Equilibrium in Physical Processes

Ice sits in a sealed flask of water at $0\,^\circ\text{C}$ for hours and the amount of ice never changes. Has everything simply stopped? Not at all — water molecules are still leaving the ice and joining it, constantly, just at exactly matched rates. That balanced, ever-active stillness is physical equilibrium, and it is the gateway idea for the whole Equilibrium unit in Class 11 Chemistry. This article expands the three classic physical equilibria — solid–liquid, liquid–vapour, solid–vapour — plus dissolution equilibria and Henry's law, staying entirely within the syllabus scope of general characteristics (no equilibrium constants yet; those come with chemical equilibrium). The goal is to make "dynamic" concrete: to see equilibrium as two rates racing each other to a tie, not as a finish line where change stops. A single proportionality, Henry's law, is derived from that rate-racing logic and then used on a numeric solubility problem.

The core insight

Physical equilibrium is a closed-system, constant-temperature state where two opposing physical processes (melting/freezing, evaporation/condensation, sublimation/deposition, dissolution/crystallisation) proceed at exactly equal, non-zero rates, so macroscopic properties stay constant while molecular change never stops.

The core idea

Four settings, one pattern. In every physical equilibrium a substance sits between two phases or states, moving both ways at once:

Solid ⇌ liquid, at the melting point: $\text{H}_2\text{O(s)} \rightleftharpoons \text{H}_2\text{O(l)}$ , true for ice–water only at $0\,^\circ\text{C}$ (1 atm).
Liquid ⇌ vapour, in a closed vessel at fixed temperature: $\text{H}_2\text{O(l)} \rightleftharpoons \text{H}_2\text{O(g)}$ , giving a steady equilibrium vapour pressure; boiling happens when this vapour pressure equals the external pressure.
Solid ⇌ vapour, for substances that sublime (iodine, camphor, solid $\text{CO}_2$ ): $\text{I}_2\text{(s)} \rightleftharpoons \text{I}_2\text{(g)}$ , no liquid stage involved.
Dissolution equilibrium, once a solvent is saturated with excess solute: $\text{Sugar(s)} \rightleftharpoons \text{Sugar(aq)}$ , dissolution matched by crystallisation.

Why "dynamic" is the right word. Take liquid–vapour as the model case. Evaporation depends only on temperature and exposed surface area, so once the vessel is sealed and the temperature is fixed, the evaporation rate is essentially constant from the first instant. Condensation, by contrast, depends on how much vapour is already present to strike the surface and be recaptured — it starts at zero and climbs as vapour accumulates. The two rates must cross, and at that crossing point they lock together and stay equal, which is exactly why the vapour pressure becomes constant rather than ever-rising or ever-falling.

\text{rate}_{\text{evaporation}} = \text{rate}_{\text{condensation}} \quad \Rightarrow \quad p_{\text{vapour}} = \text{constant (at fixed } T\text{).}

where: $\text{rate}_{\text{evaporation}}$ and $\text{rate}_{\text{condensation}}$ are molecules leaving and rejoining the liquid per unit time, and $p_{\text{vapour}}$ is the equilibrium vapour pressure, a fixed property of the liquid at that temperature — independent of how much liquid or vapour is actually present, as long as both phases coexist.

General characteristics shared by every physical equilibrium: it needs a closed system at constant temperature; observable properties (mass, pressure, concentration, colour) become constant with time; it is dynamic, never static; it is reached from either direction — starting from all-liquid or all-vapour gives the same final vapour pressure; and where the balance settles depends on temperature.

Figure

Two rate curves for evaporation and condensation crossing at the point where vapour pressure becomes constant

Derivation AHenry's law from a rate balance

Show that when a gas is shaken with a liquid solvent in a closed vessel, matching the rate at which gas molecules dissolve into the liquid against the rate at which already-dissolved gas molecules escape back out forces the equilibrium concentration of dissolved gas to be directly proportional to the pressure of gas above the liquid, which is the statement of Henry's law used throughout this unit.

Step 1. Picture the gas above the solvent constantly bombarding the liquid surface. The rate at which gas molecules cross into solution depends on how often they strike the surface, and collision frequency with a surface rises in direct proportion to the gas pressure $p$ above it (more molecules in the same space means more impacts per second). So $\text{rate}_{\text{dissolving}} \propto p$ .

Step 2. Meanwhile, molecules already dissolved in the liquid are constantly escaping back into the gas phase. This escape rate depends only on how many dissolved molecules are near the surface, which is set by the concentration $C$ of dissolved gas already in the solvent. So $\text{rate}_{\text{escaping}} \propto C$ .

Step 3. At equilibrium these two opposing rates are equal by definition (this is exactly the dynamic-equilibrium condition seen above for evaporation and condensation, applied here to a gas dissolving in a liquid instead of a liquid vaporising): $\text{rate}_{\text{dissolving}} = \text{rate}_{\text{escaping}}$ .

Step 4. Substituting the proportionalities from Steps 1 and 2 into the equality of Step 3 gives $k_1 p = k_2 C$ for some constants $k_1, k_2$ that depend on the gas, the solvent, and the temperature. Rearranging,

C = k_H\, p, \qquad k_H = \dfrac{k_1}{k_2}.

where: $C$ is the equilibrium concentration (or mass per unit volume) of dissolved gas, $p$ is the pressure of the gas above the solvent, and $k_H$ is the Henry's law constant for that gas–solvent pair at a fixed temperature. This is Henry's law: solubility of a gas in a liquid is directly proportional to the pressure of the gas above the solution, at constant temperature. Because both $k_1$ and $k_2$ grow with temperature but the escaping term ( $k_2$ , tied to molecules already vibrating in the liquid) grows faster, $k_H$ falls as temperature rises — so gas solubility decreases with rising temperature, matching the everyday observation that warm soda goes flat faster.

Worked example

A bottle of soda is sealed under carbon dioxide gas at a pressure of $4$ atm, giving a dissolved $\text{CO}_2$ concentration of $0.16\ \text{mol/L}$ at $25\,^\circ\text{C}$ . If the same solvent were instead equilibrated with $\text{CO}_2$ at only $1$ atm (roughly its partial pressure in open air) at the same temperature, what dissolved concentration would result once equilibrium is reached, and why does the soda "go flat" once opened?

Step 1. Apply Henry's law, $C = k_H p$ , at the sealed condition to find the constant: $k_H = C/p = 0.16\ \text{mol/L} \div 4\ \text{atm} = 0.04\ \text{mol/L per atm}$ .

Step 2. Since $k_H$ only depends on the gas, solvent, and temperature (all unchanged here), use the same $k_H$ at the new pressure: $C_{\text{new}} = k_H \times p_{\text{new}} = 0.04 \times 1 = 0.04\ \text{mol/L}$ .

Step 3. Compare: dropping the pressure from $4$ atm to $1$ atm quarters the equilibrium solubility, exactly matching the direct proportionality of Step 2's formula — halve the pressure and you halve the equilibrium concentration, and so on.

Step 4. Opening the bottle releases the pressurised $\text{CO}_2$ above the liquid to roughly atmospheric pressure. The dissolution–escape equilibrium is no longer balanced at $0.16\ \text{mol/L}$ ; escape now outpaces dissolving (bubbles rushing out), and the dissolved gas concentration drifts down toward the much lower equilibrium value for $1$ atm. The new equilibrium concentration is $0.04\ \text{mol/L}$ , one quarter of the sealed value — which is why the drink visibly fizzes and eventually goes flat.

Common traps

Thinking equilibrium means all change has stopped — it means forward and reverse rates are equal and non-zero, not zero.
Assuming every solid must melt before it can vaporise — sublimating solids (iodine, camphor, solid $\text{CO}_2$ ) go straight to solid–vapour equilibrium.
Believing vapour pressure depends on how much liquid or vapour is present — it does not, as long as both phases coexist at that temperature.
Mixing up "saturated" with "concentrated" — a saturated solution is in dynamic equilibrium with undissolved solute and can still be dilute (e.g. saturated limewater).

Ask the tutor: ask the AI tutor to sketch the evaporation-versus-condensation rate race for a substance of your choice, or to work a Henry's law problem with different starting numbers.

If molecules are still moving at equilibrium, why do we call macroscopic properties constant?

Because "constant" refers to what you can measure in bulk — total mass of ice, vapour pressure reading, colour intensity — not to what any single molecule is doing. Millions of individual molecules keep crossing between phases every second, but since equal numbers cross in each direction, the net change in any bulk property is zero. It is the same reason a crowded two-way revolving door can have people constantly walking through it while the number of people in the building stays fixed.

Does Henry's law apply to any gas dissolving in any liquid?

Henry's law works well for gases that do not react chemically with the solvent and are present in fairly low concentration, which covers the syllabus cases like $\text{O}_2$ or $\text{CO}_2$ in water. It breaks down for gases that ionise or react with the solvent, such as ammonia or hydrogen chloride in water, because those form new species and the simple proportionality between pressure and dissolved-gas concentration no longer holds.

Why does a fizzy drink go flat faster when it is warm, not just when it is opened?

Two separate effects push the same way here. Opening the bottle drops the pressure above the liquid, which lowers the equilibrium solubility as Henry's law predicts. Raising the temperature does something extra: it makes dissolved gas molecules escape back into the vapour phase faster (the escaping rate in the derivation above), which lowers Henry's law constant $k_H$ itself. So a warm, opened drink loses its dissolved $\text{CO}_2$ from both a lower target solubility and a smaller $k_H$ , and goes flat noticeably faster than a cold, opened one.

Equilibrium in Physical Processes

The core idea

Worked example

Common traps

Before this

Where this leads