What will be the correct order of vapour pressure of water, acetone and ether at 30°C. Given that among these

NCERT Exemplar · Q17

Q. What will be the correct order of vapour pressure of water, acetone and ether at 30°C. Given that among these compounds, water has maximum boiling point and ether has minimum boiling point?

(i) Water < ether < acetone

(ii) Water < acetone < ether

(iii) Ether < acetone < water

(iv) Acetone < ether < water

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Concept understanding — Boiling Point Trends

Boiling Point Trends (Organic Compounds)

A substance's boiling point is set by how much energy is needed to overcome the attractive forces HOLDING its molecules together in the liquid — the stronger those intermolecular forces, the higher the boiling point.

The Forces, Weakest to Strongest

Van der Waals (London dispersion) forces — present in every molecule, and they grow stronger as the molecule gets bigger (more electrons, larger surface area of contact between neighbouring molecules) and more polarisable.
Dipole–dipole forces — present in polar molecules, add an extra attraction on top of dispersion forces.
Hydrogen bonding — present when H is bonded directly to N, O, or F; much stronger than ordinary dipole–dipole attraction, and it raises the boiling point sharply compared to a similarly-sized molecule without it.

Trend 1: Down a Series of Halogens (Same Alkyl Group)

For a fixed R group, boiling point rises as the halogen gets heavier: $R{-}I > R{-}Br > R{-}Cl > R{-}F$ . This looks surprising at first, since electronegativity (and so bond polarity/dipole moment) actually DECREASES down the group — but boiling point here is dominated by the growing size and polarisability of the halogen atom (stronger dispersion forces), which outweighs the shrinking dipole contribution.

The measured values for the methyl, ethyl and propyl halides show this rise clearly:

Measured boiling points of methyl, ethyl and propyl chlorides, bromides and iodides (NCERT Fig 6.1), rising steadily from R-Cl to R-Br to R-I for every alkyl group

Trend 2: Chain Length and Branching

Longer chains (more carbons) have more surface area for van der Waals contact between neighbouring molecules, so boiling point rises with chain length within a homologous series.
Branching LOWERS boiling point compared to a straight-chain isomer of the same molecular formula — a more compact, spherical shape has less surface-to-surface contact with neighbouring molecules, weakening the dispersion forces. (E.g. neopentane boils well below n-pentane.)

Trend 3: Hydrogen Bonding Beats Molecular Mass …

Why this formula?

Boiling Point Trends: Why They Happen

Boiling point is the temperature at which a liquid's vapor pressure equals the external atmospheric pressure. To understand why boiling points follow certain trends, we must first understand what determines vapor pressure.

The Core Idea: Intermolecular Forces

A liquid boils when its molecules have enough kinetic energy to overcome the intermolecular forces (IMFs) holding them together in the liquid phase. Stronger IMFs → harder to escape → lower vapor pressure at a given temperature → higher boiling point.

There is no single "formula" for boiling point, but the relationship is captured by the Clausius–Clapeyron equation, which links vapor pressure ( $P$ ) to temperature ( $T$ ) and the enthalpy of vaporization ( $\Delta H_{\text{vap}}$ ):

\ln P = -\frac{\Delta H_{\text{vap}}}{R} \cdot \frac{1}{T} + C

Where:

$P$ = vapor pressure
$\Delta H_{\text{vap}}$ = enthalpy of vaporization (energy needed to vaporize 1 mole)
$R$ = gas constant
$T$ = absolute temperature (Kelvin)
$C$ = constant (depends on substance)

Why this formula makes sense

$\Delta H_{\text{vap}}$ is large when IMFs are strong — more energy is needed to separate molecules.
At boiling point, $P = P_{\text{atm}}$ (usually 1 atm). So a substance with larger $\Delta H_{\text{vap}}$ needs a higher $T$ to reach that pressure.

Thus, boiling point ∝ strength of intermolecular forces.

The Four Key Trends (with Reasoning)

1. Trend across a period (e.g., Period 2: CH₄ → NH₃ → H₂O → HF)

Molecule	IMFs present	Boiling point (°C)
CH₄	London dispersion only	-161
NH₃	Dispersion + H-bonding	-33
H₂O	Dispersion + H-bonding (2 per molecule)	100
HF	Dispersion + H-bonding	19

Why?

CH₄ is nonpolar — only weak London dispersion forces.
NH₃, H₂O, HF have hydrogen bonding (strongest IMF).
H₂O forms two H-bonds per molecule (donor + acceptor), while NH₃ forms one and HF forms one — hence H₂O has the highest boiling point.

Key insight: Hydrogen bonding dominates over molecular mass in small molecules.

2. Trend down a group (e.g., Halogens: F₂ → Cl₂ → Br₂ → I₂)

Molecule	Molar mass (g/mol)	Boiling point (°C)
F₂	38	-188
Cl₂	71	-34
Br₂	160	59
I₂	254	184

Why?

All are nonpolar — only London dispersion forces.
Dispersion force strength increases with number of electrons (larger molar mass → more polarizable electron cloud → stronger temporary dipoles).
So boiling point increases down the group.

Key insight: For nonpolar molecules, molar mass (electron count) is the primary factor.

3. Branching in alkanes (e.g., C₅H₁₂ isomers)

Isomer	Boiling point (°C)
n-pentane (straight chain)	36
2-methylbutane (branched)	28
2,2-dimethylpropane (highly branched)	10

Why?

All have same molecular formula — same molar mass. …

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Vapour pressure and boiling point are inversely related — the compound with the lowest boiling point has the highest vapour pressure. Since ether has the minimum boiling point and water the maximum, the order of vapour pressure is water < acetone < ether, which is option (ii).

Why Boiling Point Tells You Vapour Pressure

The key idea is simple: boiling point is the temperature at which the vapour pressure of a liquid equals the surrounding atmospheric pressure. At any fixed temperature (here, 30°C), a liquid with a lower boiling point is already closer to its boiling condition — meaning it has a higher vapour pressure at that temperature. Conversely, a liquid with a higher boiling point has stronger intermolecular forces holding its molecules together, so fewer molecules escape into the vapour phase, giving a lower vapour pressure.

So the relationship is:

\text{Boiling point} \uparrow \quad \Longrightarrow \quad \text{Vapour pressure at a given } T \downarrow

This is an inverse proportionality, not a linear one, but the direction is rock-solid.

Step-by-step reasoning

Identify the given boiling-point order. The problem states: water has the maximum boiling point, ether has the minimum boiling point. Acetone lies in between. So:

\text{Boiling point: Water} > \text{Acetone} > \text{Ether}

Apply the inverse relationship. Since vapour pressure at a fixed temperature is inversely related to boiling point, the order of vapour pressure is exactly the reverse: …

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Q.What will be the correct order of vapour pressure of water, acetone and ether at 30°C. Given that among these compounds, water has maximum boiling point and ether has minimum boiling point?

Why Boiling Point Tells You Vapour Pressure

Step-by-step reasoning

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