Adsorption isotherms and fixed-bed adsorption

Physisorption and chemisorption, common adsorbents, Henry, Langmuir, Freundlich and BET isotherms, batch adsorption balances, and fixed-bed breakthrough, mass-transfer zone, LUB scale-up and regeneration.

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Why it matters

Adsorption removes trace impurities that other operations cannot reach economically: organics and colour from water, solvent vapours from air, moisture from natural gas, and it separates air into oxygen and nitrogen by pressure-swing adsorption. Design rests on two things – the equilibrium isotherm (how much the adsorbent can hold) and the breakthrough behaviour of a fixed bed (how long it lasts before the outlet goes off-spec).

Key ideas

Adsorption is the accumulation of a solute (adsorbate) from a fluid on the surface of a solid (adsorbent), mostly on internal pore surfaces (activated carbon has 500–1500 m²/g).

  • Physical adsorption (physisorption): van der Waals forces, low heat of adsorption (comparable to condensation), fast, reversible, multilayer possible, decreases with temperature. Basis of most separations.
  • Chemisorption: chemical bonds, high heat of adsorption, monolayer only, often slow and activated, sometimes irreversible. Basis of catalysis. Both are exothermic, so equilibrium uptake falls as temperature rises; chemisorption rates may rise with temperature because they are activated.

Common adsorbents. Activated carbon (organics, hydrophobic), silica gel and activated alumina (drying), zeolites/molecular sieves (size-selective, drying, air separation), polymeric resins.

Isotherms (equilibrium at constant T).

  • Linear (Henry's law): q = K·C, valid at low concentration.
  • Langmuir: monolayer on a fixed number of identical sites, no interaction between adsorbed molecules. q rises linearly at low C and saturates at q_max.
  • Freundlich: empirical, heterogeneous surfaces; 1/n < 1 means favourable adsorption. No saturation limit.
  • BET: multilayer physisorption; used to measure surface area from N₂ adsorption at 77 K. Favourable (concave-down) isotherms give sharp, self-sharpening fronts in fixed beds; unfavourable ones give spreading fronts.

Batch (stirred-tank) adsorption. A solute balance between liquid and solid plus the isotherm gives the final concentration; repeating with fresh adsorbent in stages (cross-current) or counter-current reduces adsorbent use.

Fixed-bed adsorption. Feed flows through a packed bed. Near the inlet the adsorbent saturates; downstream a mass-transfer zone (MTZ) moves through the bed in which concentration falls from C_F to nearly zero. The outlet concentration versus time is the breakthrough curve:

  • Breakpoint t_b: outlet reaches the allowed limit (often 5–10 % of C_F); the bed is taken offline and regenerated.
  • Exhaustion: outlet ≈ C_F.
  • For an ideal (infinitely sharp) front, the bed would saturate completely at the stoichiometric time t*. In reality the MTZ leaves part of the bed unused at breakpoint – the length of unused bed (LUB). A narrow MTZ (fast mass transfer, favourable isotherm, low velocity, small particles) uses the bed better.
  • Scale-up: LUB measured in a test column stays nearly constant at the same velocity, so a longer bed adds full-capacity length.
  • Regeneration: temperature swing (TSA, hot gas or steam), pressure swing (PSA), or displacement/purge.

Formulas

Langmuir: q = q_max·K·C/(1 + K·C); linear form C/q = 1/(q_max·K) + C/q_max Freundlich: q = K_F·C^(1/n); linear form log q = log K_F + (1/n)·log C Henry: q = K_H·C

  • q – adsorbate per mass of adsorbent (mg/g or mol/kg); C – fluid concentration (mg/L or mol/m³); K – L/mg (or m³/mol); q_max – monolayer capacity.

Batch balance: V (C₀ − C_e) = m·q_e

  • V – liquid volume, L; m – adsorbent mass, g.

Fixed bed (neglecting fluid holdup in voids): Stoichiometric time: t* = q_F·ρ_b·L / (u·C_F) Length of unused bed: LUB = L·(1 − t_b/t*) Breakpoint for a new length L′ at the same velocity: t_b′ = t*′·(1 − LUB/L′), with t*′ = t*·L′/L

  • q_F – adsorbate loading in equilibrium with feed, kg/kg; ρ_b – bed bulk density, kg/m³; L – bed length, m; u – superficial velocity, m/min; C_F – feed concentration, kg/m³.

Worked examples

Example 1 – batch adsorption with a Langmuir isotherm (standard). Given: 2 L of water with C₀ = 100 mg/L; 2 g of activated carbon; Langmuir q_max = 50 mg/g, K = 0.1 L/mg.

  1. Balance: V(C₀ − C) = m·q_max·K·C/(1 + K·C) → 2(100 − C)(1 + 0.1C) = 2 × 50 × 0.1 × C = 10C.
  2. Expand: 200 + 18C − 0.2C² = 10C → 0.2C² − 8C − 200 = 0 → C² − 40C − 1000 = 0.
  3. C = 20 + √(400 + 1000) = 20 + 37.42 = 57.4 mg/L.
  4. q = 2 × (100 − 57.4)/2 = 42.6 mg/g (check: 50 × 0.1 × 57.4/(1 + 5.74) = 42.6 ✓).

C_e = 57.4 mg/L; q_e = 42.6 mg/g; 42.6 % removal. Near saturation (q close to q_max), adding more carbon helps more than longer contact.

Example 2 – fixed-bed breakthrough and scale-up (GATE level). Given: test bed L = 0.5 m, ρ_b = 500 kg/m³, feed C_F = 2 kg/m³, superficial velocity u = 0.1 m/min, equilibrium loading with feed q_F = 0.2 kg/kg. Breakthrough observed at t_b = 200 min. Predict t_b for a 1.0 m bed at the same velocity.

  1. t* = q_F·ρ_b·L/(u·C_F) = 0.2 × 500 × 0.5/(0.1 × 2) = 250 min.
  2. LUB = L(1 − t_b/t*) = 0.5 × (1 − 200/250) = 0.10 m.
  3. New bed: t*′ = 250 × 1.0/0.5 = 500 min.
  4. t_b′ = 500 × (1 − 0.10/1.0) = 450 min.

LUB = 0.10 m; the 1.0 m bed breaks through at about 450 min (bed utilisation rises from 80 % to 90 %).

Common mistakes

  • Using the Langmuir linear form with axes swapped and misreading slope and intercept.
  • Forgetting that adsorption equilibrium falls with temperature (physisorption is exothermic).
  • Treating Freundlich as having a maximum capacity.
  • Doubling the bed length and doubling breakthrough time – the unused bed length does not scale.
  • Ignoring units: mg/L with L/mg, kg/m³ with m³/kg.
  • Confusing breakthrough (outlet starts rising) with exhaustion (outlet equals feed).

For GATE CH

Expect Langmuir and Freundlich calculations (q from C, C from q, parameters from linearised data), batch adsorbent dose problems, stoichiometric time and LUB scale-up for fixed beds, and concepts – physisorption versus chemisorption, favourable isotherms, MTZ, PSA and TSA.

Quick check

  1. Which isotherm assumes a monolayer on identical sites?
  2. For Freundlich with 1/n < 1, is the isotherm favourable or unfavourable?
  3. What does LUB measure?
  4. Does equilibrium physisorption capacity rise or fall with temperature?

Answers: 1. Langmuir; 2. favourable; 3. the length of bed not used at breakthrough, caused by the finite mass-transfer zone; 4. it falls.

See it move

All Chemical animations
Adsorption Isotherms Visualization

Adjust the concentration of adsorbate and observe how the adsorption amount changes according to different isotherm models. Try different models to see how they behave.

Equations used
  • Langmuir Isotherm: q = (q_max * K * C) / (1 + K * C)
  • Freundlich Isotherm: q = K_f * C^(1/n)

Interview questions

All Mass Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What is adsorption isotherm in the context of mass transfer?Concept

    An adsorption isotherm is a curve that represents the relationship between the amount of adsorbate on the adsorbent and its concentration in the fluid phase at constant temperature. It helps in understanding how adsorbates interact with adsorbents and is crucial for designing adsorption systems.

  2. 2.Explain the difference between physisorption and chemisorption.Concept

    Physisorption involves weak van der Waals forces and is usually reversible, occurring at low temperatures. Chemisorption involves the formation of chemical bonds, is usually irreversible, and occurs at higher temperatures. Physisorption has lower enthalpy changes compared to chemisorption.

  3. 3.What is a fixed-bed adsorption process?Concept

    A fixed-bed adsorption process involves passing a fluid containing adsorbate through a packed bed of adsorbent particles. The adsorbate is captured on the surface of the adsorbent, and the process continues until the bed is saturated. It is commonly used in water treatment and air purification.

  4. 4.Why is the Langmuir isotherm model commonly used in adsorption studies?Application

    The Langmuir isotherm model is commonly used because it assumes monolayer adsorption on a surface with a finite number of identical sites, which is a good approximation for many systems. It provides a simple mathematical form that can be used to estimate adsorption capacity and affinity.

  5. 5.What happens if the flow rate of the fluid in a fixed-bed adsorption column is increased?Application

    At a higher flow rate solute is fed faster, so the bed saturates sooner and breakthrough comes earlier – roughly in proportion to the feed rate for the stoichiometric part. The shorter residence time also widens the mass-transfer zone, so the breakthrough curve is less sharp and a larger fraction of the bed (a longer unused bed) is wasted at the breakpoint. Pressure drop also rises, and very high velocities can fluidise or attrit the adsorbent in upflow.

  6. 6.How does temperature affect adsorption in a fixed-bed column?Application

    Adsorption is exothermic, so the equilibrium capacity falls as temperature rises – the basis of temperature-swing regeneration. Physisorption is fast, so its uptake simply drops with temperature. Chemisorption is also exothermic, but it is often activated, so its rate can rise with temperature over some range even though the equilibrium amount eventually falls. In large beds the heat released can create a temperature front that reduces capacity and, with reactive vapours on carbon, can be a safety concern.

  7. 7.What is the significance of breakthrough curves in fixed-bed adsorption?Application

    Breakthrough curves represent the concentration of adsorbate in the effluent as a function of time. They are significant because they help determine the saturation point of the adsorbent and the efficiency of the adsorption process, aiding in the design and optimization of adsorption systems.

  8. 8.Calculate the amount of adsorbate adsorbed per unit mass of adsorbent using the Langmuir isotherm, given q_max = 10 mg/g, K_L = 0.5 L/mg, and C_e = 2 mg/L.Numerical

    Using the Langmuir isotherm equation: q = (q_max * K_L * C_e) / (1 + K_L * C_e). Substituting the given values: q = (10 * 0.5 * 2) / (1 + 0.5 * 2) = 10 / 2 = 5 mg/g.

  9. 9.Determine the equilibrium concentration of adsorbate in the fluid phase if the amount adsorbed is 4 mg/g, q_max = 8 mg/g, and K_L = 0.25 L/mg using the Langmuir isotherm.Numerical

    Rearrange the Langmuir isotherm equation to solve for C_e: C_e = q / (K_L * (q_max - q)). Substituting the given values: C_e = 4 / (0.25 * (8 - 4)) = 4 / 1 = 4 mg/L.

  10. 10.Explain why activated carbon is often used as an adsorbent in fixed-bed adsorption processes.Application

    Activated carbon is often used because it has a high surface area, which provides more sites for adsorption. It is also versatile, effective for a wide range of adsorbates, and can be regenerated for reuse, making it cost-effective for many applications.

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