Chemistry · Solutions

Henry's Law

Henry's law is the quantitative statement of how the solubility of a gas in a liquid responds to pressure, set out in NCERT Class 12 Chemistry, Unit 1 (Section 1.3.2). It anchors a small but reliable cluster of NEET questions, where a single relation, $p = K_H x$, must be read both as an equation to compute with and as a rule to reason about. This note builds the law from the equilibrium picture, fixes the meaning of the Henry's law constant and its rise with temperature, works through the NCERT examples, and explains why soda bottles, scuba tanks and high-altitude air all obey the same line.

The Law and Its Statement

Many gases dissolve in water to differing extents. Oxygen dissolves only slightly, yet that dissolved oxygen sustains all aquatic life; hydrogen chloride, by contrast, is highly soluble. The amount that dissolves at equilibrium is governed by pressure, temperature and the nature of the gas. The dependence on pressure is captured exactly by Henry's law.

Consider a gas in contact with a liquid in a closed vessel at fixed temperature. At dynamic equilibrium the rate at which gas particles enter the solution equals the rate at which they leave it. Compressing the gas above the solution raises the number of particles per unit volume, so more strike and enter the surface per second. Solubility increases until a new equilibrium is reached at the higher pressure. The solubility of a gas therefore rises with its pressure above the liquid.

William Henry was the first to give this a quantitative form. Henry's law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the surface of the liquid or solution. Using the mole fraction of the gas in solution as the measure of solubility, the most commonly used form of the law states that the partial pressure of the gas in the vapour phase, $p$, is proportional to the mole fraction of the gas, $x$, in the solution:

$$ p = K_H\,x $$

Here $K_H$ is the Henry's law constant. The relation is linear and passes through the origin: a plot of partial pressure against the mole fraction of dissolved gas is a straight line whose slope is $K_H$. Dalton, a contemporary of Henry, independently reached the same conclusion that the solubility of a gas is a function of its partial pressure.

Figure 1

Partial pressure versus mole fraction: the Henry's law line.

mole fraction of gas, x partial pressure, p large Kₕ (low solubility) small Kₕ (high solubility) slope = Kₕ 0

Both gases give straight lines through the origin. The steeper line (larger slope, larger $K_H$) reaches a high partial pressure for only a small dissolved mole fraction, signalling a poorly soluble gas. The gentler line (smaller $K_H$) dissolves to a larger mole fraction at the same pressure.

Meaning of the Henry's Law Constant

The constant $K_H$ is the slope of the line in Figure 1, and it carries the units of pressure (bar, kbar or Pa, depending on how the data are reported). Because it equals the partial pressure required to drive the gas to a given mole fraction, $K_H$ is a measure of how hard it is to dissolve the gas. Different gases have different $K_H$ values at the same temperature, so $K_H$ is a function of the nature of the gas and of the solvent.

Rearranging the law makes the interpretation explicit. For a fixed partial pressure $p$, the mole fraction of dissolved gas is

$$ x = \frac{p}{K_H} $$

so solubility is inversely proportional to $K_H$. It follows directly that the higher the value of $K_H$ at a given pressure, the lower the solubility of the gas in the liquid. This single inference is the most heavily tested idea in the topic.

NEET Trap

A bigger Kₕ means a smaller solubility, not a larger one

The name "constant" tempts candidates to treat a large $K_H$ as "dissolves a lot". The opposite is true. Since $x = p/K_H$, the gas with the largest $K_H$ is the least soluble at the same pressure and temperature. When ranking solubilities from a table of $K_H$ values, order them by the smallest $K_H$ first.

Largest $K_H$ → least soluble. Smallest $K_H$ → most soluble. Solubility ∝ $1/K_H$.

Kₕ Values for Common Gases

The NCERT table of Henry's law constants in water lists representative values. Two temperatures are shown for the diatomic gases so that the temperature trend is visible directly. Lower $K_H$ in the table corresponds to higher solubility; note how carbon dioxide, with $K_H = 1.67\ \text{kbar}$, is far more soluble than helium at $144.97\ \text{kbar}$.

GasTemperature / KKₕ / kbarReading
He293144.97Very large Kₕ → least soluble
H₂29369.16Sparingly soluble
N₂29376.48Sparingly soluble
N₂30388.84Kₕ higher at higher T
O₂29334.86More soluble than N₂
O₂30346.82Kₕ higher at higher T
Ar29840.3Moderate solubility
CO₂2981.67Small Kₕ → highly soluble
CH₄ (methane)2980.413Notably soluble
Vinyl chloride2980.611Highly soluble
Formaldehyde2981.83×10⁻⁵Extremely soluble (reacts with water)

Formaldehyde sits far below every other entry because it does not merely dissolve; $\ce{HCHO}$ reacts with water to form methylene glycol, which keeps its effective partial pressure tiny and its $K_H$ vanishingly small. This is also a hint that Henry's law applies cleanly only when the gas does not associate, dissociate or react with the solvent.

Temperature Dependence of Kₕ

Solubility of a gas in a liquid decreases as the temperature rises. When a gas dissolves, its molecules pass into the liquid phase in a process resembling condensation, which releases heat; dissolution of a gas is therefore exothermic. Because the dissolution equilibrium obeys Le Chatelier's principle, supplying heat by raising the temperature shifts the equilibrium back towards the gaseous state, and less gas remains dissolved.

In the language of the law, falling solubility at higher temperature means $K_H$ rises with temperature. The NCERT table shows this for both nitrogen and oxygen: $K_H$ of $\ce{N2}$ climbs from $76.48\ \text{kbar}$ at 293 K to $88.84\ \text{kbar}$ at 303 K, and $K_H$ of $\ce{O2}$ from $34.86$ to $46.82\ \text{kbar}$ over the same interval. It is for this reason that aquatic species are more comfortable in cold waters than in warm waters, where dissolved oxygen is scarcer.

Figure 2

As temperature rises, the Henry line steepens (Kₕ increases) and a given pressure dissolves less gas.

fixed p x(hot) x(cold) mole fraction of gas, x partial pressure, p high T (large Kₕ) low T (small Kₕ)

At the same partial pressure the hot line meets the dashed level at a smaller mole fraction: $x(\text{hot}) < x(\text{cold})$. Heating drives gas out of solution.

Worked Examples

The two NCERT computations below show the law used in both directions: first to find the amount of dissolved gas from a known $K_H$, and then to find $K_H$ itself from a measured solubility. Throughout, recall that 1 litre of water contains about 55.5 mol of water, which dominates the denominator of the mole fraction so strongly that the dissolved gas can be neglected there.

Example · NCERT 1.4

If $\ce{N2}$ gas is bubbled through water at 293 K, how many millimoles of $\ce{N2}$ dissolve in 1 litre of water? Take the partial pressure of $\ce{N2}$ as 0.987 bar and $K_H$ for $\ce{N2}$ at 293 K as 76.48 kbar.

Apply $x = p/K_H$ with $p = 0.987\ \text{bar}$ and $K_H = 76{,}480\ \text{bar}$:

$$ x_{\ce{N2}} = \frac{0.987}{76{,}480} = 1.29\times10^{-5} $$

For $n$ moles of $\ce{N2}$ in 55.5 mol of water, $x_{\ce{N2}} = \dfrac{n}{n + 55.5} \approx \dfrac{n}{55.5}$ (since $n \ll 55.5$). Hence

$$ n = 1.29\times10^{-5}\times 55.5 = 7.16\times10^{-4}\ \text{mol} = 0.716\ \text{mmol} $$

About 0.716 millimole of nitrogen dissolves, confirming how sparingly $\ce{N2}$ dissolves at ordinary pressure.

Example · NCERT 1.6

$\ce{H2S}$, a toxic gas used in qualitative analysis, has a solubility in water of 0.195 m at STP. Calculate its Henry's law constant.

A solubility of 0.195 molal means 0.195 mol of $\ce{H2S}$ in 1 kg of water, i.e. in $1000/18 = 55.5$ mol of water. The mole fraction of the gas is

$$ x_{\ce{H2S}} = \frac{0.195}{0.195 + 55.5} \approx \frac{0.195}{55.7} = 3.50\times10^{-3} $$

At STP the partial pressure of the gas is $p = 0.987\ \text{bar}$. Therefore

$$ K_H = \frac{p}{x} = \frac{0.987}{3.50\times10^{-3}} = 282\ \text{bar} $$

The small $K_H$ (about 282 bar, i.e. 0.282 kbar) is consistent with $\ce{H2S}$ being appreciably soluble in water.

Build the foundation

Henry's law is the pressure rule for the broader idea of solubility of gases and solids in liquids — the prerequisite section in NCERT 1.3.

Example · NCERT 1.7

The Henry's law constant for $\ce{CO2}$ in water is $1.67\times10^{8}\ \text{Pa}$ at 298 K. Calculate the quantity of $\ce{CO2}$ in 500 mL of soda water when packed under 2.5 atm $\ce{CO2}$ pressure at 298 K.

Convert the partial pressure: $p = 2.5\ \text{atm} = 2.5\times1.013\times10^{5} = 2.533\times10^{5}\ \text{Pa}$. The mole fraction of dissolved $\ce{CO2}$ is

$$ x_{\ce{CO2}} = \frac{p}{K_H} = \frac{2.533\times10^{5}}{1.67\times10^{8}} = 1.517\times10^{-3} $$

500 mL of water is about 500 g, i.e. $500/18 = 27.78$ mol of water. With $n_{\ce{CO2}} \ll n_{\text{water}}$,

$$ x_{\ce{CO2}} \approx \frac{n_{\ce{CO2}}}{27.78} \;\Rightarrow\; n_{\ce{CO2}} = 1.517\times10^{-3}\times 27.78 = 0.0421\ \text{mol} $$

Mass of $\ce{CO2}$ $= 0.0421 \times 44 \approx \mathbf{1.85\ g}$. The high sealing pressure forces this much gas into a modest volume of liquid.

Henry's Law and Raoult's Law

Henry's law and Raoult's law share the same algebraic shape: each makes a component's partial pressure proportional to its mole fraction in solution. They differ only in the constant of proportionality, which encodes a different physical situation.

FeatureHenry's lawRaoult's law
Equationp = K_H · xp₁ = p₁° · x₁
Applies toDissolved gas (dilute solute)Volatile component of a solution
Proportionality constantKₕ (Henry's law constant)p₁° (vapour pressure of pure liquid)
Mole fraction usedOf the gas in solutionOf the component in solution
RelationshipRaoult's law is a special case of Henry's law in which $K_H$ becomes equal to $p_1^{\circ}$, the vapour pressure of the pure liquid.

For an ideal binary solution, the solvent obeys Raoult's law and a dilute dissolved gas obeys Henry's law; the two coincide in the limit where $K_H = p_1^{\circ}$. The deeper treatment of partial pressures of liquid mixtures belongs to Raoult's law, while the broader catalogue of solution types is covered under types of solutions.

Applications

Three applications recur in both NCERT and NEET, each a direct reading of $x = p/K_H$. The first is the soft-drink bottle: to dissolve a large amount of $\ce{CO2}$, the bottle is sealed under high pressure. The high partial pressure forces a high mole fraction of carbon dioxide into solution; on opening, the pressure drops to atmospheric, solubility falls, and the excess gas escapes as fizz.

Figure 3

Three faces of Henry's law: more pressure dissolves more gas; less pressure releases it.

soda: high p CO₂ forced in surface diver: bends N₂ bubbles on ascent low p(O₂) altitude: anoxia low blood O₂

Each panel is the same line read at a different pressure: high $p$ raises dissolved $x$ (soda, diver at depth); falling $p$ lowers $x$ and releases gas (opened bottle, ascending diver, thin mountain air).

The second application is decompression sickness in scuba divers. While breathing air at high pressure underwater, the increased pressure raises the solubility of atmospheric gases in the blood. On a too-rapid ascent the pressure falls and the dissolved gases come out of solution, forming bubbles of nitrogen that block capillaries — the painful, dangerous condition known as the bends. To reduce it, divers' tanks are filled with air diluted by helium (about 11.7% helium, 56.2% nitrogen, 32.1% oxygen).

The third application is high-altitude anoxia. At high altitudes the partial pressure of oxygen is lower than at ground level, so by Henry's law the oxygen dissolved in the blood and tissues falls. Climbers with low blood oxygen become weak and unable to think clearly — the defining symptoms of anoxia. The same law that fills a soda bottle explains why mountaineers struggle to breathe.

NEET Trap

Two independent statements that students fuse incorrectly

Keep the two trends separate. With respect to gas identity, a larger $K_H$ means lower solubility. With respect to temperature, $K_H$ of any given gas rises as temperature increases, so solubility falls. A common error is to claim that heating "increases $K_H$ and therefore increases solubility" — the higher $K_H$ decreases solubility.

Higher $K_H$ ⇒ lower solubility (always). Higher $T$ ⇒ higher $K_H$ ⇒ lower gas solubility.

Quick Recap

Henry's Law in one screen

  • Henry's law: at constant temperature, gas solubility is proportional to partial pressure, $p = K_H x$.
  • $K_H$ is the slope of the $p$ versus $x$ line and has units of pressure; it depends on the gas and the solvent.
  • Solubility $x = p/K_H$, so a larger $K_H$ means a less soluble gas — order solubilities by smallest $K_H$.
  • Dissolution of a gas is exothermic, so $K_H$ rises with temperature and gas solubility falls; cold water holds more dissolved gas.
  • Raoult's law is the special case of Henry's law with $K_H = p_1^{\circ}$.
  • Applications: high-pressure sealing of soda water, the bends in scuba divers, and high-altitude anoxia.

NEET PYQ Snapshot — Henry's Law

Henry's law is examined mainly through the Kₕ–solubility ordering rule; related solubility-of-gas concepts appear alongside.

NEET 2024

The Henry's law constant (Kₕ) values of three gases (A, B, C) in water are 145, 2×10⁻⁵ and 35 kbar respectively. The solubility of these gases in water follow the order:

  1. B > A > C
  2. B > C > A
  3. A > C > B
  4. A > B > C
Answer: (2) B > C > A

Solubility $\propto 1/K_H$. Smallest $K_H$ is most soluble: B ($2\times10^{-5}$ kbar) > C (35 kbar) > A (145 kbar). The smaller the Henry's law constant, the greater the solubility.

Concept

A gas dissolves in water with a partial pressure of 1 bar giving a mole fraction of $1.0\times10^{-4}$. What is its Henry's law constant?

  1. $1.0\times10^{-4}$ bar
  2. $1.0\times10^{2}$ bar
  3. $1.0\times10^{4}$ bar
  4. $1.0\times10^{6}$ bar
Answer: (3) $1.0\times10^{4}$ bar

$K_H = p/x = 1/(1.0\times10^{-4}) = 1.0\times10^{4}$ bar. The constant carries units of pressure and equals the slope of the $p$–$x$ line.

Concept

Which statement about the solubility of a gas in a liquid is correct?

  1. Solubility increases with rise in temperature
  2. Solubility is independent of partial pressure
  3. Solubility decreases with rise in temperature and increases with pressure
  4. Kₕ decreases as temperature rises
Answer: (3)

Dissolution of a gas is exothermic: solubility falls with temperature (Kₕ rises) and rises with partial pressure, as $x = p/K_H$. Option (4) is the reverse of the true trend.

FAQs — Henry's Law

The most common conceptual snags on Henry's law for NEET.

What is Henry's law in simple terms?
Henry's law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid. In its most commonly used form, the partial pressure of the gas in the vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution, written as p = K_H·x, where K_H is the Henry's law constant.
Does a higher Henry's law constant mean higher or lower solubility?
A higher K_H means lower solubility. Since p = K_H·x, at a fixed partial pressure p the mole fraction x equals p divided by K_H, so a larger K_H gives a smaller x. Therefore the gas with the largest Henry's law constant is the least soluble at the same pressure and temperature.
Why does the solubility of a gas decrease as temperature rises?
Dissolution of a gas in a liquid is exothermic, so by Le Chatelier's principle raising the temperature shifts the equilibrium to expel gas, lowering solubility. Equivalently, K_H increases with temperature; the NCERT table shows K_H for both N2 and O2 rising as temperature increases. This is why aquatic species are more comfortable in cold water than in warm water.
How is Henry's law related to Raoult's law?
Both laws say a component's partial pressure is proportional to its mole fraction in solution. Raoult's law is p1 = p1°·x1, where the proportionality constant is the pure-component vapour pressure; Henry's law is p = K_H·x, where the constant is K_H. Raoult's law is a special case of Henry's law in which K_H becomes equal to the pure liquid's vapour pressure p1°.
What causes the bends in scuba divers, and how is it linked to Henry's law?
At depth, divers breathe air at high pressure, so by Henry's law the solubility of atmospheric gases in the blood increases. On rapid ascent the pressure falls, the dissolved gases come out of solution, and bubbles of nitrogen form in the blood that block capillaries. This painful and dangerous condition is called the bends. Diving tanks are filled with air diluted by helium to reduce this risk.
Why is a soda bottle sealed under high pressure of carbon dioxide?
To dissolve a large amount of CO2 in the drink, the bottle is sealed under a high partial pressure of carbon dioxide. By Henry's law a high partial pressure forces a high mole fraction of CO2 into solution. When the cap is opened the pressure drops to atmospheric, the solubility falls sharply, and the excess CO2 escapes as the familiar fizz.