\) kg/kmol

Use your expression from example 3, determine the binary diffusivity \(D_{AB}\) of \(A\) in air.

Summary

Compare diffusion in gas and in liquid
\(D_{AB}\vert_{l} \ll D_{AB}\vert_{g} \\ D_{AB}\vert_{s}\)
Theories for predicting diffusivity in liquid, and when to apply them