Henry's Law

Open a warm soda and bubbles stream out for minutes. Open a cold one and it stays fizzy much longer. The difference is not magic. It is Henry's law: a dilute solute's vapor pressure is proportional to its mole fraction in solution. This single proportionality explains why fish need cold, aerated water. It also explains why deep-sea divers risk "the bends." And it explains why carbonated drinks go flat faster on a warm day. Henry's law matters because most real solutes are not Raoult's-law solvents. Raoult's law works beautifully for the solvent in a dilute solution. But the dissolved gas or trace solute experiences a completely different molecular environment. It is surrounded by solvent molecules rather than its own kind. Henry's law is the empirical fix. A separate proportionality constant, the Henry's law constant, captures how that solute actually behaves at infinite dilution. This article builds Henry's law from the vapor-pressure curve of a real binary mixture. You will see why the solute's initial slope differs from Raoult's law. Then you will derive the Henry's law constant's role as a reference state. Finally, two worked numerical examples show how to compute gas solubilities from tabulated constants. One is foundational, one is exam-level.

Key takeaways

Henry's law states that a dilute solute's partial vapor pressure is proportional to its mole fraction: $p_B = K_B x_B$ .
The Henry's law constant $K_B$ is an empirical, solute-and-solvent-specific slope, not a universal number like $p_B^*$ in Raoult's law.
Henry's law holds only in the dilute limit, $x_B \to 0$ ; Raoult's law holds for the solvent-rich limit, $x_A \to 1$ .
Gibbs–Duhem consistency guarantees that if the solvent obeys Raoult's law near $x_A = 1$ , the solute must obey a linear (Henry's) law near $x_B = 0$ .
Henry's law constants explain gas solubility trends: O₂ and CO₂ dissolve more as partial pressure rises and less as temperature rises.
Key idea: at infinite dilution, a solute's escaping tendency is proportional to its concentration, but the proportionality constant reflects solute–solvent interactions, not the pure solute's vapor pressure.

The core insight

In a dilute solution the solute molecules are so surrounded by solvent that their environment—and hence their partial vapor pressure per unit mole fraction—no longer resembles the pure solute; Henry's law captures this new, solvent-dependent proportionality, $p_B = K_B x_B$ , as the correct dilute-limit reference behavior for real solutes.

The physical picture

Picture a beaker of water with a trace of dissolved oxygen, or a champagne flute with dissolved CO₂. Each dissolved gas molecule is essentially isolated, surrounded on all sides by solvent molecules. Its tendency to escape into the vapor phase — its partial vapor pressure — depends on how strongly it interacts with those solvent neighbors, not on how it would interact with more of itself, because there essentially is no "itself" nearby. Contrast this with the solvent. In a dilute solution the solvent is still overwhelmingly surrounded by other solvent molecules, just as in the pure liquid. That is why the solvent's partial vapor pressure obeys Raoult's law, $p_A = x_A p_A^*$ , right down to $x_A \to 1$ : its local environment barely changes upon dilute mixing. If you plot the solute's actual partial vapor pressure against its mole fraction across the whole composition range, you typically get a curve that only touches the Raoult's-law line (extrapolated pure-solute vapor pressure $p_B^*$ ) at $x_B = 1$ . Near $x_B = 0$ , the real curve has a different initial slope. That slope is the Henry's law constant $K_B$ : it is what you get by drawing the tangent to the true vapor-pressure curve at $x_B = 0$ and reading off its slope.

Figure

Real solute vapor pressure curve versus Raoult's and Henry's law lines

The key geometric fact: the real curve and the Henry's tangent line coincide only as $x_B \to 0$ . Away from that limit the two diverge, which is exactly why Henry's law is stated as a dilute-solution limiting law, not an equation valid at all concentrations.

The core relations

p_B = K_B\, x_B \qquad (x_B \to 0)

where: $p_B$ is the partial vapor pressure of solute B above the solution, $x_B$ its mole fraction in the liquid, and $K_B$ the Henry's law constant (units of pressure) for that solute–solvent pair.

p_A = x_A\, p_A^{*}

where: this is Raoult's law for the solvent A, valid in the same dilute limit ( $x_A \to 1$ ), with $p_A^{*}$ the solvent's pure vapor pressure.

x_B = \dfrac{p_B}{K_B}

where: rearranged form used to compute solubility — the equilibrium mole fraction of dissolved gas from its partial pressure above the solution.

K_H = \dfrac{p_B}{c_B} \quad \text{or} \quad K_H' = \dfrac{c_B}{p_B}

where: common chemistry/engineering variants express Henry's constant using molar concentration $c_B$ or molality instead of mole fraction; always check which convention a table uses.

Derivation AFrom Gibbs–Duhem consistency to Henry's law

Goal. Show that if the solvent A obeys Raoult's law near $x_A = 1$ , thermodynamic consistency (via the Gibbs–Duhem equation) forces the solute B to obey a linear law, $p_B = K_B x_B$ , near $x_B = 0$ , and that $K_B$ need not equal $p_B^{*}$ .

Step 1: Chemical potentials from vapor pressures. Treat the vapor above the solution as ideal gas. The chemical potential of each component in solution equals its chemical potential in the vapor phase at equilibrium:

\mu_A(\text{soln}) = \mu_A^{\ominus}(g) + RT\ln\!\left(\frac{p_A}{p^{\ominus}}\right), \qquad \mu_B(\text{soln}) = \mu_B^{\ominus}(g) + RT\ln\!\left(\frac{p_B}{p^{\ominus}}\right)

Step 2: Gibbs–Duhem constrains the two components together. At constant $T$ and $p$ , the Gibbs–Duhem equation states

x_A\, d\mu_A + x_B\, d\mu_B = 0

This means the composition-dependence of $\mu_A$ and $\mu_B$ are not independent — once you know how one chemical potential varies with composition, the other is fixed.

Step 3: Insert Raoult's law for the solvent. Suppose experimentally $p_A = x_A p_A^{*}$ holds as $x_A \to 1$ (i.e. $x_B \to 0$ ). Then

d\mu_A = RT\, d\ln p_A = RT\, d\ln x_A

Substitute into Gibbs–Duhem:

x_A RT\, d\ln x_A + x_B\, d\mu_B = 0

Since $d\ln x_A = \dfrac{dx_A}{x_A} = -\dfrac{dx_B}{x_A}$ (because $x_A + x_B = 1$ ), this becomes

-RT\, dx_B + x_B\, d\mu_B = 0 \quad\Longrightarrow\quad d\mu_B = RT\,\frac{dx_B}{x_B} = RT\, d\ln x_B

Step 4: Integrate. Integrating $d\mu_B = RT\, d\ln x_B$ gives

\mu_B = RT\ln x_B + \text{constant}

Comparing with $\mu_B = \mu_B^{\ominus}(g) + RT\ln(p_B/p^{\ominus})$ shows that $\ln p_B$ must also be linear in $\ln x_B$ with unit slope, so

p_B = (\text{constant}) \times x_B

Step 5: Identify the constant as $K_B$ , not $p_B^{*}$ . The integration constant is fixed by matching the real solute behavior at $x_B \to 0$ , where the solute is surrounded entirely by solvent A. There is no thermodynamic reason this constant should equal the pure-solute vapor pressure $p_B^{*}$ — that would require the solute's environment at infinite dilution in A to resemble pure B, which it does not. The constant is therefore an independent, empirically measured quantity:

\boxed{p_B = K_B\, x_B, \qquad K_B \neq p_B^{*} \text{ in general}}

This is the promised result: linearity is guaranteed by thermodynamic consistency (Gibbs–Duhem plus Raoult's law for the solvent), but the value of the slope $K_B$ is a genuine new empirical parameter reflecting solute–solvent interactions.

Worked examples

Example 1 — Foundational: solubility of O₂ in water from the Henry's law constant. The Henry's law constant for O₂ in water at 25 °C, in mole-fraction form, is

K_{\mathrm{O_2}} = 4.40 \times 10^{4}\ \text{atm}

(This is a standard tabulated value, e.g. Atkins.) Air is about 20.9% O₂ by mole fraction, and atmospheric pressure is 1.00 atm, so the partial pressure of O₂ above the water is

p_{\mathrm{O_2}} = 0.209 \times 1.00\ \text{atm} = 0.209\ \text{atm}

Apply Henry's law in the rearranged form:

x_{\mathrm{O_2}} = \frac{p_{\mathrm{O_2}}}{K_{\mathrm{O_2}}} = \frac{0.209\ \text{atm}}{4.40\times10^{4}\ \text{atm}} = 4.75\times10^{-6}

Convert to molar concentration. One liter of water is about 55.5 mol (since $1000\text{ g}/18.02\text{ g mol}^{-1} = 55.5\ \text{mol}$ ), and at such low $x_{\mathrm{O_2}}$ the total moles per liter is essentially unchanged, so

n_{\mathrm{O_2}} \approx x_{\mathrm{O_2}} \times 55.5\ \text{mol L}^{-1} = 4.75\times10^{-6} \times 55.5 = 2.64\times10^{-4}\ \text{mol L}^{-1}

In mass terms, that's $2.64\times10^{-4}\ \text{mol L}^{-1} \times 32.0\ \text{g mol}^{-1} = 8.4\times10^{-3}\ \text{g L}^{-1} \approx 8.4\ \text{mg L}^{-1}$ .

Sanity check: typical textbook and field values for dissolved oxygen in air-saturated water at 25 °C are quoted around 8 mg/L, matching well. This concentration is exactly why fish need well-aerated water — 8 mg/L is a small reservoir, easily depleted by warm temperatures (which lower solubility further) or by biological oxygen demand.

Example 2 — Applied, harder: pressure needed to carbonate a soft drink, and how much escapes on opening. Manufacturers carbonate soda to roughly $p_{\mathrm{CO_2}} = 4$ atm inside a sealed bottle at bottling temperature (about 10 °C for this problem). The Henry's law constant for CO₂ in water near 10 °C is approximately

K_{\mathrm{CO_2}}(10\,^{\circ}\mathrm{C}) \approx 1.04\times10^{3}\ \text{atm}

(Henry's constants for CO₂ decrease at lower temperature — gases are generally more soluble in cold water — versus the value near $1.64\times10^3$ atm at 25 °C.)

Step 1 — mole fraction dissolved at bottling.

x_{\mathrm{CO_2}} = \frac{p_{\mathrm{CO_2}}}{K_{\mathrm{CO_2}}} = \frac{4.00\ \text{atm}}{1.04\times10^{3}\ \text{atm}} = 3.85\times10^{-3}

Step 2 — convert to concentration. Using 55.5 mol water per liter as before,

c_{\mathrm{CO_2}} \approx 3.85\times10^{-3} \times 55.5\ \text{mol L}^{-1} = 0.214\ \text{mol L}^{-1}

That is about $0.214 \times 44.0\ \text{g mol}^{-1} = 9.4\ \text{g L}^{-1}$ of CO₂ — consistent with commercial soda's roughly 3–4 volumes of CO₂ per volume of liquid.

Step 3 — what happens when you open the bottle. Once opened, the gas space above the drink equilibrates with atmospheric CO₂ partial pressure, only about $p_{\mathrm{CO_2}} \approx 4\times10^{-4}$ atm (CO₂ is roughly 400 ppm of air). Now use Henry's law again, at the warmer serving temperature, say 20 °C, where $K_{\mathrm{CO_2}}(20\,^{\circ}\mathrm{C}) \approx 1.51\times10^{3}$ atm:

x_{\mathrm{CO_2}}^{\text{equilibrium, open}} = \frac{4\times10^{-4}\ \text{atm}}{1.51\times10^{3}\ \text{atm}} = 2.6\times10^{-7}

This is over 10,000 times smaller than the dissolved mole fraction at bottling ( $3.85\times10^{-3}$ ). The drink is wildly supersaturated relative to the new equilibrium, which is exactly why it fizzes vigorously and continues degassing until nearly all excess CO₂ escapes — a slow process because it is diffusion- and nucleation-limited, not because equilibrium favors retaining the gas.

Sanity check: the huge mismatch (four orders of magnitude) between sealed and open equilibrium mole fractions correctly predicts that an opened soda will keep fizzing for many minutes, and that warming it (lower $K_{\mathrm{CO_2}}$ ... wait, higher $K$ at higher $T$ means lower solubility) accelerates flatness, matching everyday experience.

Limiting cases & common traps

Situation	Correct treatment
Solute mole fraction not small	Henry's law fails; need full activity coefficient treatment (real/non-ideal solution)
Solvent near $x_A = 1$	Raoult's law applies to the solvent even while Henry's law applies to the solute — both hold simultaneously in the same dilute solution
Comparing $K_B$ across solutes	Never compare $K_B$ to $p_B^{*}$ directly — they describe different physical limits and are generally unequal
Units of $K_H$	Mole-fraction-based $K_B$ (atm), concentration-based $K_H$ (atm·L/mol), and molality-based constants are all in common use; always match units before calculating
Temperature dependence	Gas solubility usually decreases as $T$ rises (van't Hoff-type behavior), i.e. $K_B$ typically increases with $T$ for dissolved gases

The chemist behind it

William Henry was a Manchester physician and chemist working in the early 1800s, a contemporary and friend of John Dalton. Henry was investigating the "absorption of gases by water and other liquids" — motivated partly by practical questions in brewing and mineral-water manufacture — when he found, in 1803, that the mass of gas dissolved at a fixed temperature is directly proportional to the pressure of that gas above the liquid. His experiments used simple glass apparatus to measure gas volumes absorbed at controlled pressures, a remarkably careful quantitative study for its era. The law bearing his name became one of the earliest quantitative statements linking a measurable macroscopic property (pressure) to a solution property (solubility), predating the molecular/thermodynamic understanding of solutions by nearly a century.

Origins · 1 minWilliam Henry and how this was discovered

William Henry

1774-1836 · English chemist who discovered the gas-solubility law bearing his name in 1803

Ask the tutor: How does the activity coefficient generalize Henry's law once the solute concentration is no longer dilute?

Is Henry's law only for gases dissolved in liquids

No. Henry's law describes any dilute solute's vapor-liquid equilibrium behavior, including volatile liquids dissolved in another liquid at low concentration. Gas solubility is simply the most common and practically important application, because gases are almost always present at trace mole fractions in liquid solutions, keeping the system safely within the dilute limit where the linear law holds.

How is the Henry's law constant actually measured

Experimentally, you equilibrate the liquid solution with the gas at a series of low partial pressures, measure the equilibrium mole fraction dissolved (via gas chromatography, manometry, or mass balance), and plot $p_B$ against $x_B$ . The slope of the best-fit line through the low-concentration data, extrapolated to $x_B \to 0$ , is $K_B$ . Modern tabulated Henry's constants for atmospheric gases are compiled from many such careful equilibrium measurements across temperatures.

Why is K_B usually much larger than p_B* for gases like O2 or N2 in water

Because water strongly hydrogen-bonds to itself but interacts only weakly with nonpolar gas molecules like O₂ or N₂. A dissolved O₂ molecule is thermodynamically "unhappy" surrounded by water, so it has a strong tendency to escape — a large $K_B$ . The extrapolated pure-liquid vapor pressure $p_B^{*}$ for O₂, by contrast, describes O₂ molecules surrounded by other O₂ molecules, a completely different (and hypothetical, since O₂ is a gas at 25 °C) reference state.

Does Henry's law connect to activities and the ideal-dilute solution model

Yes — Henry's law is the operational definition of ideal-dilute behavior for the solute. In the activity framework, the solute's chemical potential is written using $K_B$ as the reference (Henry's law) standard state, exactly parallel to how Raoult's law defines the solvent's standard state. Deviations from Henry's law at higher concentration are captured by an activity coefficient, the natural next topic after this one.