Membrane transport (Kedem–Katchalsky): Which expression correctly gives the solvent volume flux through a semi-permeable membrane in terms of membrane hydraulic permeability (Lp), applied pressure difference (Δp), osmotic pressure difference (Δπ), and reflection coefficient (σ)?

Introduction:
Coupled flow of solvent and solute across membranes is described by nonequilibrium thermodynamics (Kedem–Katchalsky). Exams routinely ask for the form of the solvent flux equation that accounts for both hydraulic pressure and osmotic forces modulated by the reflection coefficient σ.

Given Data / Assumptions:

• J denotes solvent volume flux (per unit area, e.g., m/s).
• Lp is hydraulic permeability of the membrane.
• Δp is transmembrane pressure; Δπ is osmotic pressure difference.
• σ (0 ≤ σ ≤ 1) quantifies how effectively the membrane reflects solute.

Concept / Approach:
In pressure-driven transport, solvent flux is reduced by the opposing osmotic force proportional to σΔπ. When σ = 1 (perfect rejection), osmotic pressure fully opposes Δp; when σ = 0 (solute passes freely), there is no osmotic opposition. The linear phenomenological relation is J = Lp*(Δp − σΔπ).

Step-by-Step Solution:
Start with Kedem–Katchalsky: J = Lp*(Δp − σΔπ).Interpret σ: perfect semipermeability (σ = 1) gives J = Lp*(Δp − Δπ).If Δp = σΔπ, net solvent flux is zero (osmotic equilibrium).Units check: Lp (m·Pa^-1·s^-1) × Pa = m·s^-1 ⇒ consistent.

Verification / Alternative check:
Reverse osmosis operation shows flux declines as feed osmotic pressure rises; the reduction scales with σ, matching the equation’s prediction across different salts and membranes.

Why Other Options Are Wrong:
J/A = Lp*(Δp − σΔπ): J already has area in its definition; dividing by A again is incorrect.

(Lp/σ)Δπ and Lp/Δp forms: dimensionality and physics are wrong; they omit the necessary pressure difference structure.

Common Pitfalls:

• Confusing flux (per area) with volumetric flow rate (Q = JA).
• Forgetting σ’s range and physical meaning.

Final Answer:
J = Lp * (Δp − σΔπ)