Langmuir Isotherm Calculator

Find the fraction of an adsorbent surface covered at a given pressure.

Solves coverage, K_eq or pressure, with eight pressure units and a molar-concentration scale.

Updated October 16, 2026

CreatorFrank Zhao

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Introduction

A surface has a limited number of places where a molecule can settle. Tell this page how strongly the adsorbate sticks and how much of it is in the gas or the solution, and it returns how much of the surface is occupied — as a fraction from 0 to 1 and as a percent from 0 to 100.

The relation is the Langmuir isotherm, and it ties three quantities together: the coverage $\theta$ , the adsorption equilibrium constant $K_{\mathrm{eq}}$ and the partial pressure or concentration $P$ at equilibrium. Any one of the three follows from the other two, so the same page predicts a coverage from a known constant, checks a measured coverage against a fitted constant, or extracts a constant from a single measured point.

That covers most of what an adsorption isotherm is used for in practice: coursework and lab reports on gas–solid or liquid–solid adsorption, work on activated carbon and catalyst surfaces, and reading someone else's published fit. Because the same equation is written for a gas under a partial pressure and for a dilute solution under a concentration, the pressure box accepts a molar concentration too — and the two are kept apart rather than silently converted into each other.

Fraction and percent

The coverage on both scales, from one calculation

Any of the three

Enter two of the constant, the pressure and the coverage

Eight pressures plus mol/L

Pressures convert to pascals; the molar scale is read as typed

Quick start

Choose the scale your experiment was measured on

A gas-phase measurement goes in on a pressure scale — pascals, bar, atmospheres, torr, hectopascals, kilopascals, millimetres of mercury or centimetres of water column. Adsorption from a liquid goes in on the molar scale instead. The choice is not cosmetic: it fixes what the equilibrium constant has to mean, so pick it before typing anything else. If your liquid work is a mass and a volume rather than a concentration, the molarity calculator turns it into moles per litre first.

Enter the equilibrium constant per pascal

On the pressure scales this number belongs to the pascal, not to whichever unit is on display, because the pressure you type is converted to pascals before the two are multiplied. A constant published per bar, per kilopascal or per torr therefore needs one division before it goes in — the table in the next section has the factors. The molar scale is the exception: there the constant is read per mole per litre exactly as you type it.

Enter the pressure or concentration you are testing

Any value above zero is accepted, in any of the scales. Zero is refused: it describes a surface nothing has reached, which is the $\theta = 0$ limit the equation cannot be turned around from.

Read the coverage in both forms

Take a constant of 0.002 — which is $2\ \mathrm{kPa^{-1}}$ written per pascal — and a run at 3 kPa. The product is 6, so the coverage is six parts in seven:

\theta

=

\frac{K_{\mathrm{eq}} P}{1 + K_{\mathrm{eq}} P}

=

\frac{0.002 \times 3000}{1 + 0.002 \times 3000}

=

0.857143

The empty boxes fill in as 0.857143 and 85.7143 %, and the two values you typed stay as you left them. The percent box is the same coverage on a different scale rather than a second quantity, so the two always agree.

Or start from a coverage you measured

Leave the constant or the pressure empty instead, type the measured coverage, and the page works back to it. One thing is worth knowing before you trust that number: the reverse forms divide by $1 - \theta$ , so a constant recovered near saturation is far more sensitive to a reading error than one recovered in the middle of the curve. The tips section works through the numbers.

The isotherm and its units

The Langmuir picture is simple enough to state in one line: adsorption stops at a single layer, every site behaves identically, an occupied site does not influence its neighbours, and the surface is at equilibrium with the surrounding gas or solution. Under those assumptions the coverage depends only on the product of the constant and the pressure.

\theta = \frac{K_{\mathrm{eq}}\,P}{1 + K_{\mathrm{eq}}\,P}

$\theta$ is the fraction of surface sites occupied, from 0 (bare) towards 1 (a complete layer). $K_{\mathrm{eq}}$ is the adsorption equilibrium constant — large means the surface fills at a low pressure, small means a lot of pressure is needed. $P$ is the partial pressure of the gas, or the molar concentration of the solution, once equilibrium has been reached.

Half coverage is the landmark point

Because only the product $K_{\mathrm{eq}} P$ matters, the curve always has the same shape and only its horizontal scale changes. It passes exactly through half coverage when that product is 1, and it flattens out afterwards:

K_{\mathrm{eq}} P = 1

\Rightarrow

\theta = 0.5

\qquad

K_{\mathrm{eq}} P = 10

\Rightarrow

\theta = 0.909091

\qquad

K_{\mathrm{eq}} P = 100

\Rightarrow

\theta = 0.990099

Ten times the pressure buys only about five more points of coverage once the midpoint is behind you, and a hundred times the pressure buys less than nine. That saturation is the reason a curve that keeps climbing steeply at high pressure is telling you the single-layer picture no longer holds.

The two reverse forms, and why full coverage is a wall

K_{\mathrm{eq}}

=

\frac{\theta}{P\,(1 - \theta)}

\qquad

P

=

\frac{\theta}{K_{\mathrm{eq}}\,(1 - \theta)}

Both of them divide by $1 - \theta$ . A coverage of exactly 1 would need the product $K_{\mathrm{eq}} P$ to be infinite, so there is no finite pressure and no finite constant behind a reading of 100 % — the page leaves those boxes empty and explains why rather than printing a meaningless number. Anything below 100 % works, however close it gets.

Putting a published constant on the pascal scale

A constant quoted in the literature comes with whatever pressure scale the authors worked in, and it has to be divided by the number of pascals in that unit before it means the same thing here. The solution scale is the one entry that is not converted at all.

Constant quoted per	Divide it by	What to type in the equilibrium constant box
$\mathrm{bar^{-1}}$	100 000	$1\ \mathrm{bar^{-1}} = 0.00001\ \mathrm{Pa^{-1}}$
$\mathrm{kPa^{-1}}$	1 000	$2\ \mathrm{kPa^{-1}} = 0.002\ \mathrm{Pa^{-1}}$
$\mathrm{atm^{-1}}$	101 325	$1\ \mathrm{atm^{-1}} \approx 0.0000098692\ \mathrm{Pa^{-1}}$
$\mathrm{torr^{-1}}$	133.32236842105263	$1\ \mathrm{torr^{-1}} \approx 0.0075006168\ \mathrm{Pa^{-1}}$
$\mathrm{mmHg^{-1}}$	133.322387415	$1\ \mathrm{mmHg^{-1}} \approx 0.0075006158\ \mathrm{Pa^{-1}}$
$\mathrm{cmH_2O^{-1}}$	98.0665	$1\ \mathrm{cmH_2O^{-1}} \approx 0.0101971621\ \mathrm{Pa^{-1}}$
$(\mathrm{mol/L})^{-1}$	nothing	$133.333\ \mathrm{L/mol}\ \text{is typed as } 133.333$

Torr and millimetres of mercury are close relatives but not the same number of pascals — $101325/760 = 133.32236842105263$ against the conventional $133.322387415$ — so a constant converted with one of those factors reproduces a reading on the other scale to about seven significant figures. The centimetre of water column is the conventional $98.0665$ Pa, which is the product of standard gravity and the density of water rather than a defined exact conversion.

What the equation assumes

A single adsorbed layer, chemically equivalent sites, no interaction between adsorbed molecules, an ideal gas or dilute solution, and a surface that has genuinely reached equilibrium with its surroundings. Real adsorbents are rarely that tidy: pores of different sizes and defect sites make some positions much more attractive than others, and at higher pressures most systems move on to multilayer adsorption or capillary condensation. Those are the cases where a measured isotherm bends upward away from the Langmuir curve instead of flattening.

Worked examples

Recover the constant from an isotherm midpoint

Situation: A gas-adsorption run on a powdered solid holds the pressure at 0.8 kPa until the coverage stops changing, and the midpoint of the measured curve sits at half the surface.

Entered

Equilibrium constantleft empty

Partial pressure0.8 kPa

Surface fraction0.5

Surface percentfollows the fraction

Returned

K_{\mathrm{eq}}

=

\frac{\theta}{P\,(1 - \theta)}

=

\frac{0.5}{800 \times 0.5}

=

0.00125\ \mathrm{Pa^{-1}}

That is the same surface as $1.25\ \mathrm{kPa^{-1}}$ , and the pressure box can stay on kilopascals while the constant is typed per pascal — the two entries just have to agree with each other.

In practice: Check the constant on a second point before you believe it. The same value predicts 0.789474 (78.9474 %) at 3 kPa and 0.6 (60 %) at 1.2 kPa; if the measured curve disagrees badly with either, the surface is not one layer of identical sites, however well the midpoint fitted.

Two similar numbers, two very different surfaces

Situation: A published constant reads 2 per kilopascal and the run was at 3 kPa, so both numbers go in as they stand: 2 into the constant box, 3 with kilopascals selected.

Entered

Equilibrium constant2 (meant as 2 per kPa)

Partial pressure3 kPa

Surface fractionleft empty

Surface percentleft empty

Returned

\theta

=

\frac{2 \times 3000}{1 + 2 \times 3000}

=

\frac{6000}{6001}

=

0.999833

The pressure box converted 3 kPa into 3000 Pa before multiplying, so the constant had to be per pascal for the product to mean anything. Entering 0.002 gives $\frac{6}{7} = 0.857143$ , the coverage the paper describes.

In practice: The constant has to be per pascal, and 2 per kilopascal works out to 0.002 per pascal. Typed that way, the same 3 kPa returns 0.857143 (85.7143 %) instead of 0.999833 — a difference of fourteen points of coverage from one missing conversion, and the wrong answer is the plausible-looking one.

Adsorption from solution, where the scale is moles per litre

Situation: A dye-adsorption experiment on a powdered adsorbent uses a 0.005 mol/L solution and reaches 40 % coverage once the solution and the solid have equilibrated.

Entered

Equilibrium constantleft empty

Partial pressure, molar scale0.005 mol/L

Surface percent40

Surface fractionfollows the percent

Returned

\theta

=

0.4

\qquad

K_{\mathrm{eq}}

=

\frac{0.4}{0.005 \times 0.6}

=

133.333\ \mathrm{L/mol}

The mass concentration to molar concentration converter covers the g/L and mg/L cases, and a solution prepared by mass and volume can go through the molarity route instead.

In practice: On the molar scale the constant carries litres per mole and needs no pascal conversion at all. If the analysis instead reports 2 mg/L of dye, convert it to moles per litre before typing it in — the mass concentration to molar concentration converter takes the molar mass and returns the number this box wants.

Practical tips

Measure near half coverage, not on the plateau

The sensitivity of the reverse form is $\frac{1}{\theta\,(1 - \theta)}$ , so it grows sharply as the surface fills. Read one percentage point low in the middle — 0.49 instead of 0.50 at 0.8 kPa — and the constant comes back 3.9 % smaller (0.00120098 instead of 0.00125). Read one point low on the plateau — 0.94 instead of 0.95 at 20 kPa — and it comes back 17.5 % smaller (0.000783333 instead of 0.00095). A single-point fit is worth much more from the middle of the curve.

Never write the constant without its scale

A bare $K_{\mathrm{eq}} = 0.002$ is ambiguous: per pascal or per kilopascal? The two describe surfaces that differ by a factor of a thousand. Record the scale alongside the number whenever you fit, report or reuse one — $0.002\ \mathrm{Pa^{-1}}$ and $2\ \mathrm{kPa^{-1}}$ are the same surface, and the box on this page expects the per-pascal form on every pressure scale.

Check whether you are holding a fraction or a percent

The two coverage boxes read different scales. Typing 0.5 into the fraction box says half the surface is covered; typing 0.5 into the percent box says half a percent is, and the constant it back-solves is about two hundred times smaller — 0.00000628141 instead of 0.00125 at 0.8 kPa. Decide which number your instrument or your spreadsheet column holds before it goes in.

One constant, one temperature

$K_{\mathrm{eq}}$ belongs to the adsorbate–adsorbent pair at the temperature the run was made at, which is why isotherms are measured at constant temperature in the first place. A constant fitted at 298 K carried over to a 273 K experiment gives a coverage that looks reasonable and is not the one you measured, and this page has no temperature input that could correct for it.

Two measured points are worth more than one

A single pressure and a single coverage determine a constant exactly, which also means they can never contradict it. Put the same constant back in and compare the coverage it predicts with a second measurement; that comparison, not the fit itself, is what tells you whether the surface follows this model.

Questions

Why does the pressure box offer moles per litre next to the pressures?

Because the same isotherm is used for adsorption out of a dilute solution, where the driving quantity is the equilibrium concentration rather than a partial pressure. The two scales are not convertible into each other — turning a pascal into a mole per litre needs a solubility or Henry's-law relation that this equation does not contain — so the molar entry is taken exactly as typed and the constant carries litres per mole. Every other entry in that box really is a pressure scale, converted to pascals before the multiplication.

Why is a coverage of exactly 100 % a dead end?

A full layer is a limit rather than a state you reach at a finite pressure: $\theta = \frac{K_{\mathrm{eq}} P}{1 + K_{\mathrm{eq}} P}$ stays below 1 for every finite value and only approaches 1 as the product grows without bound. Both reverse forms divide by $1 - \theta$ , so at exactly 100 % there is no finite pressure or constant to report, and the page leaves those boxes empty with an explanation instead of inventing one.

Can I compare this constant with a value from a paper?

Only once both are on the same pressure scale and the same temperature. A constant of $1\ \mathrm{bar^{-1}}$ is $0.00001\ \mathrm{Pa^{-1}}$ , while one per torr is about $0.0075\ \mathrm{Pa^{-1}}$ , so two published numbers that look wildly different can describe the same surface. If what you have is the equilibrium constant of a reaction in solution rather than a surface constant, that is a different quantity, and the equilibrium constant calculator solves it from concentrations.

Does this page give me the adsorption capacity in mg per gram?

No. The output is coverage — how much of the surface is occupied — which is not the same as how many milligrams one gram of solid holds. Converting between them needs the number of sites or the monolayer capacity of your particular material, and neither follows from a pressure and a constant alone, so that step has to come from your own measurements of the amount adsorbed.

Limitations and scope

One layer, one kind of site, no neighbours. The equation assumes a single adsorbed layer, chemically equivalent sites, no interaction between adsorbed molecules and an equilibrium that has actually been reached. Real adsorbents have pores of different sizes and more attractive defect sites, and many systems switch to multilayer adsorption at higher pressure — the point where a measured isotherm stops flattening and starts climbing again.
The constant is only as good as its scale and its temperature. A constant entered per a different pressure unit, or fitted at another temperature, gives a coverage that looks plausible and is wrong. Nothing in the box can reveal which scale a plain number came from, so that check is yours.
Coverage is not a capacity, and one point is not a fit. The page takes your constant at face value: it does not fit a curve, weigh your data points or report an uncertainty, and a single pressure with a single coverage determines a constant exactly without ever testing the model behind it.
Not a basis for process or safety decisions. Column sizing, breakthrough times, catalyst loading and anything involving hazardous or regulated materials need measured isotherm data and a material specification. Treat this page as an educational and estimating tool and verify the numbers that matter against your own measurements.

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