What is the significance of Henry's Law constant KH?

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K_H

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Concept understanding — Henrys Law

Henry's Law: The Physics of "Fizz"

Imagine you open a cold bottle of soda. You hear that familiar psshhht sound. Bubbles rush out. Now think: why were those bubbles inside the bottle in the first place? The liquid wasn't boiling. The answer is Henry's Law.

The Intuition: Gas Wants to Dissolve

Gases are just molecules flying around. When a gas touches a liquid, some of those molecules get "trapped" inside the liquid — they dissolve. But here's the key: the more you push on the gas, the more of it gets forced into the liquid.

Think of a crowded bus. If you push more people toward the door (higher pressure), more people get squeezed inside. If you let the pressure off (open the bottle), people rush out. That's exactly what happens with gas and liquid.

In the soda bottle, carbon dioxide gas is pumped in at high pressure. That pressure forces a huge amount of CO₂ to dissolve into the liquid. When you open the bottle, the pressure above the liquid drops to normal air pressure. Suddenly, the liquid can't hold all that CO₂ anymore — so it escapes as bubbles. That's the fizz.

The Precise Statement

Henry's Law says:

C = k_H \cdot P

Where:

$C$ = concentration of the dissolved gas in the liquid (usually mol/L or g/L)
$P$ = partial pressure of that gas above the liquid (usually atm or kPa)
$k_H$ = Henry's law constant — a number that depends on the specific gas, the liquid, and the temperature

In words: At a constant temperature, the amount of gas that dissolves in a liquid is directly proportional to the partial pressure of that gas above the liquid.

What the Constant $k_H$ Tells You

$k_H$ is not universal. It's different for every gas-liquid pair. For example:

CO₂ in water has a certain $k_H$
O₂ in water has a different $k_H$ (smaller — oxygen doesn't dissolve as easily)

Temperature matters too. Higher temperature means lower $k_H$ — gases become less soluble in hot liquids. That's why a warm soda goes flat faster than a cold one.

Watch out

Henry's Law works only for dilute solutions and non-reacting gases. If the gas reacts chemically with the liquid (like HCl gas dissolving in water to form hydrochloric acid), Henry's Law does not apply — the concentration will be much higher than predicted.

Real-Life Examples

Situation	What Henry's Law explains
Soda fizz	High pressure forces CO₂ in; releasing pressure lets it out

Why this formula?

Henry's Law: Why the Formula Holds

Henry's Law describes the solubility of a gas in a liquid at a constant temperature. The key formula is:

P = k_H \cdot x

Where:

$P$ = partial pressure of the gas above the liquid
$x$ = mole fraction of the gas dissolved in the liquid
$k_H$ = Henry's constant (depends on gas, liquid, and temperature)

Why This Linear Relationship Exists

1. Dynamic Equilibrium at the Interface

Imagine a gas above a liquid. At the molecular level:

Gas molecules constantly strike the liquid surface and dissolve
Dissolved molecules constantly escape back into the gas phase

At equilibrium, the rate of dissolution equals the rate of escape. This is a dynamic balance, not a static one.

2. The Driving Force for Dissolution

The rate at which gas molecules enter the liquid depends on:

How many gas molecules hit the surface — this is proportional to the partial pressure $P$ of the gas
How easily they dissolve — this is captured by $k_H$

So:

\text{Rate}_{\text{dissolve}} \propto P

3. The Driving Force for Escape

The rate at which dissolved molecules leave the liquid depends on:

How many dissolved molecules are near the surface — this is proportional to the mole fraction $x$ of the gas in the liquid
How easily they escape — also captured by $k_H$

So:

\text{Rate}_{\text{escape}} \propto x

4. Equating the Two Rates

At equilibrium:

\text{Rate}_{\text{dissolve}} = \text{Rate}_{\text{escape}}

Therefore:

P \propto x

Introducing the proportionality constant $k_H$ :

P = k_H \cdot x

Why It's Linear (Not Exponential or Logarithmic)

The linearity arises because:

No saturation effects at low concentrations — the molecules don't "crowd" each other
Ideal behavior is assumed — gas molecules don't interact strongly with each other or with the solvent
Temperature is constant — $k_H$ doesn't change …

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Henry’s Law constant $K_H$ tells us how soluble a gas is in a liquid at a given temperature — a larger $K_H$ means lower solubility, and vice versa. It is the proportionality constant in $p = K_H x$ , where $p$ is the partial pressure of the gas above the liquid and $x$ is its mole fraction in the solution.

Henry’s Law is the go-to principle for understanding how gases dissolve in liquids. It states that at constant temperature, the amount of gas dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid. Mathematically:

p = K_H \, x

Here, $p$ is the partial pressure of the gas (in atm or bar), $x$ is its mole fraction in the solution, and $K_H$ is the Henry’s Law constant. The units of $K_H$ are the same as pressure (e.g., atm, bar, Pa).

The key insight: $K_H$ is not a universal constant — it depends on the gas, the solvent, and the temperature. Its value carries a direct physical meaning about solubility.

Step-by-step reasoning

What $K_H$ actually measures

Rearranging the law: $x = \frac{p}{K_H}$ . For a fixed partial pressure $p$ , a larger $K_H$ gives a smaller mole fraction $x$ — meaning less gas dissolves. So $K_H$ is inversely related to solubility. A small $K_H$ means the gas dissolves readily; a large $K_H$ means it barely dissolves.
Temperature dependence

Dissolving a gas is generally exothermic (heat is released). According to Le Chatelier’s principle, raising temperature shifts equilibrium toward the gas phase, decreasing solubility. This shows up as an increase in $K_H$ with temperature. So $K_H$ is not fixed — it rises when you heat the solution.
Comparing different gases

For the same solvent and temperature, gases with weaker intermolecular forces (like helium) have high $K_H$ values and low solubility. Gases that react with the solvent (like $\text{CO}_2$ in water, forming carbonic acid) have very low $K_H$ values because they dissolve extensively.
Practical significance
- In carbonated drinks, $\text{CO}_2$ is bottled under high pressure. When opened, pressure drops, and $p < K_H x$ — so gas escapes (fizzing). …

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Here are the common mistakes students make regarding the significance of Henry’s Law constant ( $K_H$ ) and how to avoid each.

Mistake 1: Confusing $K_H$ with Solubility

The error:

Students often think a high $K_H$ means high solubility. This is the most frequent trap.

Why it happens:

Henry’s Law is usually written as:

P = K_H \cdot x

where $P$ is the partial pressure of the gas above the liquid and $x$ is its mole fraction in the solution.

The correct interpretation:

$K_H$ is a proportionality constant linking pressure to mole fraction.
A high $K_H$ means that for a given pressure, the mole fraction ( $x$ ) is small — i.e., the gas is less soluble.
A low $K_H$ means the gas is more soluble.

How to avoid:

Always think: “ $K_H$ is the pressure needed to achieve a certain concentration.”

High $K_H$ → needs high pressure to dissolve → low solubility.
Low $K_H$ → dissolves easily at low pressure → high solubility.

Mistake 2: Ignoring the Temperature Dependence

The error:

Treating $K_H$ as a constant for a given gas-liquid pair, regardless of temperature.

Why it happens:

The word “constant” in “Henry’s Law constant” misleads students into thinking it never changes.

The correct fact:

$K_H$ is strongly temperature-dependent.

For most gases, $K_H$ increases with temperature → solubility decreases as temperature rises.
This is why a cold soda fizzes more than a warm one.

How to avoid:

Always note: “ $K_H$ is constant only at a fixed temperature.”

In exam problems, if temperature changes, $K_H$ changes — you cannot use the same value.

Mistake 3: Misinterpreting the Units of $K_H$

The error:

Using the wrong units or ignoring unit consistency when applying Henry’s Law.

Why it happens:

$K_H$ can be expressed in different forms:

$K_H$ (pressure/concentration): e.g., $\text{atm} / \text{mole fraction}$
$K_H$ (concentration/pressure): e.g., $\text{mol} / (\text{L} \cdot \text{atm})$

Students mix these up and get the inverse relationship wrong.

The correct approach:

Always check the definition given in the problem.

If $P = K_H \cdot x$ , then $K_H$ has units of pressure.
If $c = K_H \cdot P$ , then $K_H$ has units of concentration/pressure.

How to avoid:

Write the exact form of Henry’s Law you are using.
Convert units carefully — especially when dealing with $\text{bar}$ , $\text{atm}$ , or $\text{Pa}$ .

Mistake 4: Thinking $K_H$ Applies to All Gases Equally

The error:

Assuming $K_H$ values for different gases are similar or that the same gas has the same $K_H$ in all solvents.

Why it happens:

Students memorize the formula but skip the physical chemistry behind it.

The correct fact:

$K_H$ depends on:

The nature of the gas (e.g., $\text{CO}_2$ is much more soluble than $\text{O}_2$ in water).
The nature of the solvent (e.g., $\text{O}_2$ dissolves better in blood than in pure water due to hemoglobin).

How to avoid:

Remember: $K_H$ is a specific property of a gas-solvent pair. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
$K_H$
›Reveal solutionSolution
Applying Henry's law to find methane's mole fraction, then converting to moles dissolved in 1 L of water, gives about $1.38\times10^{-1}$ mol — option (D).

Concept and Intuition

Henry's law states that the partial pressure of a gas above a solution is proportional to its mole fraction in the solution: $p = K_H x$ . A large $K_H$ (here 400 bar) means the gas is not very soluble, so only a small mole fraction dissolves under a given pressure. Once we know the mole fraction, and knowing that water vastly outnumbers the dissolved gas molecules, we can approximate moles of water as the total moles in solution to extract the actual moles of dissolved methane.

Step-by-Step Solution

Henry's law: $p = K_H x \Rightarrow x = \dfrac{p}{K_H} = \dfrac{1\ \text{bar}}{0.4\ \text{kbar}} = \dfrac{1}{400} = 2.5\times10^{-3}$ .

Moles of water in 1 L (mass ≈ 1000 g, taking density ≈ 1 g/mL): $n_{water} = \dfrac{1000}{18} = 55.56$ mol. …
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AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
$\mathrm{N_2}$
›Reveal solutionSolution
Henry's law constant increases as solubility decreases; helium, being the least soluble of the four gases in water, has the highest $K_H$ — option (C).

Concept and Intuition

Henry's law states $p = K_H \cdot x$ , where $p$ is the partial pressure of the gas and $x$ is its mole fraction dissolved in the liquid. A larger $K_H$ means a smaller mole fraction dissolves for the same partial pressure — i.e., higher $K_H$ = lower solubility. Noble gases like helium have very weak intermolecular (van der Waals) interactions with water and are notoriously poorly soluble, which is why divers use helium-based gas mixtures (to reduce the amount of gas — like nitrogen — that dissolves in blood and causes decompression sickness).

Step-by-Step Solution

Recall the qualitative solubility order in water for these gases: $\mathrm{O_2}$ is somewhat more soluble than $\mathrm{N_2}$ , and both are more soluble than the very inert, small, monoatomic He.

Since $K_H \propto \dfrac{1}{\text{solubility}}$ , the least soluble gas has the largest $K_H$ . …
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AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
$\mathrm{N_2}$
›Reveal solutionSolution
Henry's law with equal partial pressures gives a mole-fraction ratio inversely proportional to the Henry's law constants, $\approx0.45$ .

Concept and Intuition

Henry's law states that the partial pressure of a gas above a liquid is directly proportional to its mole fraction dissolved in the liquid: $p=K_H\,x$ . A larger $K_H$ means the gas is less soluble (needs a higher partial pressure to dissolve the same mole fraction), so for two gases at the same partial pressure, the one with the larger $K_H$ ends up with the smaller mole fraction dissolved.

Step-by-Step Solution

Write Henry's law for each gas: $p_{N_2}=K_{H,N_2}\,x_{N_2}$ and $p_{O_2}=K_{H,O_2}\,x_{O_2}$ .

Given $p_{N_2}=p_{O_2}$ (equal partial pressures), divide the two equations: $\dfrac{x_{N_2}}{x_{O_2}}=\dfrac{K_{H,O_2}}{K_{H,N_2}}$ . …
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AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQ
$4.27\times10^5$
›Reveal solutionSolution
This is a direct application of Henry's law relating partial pressure of a gas above a liquid to its mole fraction dissolved in it.

Concept and Intuition

Henry's law states $p = K_H\,x$ , where $p$ is the partial pressure of the gas, $x$ is its mole fraction in solution, and $K_H$ is the Henry's law constant (here expressed in mm Hg, so the pressure must also be converted to mm Hg for consistent units).

Step-by-Step Solution

Convert pressure to mm Hg: $2\text{ atm} = 2\times760 = 1520$ mm Hg.

Apply Henry's law: $x = \dfrac{p}{K_H} = \dfrac{1520}{4.27\times10^5}$ . …
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AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
$\mathrm{CO_2}$
›Reveal solutionSolution
Using Henry's law $p = K_H x$ to get the mole fraction of dissolved $\mathrm{CO_2}$ , then converting mole fraction to molarity via the moles of water in 200 mL, gives $\approx 1.11\times10^{-1}\ \mathrm{mol\,L^{-1}}$ .

Concept and Intuition

Henry's law states that the partial pressure of a gas above a solution is proportional to its mole fraction in the solution: $p = K_H\,x_{gas}$ . This lets us find the mole fraction of dissolved gas directly from the applied pressure and the Henry's law constant. Since the mole fraction of a dilute solute is tiny, we can approximate the total moles of solution as just the moles of solvent (water), and convert mole fraction to concentration using the known amount of solvent present.

Step-by-Step Solution

Apply Henry's law: $x_{CO_2} = \dfrac{p}{K_H} = \dfrac{3.4}{1.7\times10^3} = 2\times10^{-3}$ .

Find moles of water in 200 mL of soda water (density $\approx 1\ \mathrm{g/mL}$ , so mass $= 200\ \mathrm{g}$ ): $n_{H_2O} = \dfrac{200}{18} = 11.11\ \mathrm{mol}$ . …
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AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
$K_H$
›Reveal solutionSolution
This is a direct Henry's law + unit-conversion problem: convert the given partial pressure into a mole fraction, then into ppm (mg solute per kg solvent). The answer is 35.50 ppm.

Concept and Intuition

Henry's law states that for a gas dissolved in a liquid at a given temperature, the partial pressure of the gas above the solution is proportional to the mole fraction of the gas dissolved in the liquid: $p = K_H\,x$ . A large $K_H$ (like 50 kbar here) means the gas is poorly soluble — it takes a huge partial pressure to force even a tiny mole fraction into solution, which is exactly the situation for $O_2$ in water. Once we know the mole fraction, we can convert it to ppm because for very dilute solutions the mole fraction is essentially the ratio of moles of solute to (moles of solute + moles of solvent) ≈ moles of solute / moles of solvent.

Step-by-Step Solution

Apply Henry's law: $x_{O_2} = \dfrac{p}{K_H} = \dfrac{1\ \text{bar}}{50000\ \text{bar}} = 2\times10^{-5}$ .

Since the solution is dilute, $x_{O_2} \approx \dfrac{n(O_2)}{n(H_2O)}$ .

In 1 L of water, mass $\approx 1000$ g, so $n(H_2O) = \dfrac{1000}{18} = 55.56$ mol.

$n(O_2) = x_{O_2}\times n(H_2O) = 2\times10^{-5}\times 55.56 = 1.111\times10^{-3}$ mol. …
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AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
$K_H$
›Reveal solutionSolution
Since solubility varies inversely with Henry's law constant, ranking the given KH values from largest to smallest and reversing gives the increasing-solubility order Ar<CO2<CH4<HCHO. Answer (D).

Concept and Intuition

Henry's law states $p=K_H x$ , where $p$ is the gas's partial pressure above the solution and $x$ is its dissolved mole fraction. For a FIXED partial pressure, a larger $K_H$ forces a SMALLER equilibrium mole fraction $x=p/K_H$ — meaning the gas is less soluble. So solubility and $K_H$ are inversely related.

Step-by-Step Solution

List the given $K_H$ values: $Ar=40.39$ , $CO_2=1.67$ , $HCHO=1.83\times10^{-5}$ , $CH_4=0.413$ .

Sort from largest $K_H$ (least soluble) to smallest $K_H$ (most soluble): $Ar(40.39)>CO_2(1.67)>CH_4(0.413)>HCHO(1.83\times10^{-5})$ . …
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AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
$Q.A gas X is dissolved in water at 2 bar pressure. Its mole fraction in the solution is 0.02. Find the mole fraction of water in the solution when the pressure of the gas is doubled at the same temperature. (A) 0.04 (B) 0.98 (C) 0.96 (D) 0.02$
›Reveal solutionSolution
This tests Henry's law ( $p = K_H x$ ): doubling the gas pressure doubles the gas's mole fraction in solution, so the water's mole fraction drops from $0.98$ to $0.96$ .

Concept and Intuition

Henry's law says the partial pressure of a gas above a dilute solution is directly proportional to its mole fraction in the solution: $p = K_H \cdot x_{\text{gas}}$ , with $K_H$ constant at a given temperature. So if pressure doubles, the mole fraction of dissolved gas also doubles (as long as $K_H$ doesn't change, i.e., temperature is constant, as stated). The mole fraction of water then adjusts to keep the two mole fractions summing to 1 (a binary solution of gas + water).

Step-by-Step Solution

Given: at $p_1 = 2$ bar, mole fraction of gas $x_1 = 0.02$ .

Apply Henry's law to find $K_H$ : $K_H = \dfrac{p_1}{x_1} = \dfrac{2}{0.02} = 100$ bar.

At the same temperature, $K_H$ is unchanged. New pressure $p_2 = 2 \times 2 = 4$ bar.

New mole fraction of gas: $x_2 = \dfrac{p_2}{K_H} = \dfrac{4}{100} = 0.04$ .

Since gas + water make up the whole solution, mole fraction of water $= 1 - x_2 = 1 - 0.04 = 0.96$ . …
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AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
$CO_2$
›Reveal solutionSolution
Apply Henry's law to get the mole fraction of dissolved CO₂, then convert to mass using the moles of water present in 500 mL. Answer: 3.7 g.

Concept and Intuition

Henry's law states that the partial pressure of a dissolved gas is proportional to its mole fraction in solution: $p = K_H \cdot x$ . Once the (small) mole fraction of dissolved gas is known, it can be converted to moles using the (much larger) known amount of solvent, since $x \approx n_{gas}/n_{solvent}$ when the gas is dilute.

Step-by-Step Solution

Convert pressure: $P = 5\ \text{atm} = 5\times101325\ \text{Pa} = 506{,}625$ Pa.

Henry's law: $x_{CO_2} = \dfrac{P}{K_H} = \dfrac{506{,}625}{1.67\times10^8} = 3.034\times10^{-3}$ .

Moles of water in 500 mL (density $\approx 1$ g/mL): $n_{water} = \dfrac{500}{18} = 27.78$ mol. …
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Q.What is the significance of Henry's Law constant KHK_HKH​?

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Q.What is the significance of Henry's Law constant $K_H$ ?

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