At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is _____. (i) less than

NCERT Exemplar · Q3

Q. At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is __________.

(i) less than the rate of crystallisation

(ii) greater than the rate of crystallisation

(iii) equal to the rate of crystallisation

(iv) zero

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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
Scuba diving	At depth, high pressure forces more N₂ into blood; rising too fast causes decompression sickness ("the bends")
Fish breathing	Oxygen dissolves in water at the surface (where partial pressure is highest); deeper water has less dissolved O₂
Altitude sickness	At high altitude, lower atmospheric pressure means less O₂ dissolves in your blood

The Key Takeaway

Important

Henry's Law is a proportionality: double the pressure above the liquid → double the gas dissolved in the liquid (at constant temperature). It's why carbonated drinks are bottled under pressure, why deep-sea divers must ascend slowly, and why a warm drink loses its carbonation faster.

The law is simple, but its consequences are everywhere — from the soda in your hand to the air you breathe at different altitudes.

Henry's law is a key quantitative concept in the NCERT/CBSE Class 12 Chemistry chapter on Solutions, and ‘Henry's law formula’ or ‘Henry's law numericals’ are common important-question searches for board exams, JEE Main and NEET. Its real-world applications, like gas solubility in carbonated drinks and blood at altitude, make it a favourite for application-based competitive-exam questions.

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

This is analogous to Raoult's Law for ideal solutions, but for a solute gas rather than a solvent.

Key Exam Points

Henry's Law works best for dilute solutions (low $x$ )
$k_H$ increases with temperature — gases become less soluble as temperature rises
$k_H$ is different for each gas-liquid pair — e.g., $CO_2$ in water vs $O_2$ in water
The law fails if the gas reacts chemically with the solvent (e.g., $HCl$ in water)

Quick Example

If $k_H = 3.0 \times 10^4 \text{ atm}$ for $O_2$ in water at 25°C, and the partial pressure of $O_2$ in air is 0.21 atm:

x = \frac{P}{k_H} = \frac{0.21}{3.0 \times 10^4} = 7.0 \times 10^{-6}

This tiny mole fraction explains why fish need gills to extract enough oxygen from water!

Bottom line: Henry's Law is a direct consequence of dynamic equilibrium at the gas-liquid interface, where the rates of dissolution and escape balance each other linearly.

At equilibrium, opposing processes occur at equal rates; dissolution and crystallisation balance perfectly, giving (iii).

Understanding Dynamic Equilibrium

Equilibrium in chemistry is not a static, frozen state - it's a dynamic balance. When a solid dissolves in a liquid, two processes compete:

Dissolution: solid particles leave the crystal lattice and enter the solution
Crystallisation: dissolved particles return to the solid phase

Initially, only dissolution occurs. As concentration rises, crystallisation begins too.

Reaching Equilibrium

Early stage: Rate of dissolution > rate of crystallisation - net dissolution continues.
Equilibrium: Rate of dissolution = rate of crystallisation - the solution becomes saturated; concentration stays constant, but particles continuously exchange between phases.
The key insight: equilibrium does not mean nothing is happening - forward and reverse processes proceed at identical rates, so no net change occurs.

Watch out

A common mistake is thinking equilibrium means "everything stops." In reality both dissolution and crystallisation continue - they just cancel out macroscopically.

\text{Rate}_{\text{dissolution}} = \text{Rate}_{\text{crystallisation}}

✓Final answer

The correct option is (iii) : equal to the rate of crystallisation.

Common Mistakes & How to Avoid Them

Mistake 1: Confusing “equilibrium” with “no change” → Choosing (iv) zero

Why it happens:

Students often think “at equilibrium, nothing happens.” They see the word equilibrium and assume the rate must be zero.

How to avoid:

Remember: Equilibrium is dynamic, not static.

At equilibrium, the net change is zero, but the forward and reverse processes continue at equal rates.
For dissolution: solid particles leave the surface (dissolve) and dissolved particles return to the surface (crystallise) at the same speed.
So the rate is not zero — it is equal to the rate of crystallisation.

Correct choice: (iii) equal to the rate of crystallisation.

Mistake 2: Thinking dissolution stops when solution is saturated

Why it happens:

Students believe that once a solution is saturated, no more solid can dissolve, so the dissolution rate becomes zero.

How to avoid:

Saturation means the concentration of dissolved solute is at its maximum at that temperature.
But molecules are still moving: some solid leaves the surface, some dissolved solute returns.
At saturation, the two rates are equal — dissolution continues, but crystallisation matches it exactly.

Key takeaway:

“Saturated” ≠ “dissolution stopped.” It means dissolution rate = crystallisation rate.

Mistake 3: Misreading “volatile liquid solvent” and overcomplicating

Why it happens:

The phrase “volatile liquid solvent” distracts students. They think volatility changes the equilibrium behaviour.

How to avoid:

Volatility of the solvent affects vapour pressure and boiling, but not the dissolution–crystallisation equilibrium of a solid solute.
The principle of dynamic equilibrium for dissolution is the same regardless of solvent volatility.
Ignore the “volatile” label — it’s a red herring. Focus on the solid–solution interface.

Mistake 4: Picking (i) or (ii) — thinking one rate is always higher

Why it happens:

Students confuse the direction of net change before equilibrium with the state at equilibrium.

How to avoid:

Before equilibrium (unsaturated solution): dissolution rate > crystallisation rate → net dissolving.
At equilibrium: rates are equal.
After equilibrium (supersaturated): crystallisation rate > dissolution rate → net crystallisation.

The question asks at equilibrium — so only (iii) is correct.

Quick Summary Table

Mistake	Wrong choice	Why it’s wrong	Correct reasoning
Equilibrium = no activity	(iv) zero	Equilibrium is dynamic	Rates are equal, not zero
Saturation = dissolution stops	(iv) zero	Saturation is dynamic	Dissolution continues at same rate as crystallisation
Distracted by “volatile”	Any	Volatility irrelevant here	Focus on solid–solution equilibrium
Confusing before/at equilibrium	(i) or (ii)	Those describe net change before equilibrium	At equilibrium, rates are equal

Final answer: (iii) equal to the rate of crystallisation.

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

Showing the 12 most recent of 21 on this concept.

CBSE 2026Set V11 markMCQ
$Q. The percentage of helium filled in the tanks used by most scuba divers to dilute air in deep dives$
(a) 32.1
(b) 11.7
(c) 74.2
(d) 56.2
›Reveal solutionSolution
The air in a deep-diving scuba tank is diluted with about 11.7% helium.

At the high pressures experienced in deep dives, the solubility of atmospheric gases in blood increases (Henry's law, $p = K_H\,x$ ). Nitrogen in particular dissolves and, on rapid ascent, is released as bubbles causing the painful and dangerous condition known as bends. To minimise this, the air supplied to divers is diluted with the sparingly-soluble, chemically inert gas helium.

The standard diluted mixture used is approximately:

Helium $\approx 11.7\%$

Nitrogen $\approx 56.2\%$

Oxygen $\approx 32.1\%$

So the percentage of helium is 11.7%.

✓Final answer
$(b) 11.7$
CBSE 2026Set ANNUAL1 mark
$Q. On increasing temperature, solubility of gases in liquids ______ (fill in the blank).$
›Reveal solutionSolution

Dissolution of a gas in a liquid is an exothermic process, so by Le Chatelier's principle, raising the temperature shifts the equilibrium back towards the gas phase, lowering solubility.

This is why, for example, carbonated drinks lose dissolved CO2 (go 'flat') faster when warm, and why aquatic life is more stressed by low dissolved-oxygen levels in warm water.

✓Final answer
$Decreases with increase in temperature.$
CBSE 2025Set 56/4/11 markMCQ
$K_H$
›Reveal solutionSolution
Henry’s constant $K_H$ is inversely related to gas solubility — a higher $K_H$ means lower solubility. So the correct option is (B).

Why Henry’s Law works this way

Henry’s Law describes the relationship between the partial pressure of a gas above a liquid and its concentration in the liquid. The law is written as:
$p = K_H \cdot x$
where $p$ is the partial pressure of the gas, $x$ is its mole fraction in the solution, and $K_H$ is Henry’s constant.

The key intuition: $K_H$ is essentially a resistance to dissolution. A gas that dissolves easily (high solubility) will need only a small partial pressure to achieve a given concentration — so $K_H$ is small. Conversely, a gas that barely dissolves (low solubility) needs a large partial pressure to force even a tiny amount into solution — so $K_H$ is large.
$p = K_H \cdot x \quad \Rightarrow \quad K_H = \frac{p}{x}$
Step-by-step reasoning

Interpret the equation For a fixed partial pressure $p$ , the mole fraction $x$ of the dissolved gas is $x = p / K_H$ . Since $p$ is constant, $x$ (which measures solubility) is inversely proportional to $K_H$ :

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

Relate $K_H$ to solubility

High solubility → large $x$ → small $K_H$

Low solubility → small $x$ → large $K_H$

Check the options

(A) says $K_H$ is greater for gases with higher solubility — this is the opposite of what we just found.

(B) says $K_H$ is greater for gases with lower solubility — this matches the inverse relationship.

(C) says $K_H$ is constant for all gases — false, because $K_H$ depends on the gas–solvent pair and temperature.

(D) says $K_H$ is not related to solubility — false, as shown above.

Watch out

A common mistake is to read $p = K_H x$ and think “bigger $K_H$ means bigger $x$ ” — but $p$ is the independent variable here. For the same pressure, a larger $K_H$ actually gives a smaller $x$ .

Tip

Think of $K_H$ like a “price per unit” of gas dissolved. A gas that is “cheap” to dissolve (high solubility) has a low $K_H$ ; a gas that is “expensive” to dissolve (low solubility) has a high $K_H$ .

✓Final answer
$K_H$
CBSE 2025Set 56/5/11 markMCQ
$K_H$
›Reveal solutionSolution
Henry’s law constant $K_H$ actually increases with temperature, not decreases — so Assertion (A) is false. The Reason (R) is true: gas solubility does fall as temperature rises. The correct choice is (D).

Why this question trips students up

The trap here is subtle. Most of us remember that “solubility of gases decreases with temperature” — that’s drilled into us from everyday life (a cold soda fizzes more than a warm one). So Reason (R) feels solid.

But the Assertion talks about Henry’s law constant $K_H$ , not solubility directly. The two are inversely related, and that inverse relationship flips the temperature dependence. Let’s walk through it carefully.

Step-by-step reasoning

1. Recall Henry’s law

Henry’s law states that at constant temperature, the partial pressure of a gas above a liquid is proportional to its mole fraction in the liquid:
$p = K_H \cdot x$
Here $K_H$ is the Henry’s law constant. A larger $K_H$ means that for the same partial pressure, the gas dissolves less (smaller $x$ ). So $K_H$ is a measure of resistance to dissolution — it’s the opposite of solubility.
$p = K_H \cdot x \quad \Rightarrow \quad x = \frac{p}{K_H}$
2. What happens to solubility when temperature rises?

Dissolving a gas in a liquid is generally exothermic (heat is released). Le Chatelier’s principle tells us that raising temperature shifts equilibrium to favour the endothermic direction — which is the gas coming out of solution. So solubility $x$ decreases.

That makes Reason (R) true.

3. Now connect $K_H$ to temperature

From $x = p/K_H$ , if $x$ decreases at fixed $p$ , then $K_H$ must increase. So $K_H$ rises with temperature — exactly the opposite of what Assertion (A) claims.

Watch out

A common mistake is to think “solubility decreases → $K_H$ decreases”. But $K_H$ is in the denominator: smaller solubility means larger $K_H$ .

4. Check the Assertion directly

Assertion (A) says: “ $K_H$ decreases with increase in temperature.” That’s false. $K_H$ increases with temperature. So Assertion (A) is false.

5. Evaluate the options

(A) Both true, R explains A — no, A is false.

(B) Both true, R doesn’t explain A — no, A is false.

(C) A true, R false — no, A is false and R is true.

(D) A false, R true — this matches.

✓Final answer
$The correct option is (D) : Assertion (A) is false, but Reason (R) is true.$
CBSE 2025Set ANNUAL1 mark
$Q. State Henry's law.$
›Reveal solutionSolution

Henry's law states that at constant temperature, the solubility (mole fraction) of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid.

Statement: The partial pressure of a gas in the vapour phase (p) is directly proportional to the mole fraction of the gas (x) dissolved in the solution, at a given temperature:

$p = K_H \cdot x$

where $K_H$ is the Henry's law constant, which is specific to the gas-solvent pair and depends on temperature.

A higher value of $K_H$ at a given pressure means lower solubility of the gas in that solvent. Henry's law is used, for example, to explain the solubility of O2/N2 in blood at different pressures (e.g. in deep-sea diving or at high altitude).

✓Final answer
$Henry's law: p = KH x - the partial pressure of a gas is directly proportional to its mole fraction in solution.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Dissolution of a gas in liquid is a process$
(a) with increase in enthalpy
(b) with no change in enthalpy
(c) with decrease in enthalpy
(d) for which enthalpy change cannot be predicted
›Reveal solutionSolution

Dissolving a gas in a liquid is an exothermic process, so it occurs with a decrease in enthalpy.

When a gas dissolves in a liquid, gas molecules that were freely moving and widely separated come close to the solvent molecules and get surrounded by them (solvation). New, relatively strong solute–solvent (gas–liquid) interactions form as the gas molecules are trapped between solvent molecules. Energy is released when these new attractive interactions form, exactly as heat is released when a gas condenses.

Because bond/interaction formation releases energy, ΔsolH for dissolving a gas in a liquid is negative — the enthalpy of the system decreases. This is also why the solubility of a gas in a liquid decreases with increasing temperature (Le Chatelier's principle applied to an exothermic equilibrium: heating shifts it back towards the undissolved gas), as seen practically in the reduced solubility of O2/CO2 in warm water.

✓Final answer
$(c) Dissolution of a gas in a liquid occurs with a decrease in enthalpy — the process is exothermic.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The Kₕ values of Ar(g), CO2(g), HCHO(g) and CH4(g) are 40.39, 1.67, 1.83 \times 10⁻⁵ and 0.413 respectively. The correct order of their solubility is...$
(a) HCHO < CH4 < CO2 < Ar
(b) HCHO < CO2 < CH4 < Ar
(c) Ar < CO2 < CH4 < HCHO
(d) Ar < CH4 < CO2 < HCHO
›Reveal solutionSolution

Henry's law constant $K_H$ and gas solubility are inversely related — a HIGHER $K_H$ means LOWER solubility.

By Henry's law, $p = K_H \cdot x$ , where $x$ is the mole fraction of dissolved gas. For a fixed partial pressure $p$ , a larger $K_H$ forces a smaller mole fraction $x$ to dissolve — so solubility falls as $K_H$ rises. Given: $K_H$ (Ar) = 40.39, $K_H$ (CO2) = 1.67, $K_H$ (CH4) = 0.413, $K_H$ (HCHO) = 1.83×10⁻⁵. Ranking $K_H$ from highest to lowest: Ar > CO2 > CH4 > HCHO. Solubility is therefore in the REVERSE order, lowest to highest: Ar < CO2 < CH4 < HCHO.

✓Final answer
$The correct increasing solubility order is Ar < CO2 < CH4 < HCHO (option c).$
CBSE 2024Set B1 mark
$Q. Fill in the blank: The solubility of gas in a liquid is determined by ______ law.$
›Reveal solutionSolution

Henry's law states that at constant temperature, the partial pressure of a gas above a solution is directly proportional to the mole fraction of the gas dissolved in the liquid.

Henry's law: $p = K_H \cdot x$ , where p is the partial pressure of the gas above the solution, x is the mole fraction of the dissolved gas in the liquid, and $K_H$ is the Henry's law constant (specific to the gas-solvent pair and temperature). A higher $K_H$ means lower solubility for a given pressure. This explains phenomena like the 'bends' in deep-sea divers (dissolved N2 in blood) and the fizzing of carbonated drinks when a bottle is opened (reduced pressure lowers CO2 solubility).

✓Final answer
$Henry's law.$
CBSE 2024Set ANNUAL1 mark
$Q. The mathematical form of Henry's Law is ______.$
›Reveal solutionSolution

Henry's Law states that the partial pressure of a gas in the vapour phase is directly proportional to its mole fraction dissolved in the solution.

Mathematically, Henry's Law is written as: p = KH . x

where p is the partial pressure of the gas above the solution, x is the mole fraction of the dissolved gas in the liquid, and KH is the Henry's law constant (which depends on the nature of the gas and the temperature).

A higher value of KH at a given partial pressure means lower solubility of that gas.

✓Final answer
$p = KH . x (Henry's Law constant KH times mole fraction x of the dissolved gas).$
CBSE 2024Set ANNUAL1 markMCQ
$Q. The value of Henry's constant K_H :$
(a) Increases with increase in temperature
(b) Decreases with increase in temperature
(c) Remains constant
(d) First increases, then decreases
›Reveal solutionSolution

Gas solubility falls as temperature rises, and since $K_H$ is inversely related to solubility, $K_H$ rises with temperature.

Henry's law states $p = K_H \cdot x$ , where $p$ is the partial pressure of the gas above the solution and $x$ is its mole fraction dissolved. A larger $K_H$ means a smaller $x$ dissolves at the same pressure, i.e. lower solubility.

Dissolution of a gas in a liquid is an exothermic process. As temperature rises, by Le Chatelier's principle the equilibrium shifts to release dissolved gas back to the vapour phase, so solubility falls. Since $K_H$ varies inversely with solubility, $K_H$ increases as temperature increases. This is also why dissolved oxygen in water bodies falls in warmer months, and why soft-drink bottles are chilled before capping.

✓Final answer
$K_H$
CBSE 2023Set 56/1/11 markMCQ
$Q. Low concentration of oxygen in the blood and tissues of people living at high altitude is due to : (A) high atmospheric pressure (B) low temperature (C) low atmospheric pressure (D) both low temperature and high atmospheric pressure$
›Reveal solutionSolution
Henry’s Law states that the solubility of a gas in a liquid is directly proportional to its partial pressure. At high altitude, low atmospheric pressure reduces the partial pressure of oxygen, lowering its concentration in blood and tissues. The correct answer is (C).

Why Henry’s Law is the key

The question is about oxygen concentration in blood — not about how much oxygen is in the air, but how much dissolves into the bloodstream. That’s governed by Henry’s Law, which describes the solubility of a gas in a liquid.
$C = k_H \cdot P$
For oxygen in blood, $k_H$ is fixed at body temperature. So the concentration of dissolved oxygen depends only on the partial pressure of oxygen in the air you breathe.

Step-by-step reasoning

At sea level, atmospheric pressure is about 1 atm. Oxygen makes up ~21% of air, so its partial pressure is $P_{O_2} \approx 0.21 \times 1 = 0.21$ atm. Henry’s Law then gives a normal dissolved oxygen concentration in blood.

At high altitude (e.g., 3000–5000 m), total atmospheric pressure drops significantly — to about 0.7 atm at 3000 m, and even lower higher up. The fraction of oxygen in air stays 21%, but the partial pressure becomes $P_{O_2} \approx 0.21 \times 0.7 = 0.147$ atm.

Applying Henry’s Law: Since $C = k_H \cdot P_{O_2}$ , a lower $P_{O_2}$ directly gives a lower $C$ . That’s why people at high altitude have low oxygen concentration in blood and tissues — not because the air has less oxygen percentage, but because the lower total pressure reduces oxygen’s partial pressure.

Check the options:

(A) High atmospheric pressure — Wrong. High pressure would increase oxygen solubility, not decrease it.

(B) Low temperature — Temperature affects $k_H$ (solubility increases as temperature drops), but the primary cause at altitude is pressure, not temperature. Also, high-altitude temperatures are lower, which would slightly increase solubility — opposite to the observed effect.

(C) Low atmospheric pressure — Correct. This reduces $P_{O_2}$ , lowering dissolved oxygen.

(D) Both low temperature and high atmospheric pressure — Wrong on both counts.

Watch out

A common mistake is to think that “low oxygen in air” means the percentage of oxygen drops at altitude. It doesn’t — the percentage stays ~21%. What drops is the total pressure, and therefore the partial pressure. Always apply Henry’s Law: solubility depends on partial pressure, not mole fraction alone.

Tip

For exam problems, remember: Henry’s Law is the go-to whenever a question links gas concentration in a liquid to altitude, depth (scuba diving), or carbonated drinks. The constant $k_H$ is temperature-dependent, but unless the problem mentions temperature change, assume it’s fixed.

✓Final answer
$The correct option is (C) low atmospheric pressure .$
CBSE 2023Set 56/2/11 markMCQ
$K_H$
›Reveal solutionSolution
Henry’s constant $K_H$ measures gas solubility in a liquid. Since solubility decreases when temperature rises, $K_H$ must increase with temperature. The correct option is (C).

Why Henry’s constant behaves this way

Henry’s law states that at a constant temperature, the partial pressure of a gas above a liquid is proportional to its mole fraction in the liquid:
$p = K_H \cdot x$
Here $K_H$ is the Henry’s law constant. A larger $K_H$ means that for the same mole fraction $x$ , the gas exerts a higher partial pressure — which implies the gas is less soluble (it prefers to stay in the vapour phase). Conversely, a smaller $K_H$ means the gas dissolves more readily.

So $K_H$ is an inverse measure of solubility: high $K_H$ → low solubility; low $K_H$ → high solubility.

How temperature affects solubility — and therefore $K_H$

Think about dissolving a gas. When a gas dissolves in a liquid, the process is generally exothermic (heat is released). This is because gas molecules must be surrounded by solvent molecules, and the new intermolecular attractions release energy.

Le Chatelier’s principle. For an exothermic dissolution, heat is a product:

$\text{Gas} + \text{solvent} \rightleftharpoons \text{solution} + \text{heat}$
Raising the temperature adds heat, which shifts the equilibrium backward — toward the undissolved gas. So solubility decreases as temperature rises.

Connect to $K_H$ . Since $K_H$ is inversely related to solubility, a decrease in solubility means $K_H$ must increase. Every textbook example confirms this: for oxygen in water, $K_H$ roughly doubles when going from 0 °C to 40 °C.

Watch out

A common mistake is to confuse Henry’s constant with the equilibrium constant of dissolution. While both follow Le Chatelier’s principle, $K_H$ is defined with pressure in the numerator — so its temperature dependence is opposite to what you might guess if you think “constant” means unchanging. It is not constant with temperature.

Check the options.

(A) increases with decrease in temperature — wrong, because lower temperature means higher solubility, so $K_H$ is smaller.

(B) decreases with increase in temperature — wrong, that would mean solubility increases with temperature, which is false for most gases.

(C) increases with increase in temperature — correct, as explained.

(D) remains constant — wrong, $K_H$ is strongly temperature-dependent.

Tip

For quick recall: “Hot liquids lose their fizz.” A warm soda goes flat faster because $K_H$ is larger at higher temperature — the CO₂ is less soluble and escapes more readily.

✓Final answer
$K_H$

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Q.At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is __________.

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

The Key Takeaway

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)

Key Exam Points

Quick Example

Understanding Dynamic Equilibrium

Reaching Equilibrium

Concept: Dynamic Equilibrium in Solutions

Method: Dynamic Equilibrium Principle

Steps:

Common Mistakes & How to Avoid Them

Mistake 1: Confusing “equilibrium” with “no change” → Choosing (iv) zero

Mistake 2: Thinking dissolution stops when solution is saturated

Mistake 3: Misreading “volatile liquid solvent” and overcomplicating

Mistake 4: Picking (i) or (ii) — thinking one rate is always higher

Quick Summary Table

Why Henry’s Law works this way

Step-by-step reasoning

Why this question trips students up

Step-by-step reasoning

Why Henry’s Law is the key

Step-by-step reasoning

Why Henry’s constant behaves this way

How temperature affects solubility — and therefore $K_H$

Q.At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is __________.

Understanding Dynamic Equilibrium

Reaching Equilibrium

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