$K_H$ value for Ar(g), $CO_2$(g), HCHO(g) and $CH_4$(g) are 40.39, 1.67, $1.83\times10^{-5}$ and 0.413 respectively. Arrange these gases in the order of their increasing solubility. (i) $HCHO < CH_4 < CO_2 < Ar$ (ii) $HCHO < CO_2 < CH_4 < Ar$ (iii) $Ar < CO_2 < CH_4 < HCHO$ (iv) $Ar < CH_4 < CO_2 < HCHO$

KH value for Ar(g), CO₂(g), HCHO(g) and CH₄(g) are 40.39, 1.67, 1.83×10⁻⁵ and 0.413 respectively. Arrange thes

NCERT Exemplar · Q26

K_H

(i)

HCHO < CH_4 < CO_2 < Ar

(ii)

HCHO < CO_2 < CH_4 < Ar

(iii)

Ar < CO_2 < CH_4 < HCHO

(iv)

Ar < CH_4 < CO_2 < HCHO

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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 $K_H$ is inversely proportional to solubility: a higher $K_H$ means lower solubility. Ranking the gases by increasing $K_H$ gives the order of increasing solubility: Ar < CO₂ < CH₄ < HCHO.

Understanding Henry's Law

Henry's law describes how gases dissolve in liquids at equilibrium. The relationship is captured by:

p = K_H \cdot x

where $p$ is the partial pressure of the gas above the solution, $x$ is its mole fraction in the liquid, and $K_H$ is Henry's law constant.

The key insight: for a fixed partial pressure, if $K_H$ is large, the mole fraction $x$ must be small to satisfy the equation. In other words, a high $K_H$ means the gas is reluctant to dissolve — it prefers the gas phase. Conversely, a low $K_H$ means the gas readily dissolves into the liquid.

Think of $K_H$ as a measure of how much "pressure" you need to force a given amount of gas into solution. The harder it is to dissolve (higher $K_H$ ), the less soluble the gas.

Step-by-Step Solution

List the given Henry's constants

Gas $K_H$ value
Ar 40.39
CO₂ 1.67
HCHO $1.83 \times 10^{-5}$
CH₄ 0.413
Recognize the inverse relationship

Solubility $\propto \frac{1}{K_H}$ . The gas with the smallest $K_H$ is the most soluble; the gas with the largest $K_H$ is the least soluble.
Rank the $K_H$ values in increasing order

Arranging from smallest to largest:

Gas	$K_H$ value
Ar	40.39
CO₂	1.67
HCHO	$1.83 \times 10^{-5}$
CH₄	0.413

1.83 \times 10^{-5} < 0.413 < 1.67 < 40.39

This corresponds to:

K_H(\text{HCHO}) < K_H(\text{CH}_4) < K_H(\text{CO}_2) < K_H(\text{Ar})

Invert to get increasing solubility …

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🔍 The Core Concept (Why First)

Henry’s Law states:

P = K_H \cdot x

Where:

$P$ = partial pressure of gas above the liquid
$x$ = mole fraction of gas dissolved in the liquid
$K_H$ = Henry’s Law constant

Key relationship:

Higher $K_H$ → lower solubility (gas escapes more easily)
Lower $K_H$ → higher solubility (gas dissolves more easily)

So solubility is inversely proportional to $K_H$ .

✗ Mistake #1: Thinking higher $K_H$ means higher solubility

Why it happens:

Students see a large $K_H$ and assume “big number = more dissolved.” This is the most common error.

How to avoid:

Always remember:

$K_H$ is a measure of escaping tendency — not dissolving power.

Write this rule on your formula sheet:

Solubility $\propto \frac{1}{K_H}$

✗ Mistake #2: Forgetting units or comparing $K_H$ values without checking scale

Why it happens:

Here, $K_H$ values are given in different orders of magnitude (40.39, 1.67, $1.83\times10^{-5}$ , 0.413). Students sometimes ignore the exponent and just compare the digits.

How to avoid:

Always write all $K_H$ values in scientific notation with the same exponent for easy comparison:

Ar: $40.39$ → $4.039 \times 10^1$
CO₂: $1.67$ → $1.67 \times 10^0$
CH₄: $0.413$ → $4.13 \times 10^{-1}$
HCHO: $1.83 \times 10^{-5}$

Now clearly:

$1.83 \times 10^{-5} < 4.13 \times 10^{-1} < 1.67 < 40.39$

✗ Mistake #3: Arranging in order of $K_H$ instead of solubility

Why it happens:

Students list gases from smallest $K_H$ to largest $K_H$ and think that’s the solubility order — but they forget to invert.

How to avoid:

Do it in two clear steps:

Arrange $K_H$ in increasing order:

HCHO ( $1.83\times10^{-5}$ ) < CH₄ ( $0.413$ ) < CO₂ ( $1.67$ ) < Ar ( $40.39$ )
Invert for solubility: …

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Q. $K_H$ value for Ar(g), $CO_2$ (g), HCHO(g) and $CH_4$ (g) are 40.39, 1.67, $1.83\times10^{-5}$ and 0.413 respectively. Arrange these gases in the order of their increasing solubility.

Concept understanding — Henrys Law

Henry's Law: The Physics of "Fizz"

The Intuition: Gas Wants to Dissolve

The Precise Statement

What the Constant $k_H$ Tells You

Real-Life Examples

Why this formula?

Henry's Law: Why the Formula Holds

Why This Linear Relationship Exists

1. Dynamic Equilibrium at the Interface

2. The Driving Force for Dissolution

3. The Driving Force for Escape

4. Equating the Two Rates

Why It's Linear (Not Exponential or Logarithmic)

Understanding Henry's Law

Step-by-Step Solution

Method: Henry’s Law — Relating $K_H$ to Solubility

Steps to solve:

🔍 The Core Concept (Why First)

✗ Mistake #1: Thinking higher $K_H$ means higher solubility

✗ Mistake #2: Forgetting units or comparing $K_H$ values without checking scale

✗ Mistake #3: Arranging in order of $K_H$ instead of solubility

Q.KHK_HKH​ value for Ar(g), CO2CO_2CO2​(g), HCHO(g) and CH4CH_4CH4​(g) are 40.39, 1.67, 1.83×10−51.83\times10^{-5}1.83×10−5 and 0.413 respectively. Arrange these gases in the order of their increasing solubility.

Understanding Henry's Law

Step-by-Step Solution

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Q. $K_H$ value for Ar(g), $CO_2$ (g), HCHO(g) and $CH_4$ (g) are 40.39, 1.67, $1.83\times10^{-5}$ and 0.413 respectively. Arrange these gases in the order of their increasing solubility.

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