Langmuir Adsorption Isotherm Calculator
Calculate fractional surface coverage θ and adsorbed amount q from the Langmuir isotherm. Enter the equilibrium concentration or pressure, the affinity constant KL, and optionally the maximum capacity qmax.
Estimate Surface Coverage from Concentration and Affinity
Langmuir Adsorption Isotherm Formula
Where:
- \(\theta\) — fractional surface coverage (dimensionless, 0 to 1)
- \(K_L\) — Langmuir affinity constant \(= k_a / k_d\), with units reciprocal to those of \(C\)
- \(C\) — equilibrium concentration (or partial pressure for gases) of the adsorbate
- \(q_{max}\) — maximum adsorption capacity, the monolayer amount when every site is occupied
- \(q\) — adsorbed amount at equilibrium
The formula expresses the equilibrium between adsorption and desorption on a surface with a fixed number of identical, independent sites: the fraction of occupied sites rises hyperbolically with concentration and approaches 1 (complete monolayer) only asymptotically.
Langmuir Adsorption Isotherm – Calculation Example
Example: an adsorbent with a monolayer capacity of 100 mg/g is equilibrated in a solution containing 0.3 mmol/L of adsorbate, and the Langmuir affinity constant is \(K_L = 5\) L/mmol.
\[ K_L \, C = 5 \ \tfrac{\text{L}}{\text{mmol}} \times 0.3 \ \tfrac{\text{mmol}}{\text{L}} = 1.5 \]
\[ \theta = \frac{1.5}{1 + 1.5} = 0.6 \]
\[ q = q_{max} \cdot \theta = 100 \ \tfrac{\text{mg}}{\text{g}} \times 0.6 = 60 \ \tfrac{\text{mg}}{\text{g}} \]
Result: θ = 0.6 (60% of the adsorption sites are occupied) and the adsorbed amount is q = 60 mg/g, with 40 mg/g of capacity still unused.
How to Use the Calculator
- Enter the equilibrium Concentration / Pressure (C) of the adsorbate and choose its unit — molar or mass concentration for solutions, or a pressure unit (atm, kPa, bar, Pa, mmHg) for gas adsorption.
- Enter the Langmuir Affinity Constant (KL) and pick the matching reciprocal unit (e.g. mmol/L goes with L/mmol, atm with 1/atm). The unit of \(K_L\) must be the inverse of the unit of \(C\).
- If you know the monolayer capacity, keep the checkbox "I know the maximum adsorption capacity (qmax)" ticked and enter Maximum Adsorption Capacity (qmax) with its unit; otherwise untick it and only the coverage θ will be computed.
- Read the Fractional Surface Coverage (θ) and the Adsorbed Amount (q). The results also include a table of coverage at other concentrations and a step-by-step calculation.
What Is the Langmuir Adsorption Isotherm?
The Langmuir isotherm, published by Irving Langmuir in 1918, is the simplest quantitative model of adsorption at equilibrium. It describes a surface composed of a fixed number of identical, independent sites, each of which can bind at most one molecule. Adsorption proceeds at a rate \(k_a C (1-\theta)\) onto empty sites, while desorption proceeds at a rate \(k_d\,\theta\); at equilibrium the two rates are equal, which gives \(K_L = k_a/k_d\) and the hyperbolic coverage expression \(\theta = K_L C/(1 + K_L C)\).
The parameter \(K_L\) measures the affinity of the adsorbate for the surface. A large \(K_L\) means strong binding: half of the sites are already occupied when \(C = 1/K_L\) (the half-saturation point), and high coverage is achieved at low concentrations. The capacity \(q_{max}\) represents complete monolayer coverage; it is approached only asymptotically, so a surface is never perfectly "full" at finite concentration.
The same mathematics describes saturation curves elsewhere in chemistry — the Langmuir isotherm is formally identical to the Michaelis–Menten rate equation — and it is the foundation on which multilayer models such as the BET surface area method are built. In practice, \(K_L\) and \(q_{max}\) are obtained by fitting measured equilibrium data, either by non-linear regression or by linearised plots such as \(C/q\) versus \(C\), whose slope is \(1/q_{max}\) and whose intercept is \(1/(K_L q_{max})\).
Assumptions, Limitations and Practical Tips
- Monolayer adsorption: each site holds one molecule at most. If multilayer adsorption or capillary condensation occurs, the Langmuir model underestimates the uptake.
- Homogeneous surface: all sites are assumed energetically identical. Real sorbents (activated carbons, clays, biomaterials) are often heterogeneous, and the Freundlich, Sips or Toth models usually fit such data better.
- No lateral interactions: adsorbed molecules do not affect one another; the heat of adsorption is constant and independent of coverage.
- Equilibrium and ideal behaviour: the system must have reached equilibrium, the solution must be dilute, and gases must behave ideally at the pressures used.
- Temperature dependence: \(K_L\) is an equilibrium constant, so it changes with temperature — for the usual exothermic adsorption it decreases as temperature rises. Always use a \(K_L\) measured at the temperature of interest.
- Unit consistency: because the product \(K_L C\) must be dimensionless, check that the units of \(C\) and \(K_L\) are reciprocals before trusting any published value.
A quick sanity check when interpreting results: if \(K_L C \ll 1\) you are in the linear Henry regime (\(\theta \approx K_L C\)), where coverage grows in direct proportion to concentration; if \(K_L C \gg 1\) the surface is nearly saturated and further increases in \(C\) barely change \(q\). The calculator's coverage table makes it easy to see which regime you are in at a glance.
Frequently Asked Questions
How do I calculate the Langmuir isotherm by hand?
First compute the dimensionless product \(K_L C\). Then divide it by one plus itself: \(\theta = K_L C/(1 + K_L C)\). For example, with \(K_L = 4\) L/mmol and \(C = 0.25\) mmol/L, you get \(K_L C = 1\), so \(\theta = 1/2 = 0.5\). If you need the adsorbed amount, multiply θ by \(q_{max}\). A quick sanity check: θ can never exceed 1, whatever the concentration.
What is a good or typical value of the Langmuir constant KL?
There is no universal "good" value — \(K_L\) depends entirely on the adsorbate–adsorbent pair, the solvent or gas, and the temperature. It always carries units reciprocal to those of C (e.g. L/mmol, L/mg, or 1/atm). What matters practically is the ratio \(C/(1/K_L)\): if C is well below \(1/K_L\) you are in the linear regime, if C is well above it the surface is nearly saturated. Compare \(K_L\) values only for the same system and temperature.
Why must the units of KL be the inverse of the units of C?
Because the product \(K_L C\) must be dimensionless for the formula \(\theta = K_L C/(1 + K_L C)\) to make sense — you cannot add 1 to a quantity with units. So if C is in mmol/L, \(K_L\) must be in L/mmol; if C is a partial pressure in atm, \(K_L\) is in 1/atm. A value reported as "L/mg" is not interchangeable with one in "L/mmol"; convert carefully using the molar mass before comparing.
What is the difference between the Langmuir and Freundlich isotherms?
Langmuir assumes a homogeneous surface with identical, independent sites and a finite monolayer capacity \(q_{max}\), so uptake saturates at high concentration. Freundlich is an empirical power law, \(q = K C^{1/n}\), with no maximum capacity; it describes heterogeneous surfaces but keeps rising with C. If your data plateau at high concentration, Langmuir (or the Sips/Toth hybrids) usually fits better; if they never saturate, Freundlich may be the better choice.
Can I use the Langmuir isotherm with gas pressure instead of concentration?
Yes. For gas adsorption, simply replace C with the partial pressure of the adsorbate and express \(K_L\) in reciprocal pressure units (1/atm, 1/kPa, etc.). The ideal-gas assumption must hold at the pressures used. Note that this pressure form is sometimes written with a dimensionless constant b multiplied by a reference pressure — the mathematics is identical once the units are made consistent.
How do I find KL and qmax from experimental data?
Measure the adsorbed amount q at several equilibrium concentrations, then fit either by non-linear regression (preferred, less biased) or by linearisation: plot \(C/q\) versus \(C\). The slope of that line equals \(1/q_{max}\) and the intercept equals \(1/(K_L q_{max})\), so \(K_L = \text{intercept}^{-1} \cdot \text{slope}^{-1} \cdot \) … solve as \(K_L = \text{slope}/\text{intercept}\). Linearised fits can distort errors, so validate with a non-linear fit when possible.
Why doesn't the adsorbed amount q ever reach qmax exactly?
Because the Langmuir hyperbola approaches complete monolayer coverage only asymptotically. At \(C = 1/K_L\) you reach 50% of \(q_{max}\); at ten times that concentration you reach about 91%; at a hundred times, about 99%. So a surface is never perfectly "full" at any finite concentration. If your measured uptake exceeds \(q_{max}\) or keeps growing, multilayer adsorption or condensation is likely occurring and another model (e.g. BET) is needed.