Stokes-Einstein Equation: Diffusion and Particle Radius
The Stokes-Einstein equation is a fundamental bridge between microscopic thermal fluctuations and macroscopic transport phenomena in solutions. In postgraduate and competitive examinations like CSIR NET, GATE, and IIT-JAM, transport properties appear repeatedly in physical chemistry papers. Many candidates lose valuable marks not because the conceptual theory is difficult, but because they stumble over SI unit conversions, dynamic viscosity units such as poise and centipoise, or scaling factors with nanometre-scale hydrodynamic radii. Mastering this formula allows you to quickly solve questions relating diffusion coefficients, molecular sizes, and temperature-dependent solvent viscosities.
Physical Basis of the Stokes-Einstein Relation
When a solute particle moves through a liquid, it experiences two balancing forces. First, thermal kinetic energy drives random Brownian motion, causing particles to spread from regions of higher concentration to lower concentration. Macroscopically, this motion is governed by Fick's laws of diffusion through the diffusion coefficient, D. Second, the surrounding solvent resists particle displacement through viscous drag. At low velocities and small length scales typical of molecular and colloidal systems, the Reynolds number is very small, meaning viscous forces dominate inertial forces.
Under these laminar conditions, the hydrodynamic frictional drag force F opposing motion is proportional to the drift velocity v of the particle: F = f × v, where f is the frictional coefficient. George Gabriel Stokes showed that for a rigid sphere of radius r moving smoothly through a continuous fluid of dynamic viscosity η under stick boundary conditions, the frictional coefficient is f = 6 × π × η × r. Albert Einstein connected this hydrodynamic friction to thermal kinetic energy through the fluctuation-dissipation theorem: D = (k_B × T) / f. Combining these relationships yields the classical Stokes-Einstein equation.
The Stokes-Einstein Formula, Variables, and Units
Depending on whether calculations are performed on a single-molecule basis or a molar basis, the equation is written in two mathematically identical forms:
Stokes-Einstein Equation (Stick boundary condition):
D = (k_B × T) / (6 × π × η × r) = (R × T) / (6 × π × η × r × N_A)
Where:
D = translational diffusion coefficient in square metres per second (m2 s-1)
k_B = Boltzmann constant = 1.380649 × 10-23 J K-1 (kg m2 s-2 K-1)
R = universal gas constant = 8.31446 J mol-1 K-1 (kg m2 s-2 mol-1 K-1)
N_A = Avogadro constant = 6.02214 × 1023 mol-1
T = absolute temperature in Kelvin (K)
η = dynamic viscosity of the solvent in Pascal-seconds (Pa s = kg m-1 s-1 = N s m-2)
r = hydrodynamic radius (Stokes radius) of the spherical particle in metres (m)
π = mathematical constant ≈ 3.14159
Standard physical chemistry textbooks also address the boundary condition between the particle surface and solvent molecules. When the diffusing particle is substantially larger than the solvent molecules (such as a protein, polymer, or colloidal particle in water), the solvent adheres tightly to the surface, creating zero relative velocity at the interface. This corresponds to the standard stick (no-slip) condition with f = 6 × π × η × r. When the solute particle has a similar size to the solvent molecules, slip boundary conditions may apply, where tangential viscous shear vanishes at the boundary, yielding f = 4 × π × η × r and D = (k_B × T) / (4 × π × η × r). In competitive exams, always assume stick boundary conditions unless the question specifically directs otherwise.
Detailed Worked Examples
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: CSIR NET Chemical Sciences, GATE Chemistry, IIT-JAM Chemistry, CUET-PG
- Topic: Solutions and Transport Properties — Diffusion and Hydrodynamic Radius
- Difficulty: Advanced Postgraduate Level
Question: A spherical globular protein diffuses in an aqueous buffer solution at 298.15 K. The dynamic viscosity of the aqueous medium at this temperature is 0.891 cP (0.891 × 10-3 Pa s). Dynamic light scattering measurements establish that the hydrodynamic radius of the protein is 2.50 nm (2.50 × 10-9 m). Calculate the translational diffusion coefficient D of this protein in m2 s-1, assuming standard stick boundary conditions. Use k_B = 1.38065 × 10-23 J K-1.
Step-by-step Solution:
Step 1: Write down the primary formula for diffusion under stick conditions:
D = (k_B × T) / (6 × π × η × r)
Step 2: Collect all given parameters in coherent SI units:
Temperature T = 298.15 K
k_B = 1.38065 × 10-23 kg m2 s-2 K-1
η = 0.891 cP = 0.891 × 10-3 Pa s = 8.91 × 10-4 kg m-1 s-1
r = 2.50 nm = 2.50 × 10-9 m
Step 3: Evaluate the thermal energy in the numerator:
Numerator = (1.38065 × 10-23 J K-1) × 298.15 K = 4.11641 × 10-21 J
Step 4: Compute the frictional denominator (6 × π × η × r):
6 × π ≈ 6 × 3.14159265 = 18.84956
Denominator = 18.84956 × (8.91 × 10-4 kg m-1 s-1) × (2.50 × 10-9 m)
Denominator = 18.84956 × 2.22750 × 10-12 kg s-1 = 4.19874 × 10-11 kg s-1
Step 5: Divide numerator by denominator:
D = (4.11641 × 10-21 kg m2 s-2) / (4.19874 × 10-11 kg s-1) = 9.8039 × 10-11 m2 s-1
Step 6: Double-check the calculation using the molar gas constant route:
Numerator = R × T = 8.31446 × 298.15 = 2478.96 J mol-1
Denominator = 6 × π × η × r × N_A = (4.19874 × 10-11) × (6.02214 × 1023) = 2.52854 × 1013
D = 2478.96 / (2.52854 × 1013) = 9.8039 × 10-11 m2 s-1. Both routes match identically.
Answer: 9.80 × 10-11 m2 s-1
Question: An uncharged organic dye molecule is dissolved in pure ethanol at 293.15 K (20.0 °C). An electrochemical experiment determines its translational diffusion coefficient to be D = 5.25 × 10-10 m2 s-1. The dynamic viscosity of ethanol at 293.15 K is 1.20 cP (1.20 × 10-3 Pa s). Assuming stick boundary conditions and a spherical geometry, compute the hydrodynamic radius r of the dye molecule in nanometres. Use k_B = 1.38065 × 10-23 J K-1.
Step-by-step Solution:
Step 1: Rearrange the Stokes-Einstein equation to solve for hydrodynamic radius r:
r = (k_B × T) / (6 × π × η × D)
Step 2: Calculate the numerator (k_B × T):
Numerator = (1.38065 × 10-23 J K-1) × 293.15 K = 4.04738 × 10-21 J
Step 3: Evaluate the denominator (6 × π × η × D):
6 × π ≈ 18.84956
η = 1.20 × 10-3 Pa s
D = 5.25 × 10-10 m2 s-1
η × D = (1.20 × 10-3) × (5.25 × 10-10) = 6.30000 × 10-13 kg m s-2
Denominator = 18.84956 × 6.30000 × 10-13 = 1.18752 × 10-11 kg m s-2
Step 4: Compute the radius in metres:
r = (4.04738 × 10-21 kg m2 s-2) / (1.18752 × 10-11 kg m s-2) = 3.40826 × 10-10 m
Step 5: Convert metres to nanometres (1 nm = 10-9 m):
r = (3.40826 × 10-10 m) / (10-9 m nm-1) = 0.3408 nm ≈ 0.341 nm
Step 6: Double-check by substituting r = 3.40826 × 10-10 m back into the diffusion expression:
D = (4.04738 × 10-21) / [6 × π × (1.20 × 10-3) × (3.40826 × 10-10)] = 4.04738 × 10-21 / 7.70929 × 10-12 = 5.25 × 10-10 m2 s-1. The result verifies completely.
Answer: 0.341 nm (3.41 × 10-10 m)
Question: A colloidal silica nanoparticle of invariant radius r = 15.0 nm is suspended in water. At initial temperature T_1 = 298.15 K, water has a dynamic viscosity of 1.000 × 10-3 Pa s, and the particle exhibits diffusion coefficient D_1. The dispersion is heated to T_2 = 348.15 K, where the dynamic viscosity of water falls to 0.378 × 10-3 Pa s. Assuming the particle radius does not change with temperature, calculate the ratio of the diffusion coefficients D_2 / D_1.
Step-by-step Solution:
Step 1: Set up the ratio using the proportionalities of the Stokes-Einstein equation:
Since k_B, 6, π, and r are constant, D is directly proportional to T / η:
D_2 / D_1 = (T_2 / T_1) × (η_1 / η_2)
Step 2: Compute the absolute temperature ratio:
T_2 / T_1 = 348.15 K / 298.15 K = 1.16770
Step 3: Compute the inverse viscosity ratio:
η_1 / η_2 = (1.000 × 10-3 Pa s) / (0.378 × 10-3 Pa s) = 1.000 / 0.378 = 2.64550
Step 4: Multiply both ratios to obtain D_2 / D_1:
D_2 / D_1 = 1.16770 × 2.64550 = 3.08915 ≈ 3.09
Step 5: Verify by direct numerical evaluation of both diffusion constants:
D_1 = (1.38065 × 10-23 × 298.15) / [6 × π × (1.000 × 10-3) × (15.0 × 10-9)] = (4.11641 × 10-21) / (2.82743 × 10-10) = 1.45589 × 10-11 m2 s-1
D_2 = (1.38065 × 10-23 × 348.15) / [6 × π × (0.378 × 10-3) × (15.0 × 10-9)] = (4.80674 × 10-21) / (1.06877 × 10-10) = 4.49745 × 10-11 m2 s-1
Ratio = (4.49745 × 10-11) / (1.45589 × 10-11) = 3.08914 ≈ 3.09. The check is completely consistent.
Answer: 3.09
Common Mistakes and Exam Pitfalls
1. Viscosity Unit Conversions: The CGS unit for dynamic viscosity is the poise (P), where 1 P = 0.1 Pa s. Exam questions regularly state solvent viscosity in centipoise (cP), where 1 cP = 10-2 P = 10-3 Pa s = 1 mPa s. Inserting viscosity directly in cP without multiplying by 10-3 causes the calculated diffusion coefficient to be incorrect by three orders of magnitude.
2. Leaving Radii in Nanometres or Angstroms: Hydrodynamic radii are typically given in nanometres (1 nm = 10-9 m) or angstroms (1 Å = 10-10 m). You must convert r into metres before performing arithmetic operations in SI units.
3. Using Celsius Instead of Kelvin: Temperature must strictly be in Kelvin. Entering 20 instead of 293.15 K will alter your thermal energy numerator by more than an order of magnitude.
4. Mixing Molar and Microscopic Constants: Do not combine R with single-particle frictional coefficients without dividing by Avogadro's constant N_A. Always verify that units cancel cleanly into m2 s-1.
5. Neglecting Solvent Viscosity Changes with Temperature: When temperature increases, diffusion speeds up not only because T increases linearly, but also because liquid viscosity η drops sharply. Both factors must be included in ratio problems.
Exam Relevance and Question Patterns
The Stokes-Einstein equation appears across physical chemistry curricula in both theoretical and practical contexts. The table below outlines how this concept is examined across major postgraduate entrance tests:
| Competitive Examination | Question Focus and Typical Context | Mathematical Format |
|---|---|---|
| CSIR NET Chemical Sciences | Hydrodynamic radius determination of macromolecules, linkage to electrochemical diffusion currents, and temperature-viscosity ratio problems | Multiple-choice conceptual and numerical problem sets |
| GATE Chemistry | Absolute numerical evaluation of D, particle sizing from dynamic light scattering data, and stick versus slip friction factors | Numerical Answer Type (NAT) requiring exact decimal rounding |
| IIT-JAM Chemistry | Proportionality relationships between D, T, η, and solute radius; ratio problems between different solvent conditions | Multiple Choice Questions (MCQ) and Numerical Answer Type (NAT) |
| CUET-PG Chemistry | Direct formula identification, fundamental unit analysis, and straightforward single-variable substitutions | Standard single-correct multiple-choice questions |
Previous-year CSIR-NET questions and timed quizzes to check what you just read.
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