Liquid Properties and Hydrogen Bonding
Intermolecular Forces
The Kinetic Molecular Theory explains that molecules in a liquid are in constant motion but are held together by intermolecular forces — attractive forces between neighbouring molecules that are much weaker than covalent bonds within a molecule. These forces determine whether a substance is a gas, liquid, or solid at a given temperature, and they control properties like boiling point, vapour pressure, and viscosity.
Four types of intermolecular forces exist: dipole-dipole forces, ion-dipole forces, dipole-induced dipole forces, and London dispersion forces (instantaneous dipole-induced dipole forces).
Strength comparison: covalent bonds >> hydrogen bonds > dipole-dipole > London dispersion forces.
Physical origin: Intermolecular forces arise from electrostatic attractions between partial charges on molecules — they have nothing to do with valence electrons, unlike covalent or ionic bonding.
Types of Intermolecular Forces (Weakest to Strongest)
1
London dispersion forces — present in all molecules
2
Dipole-induced dipole forces (Debye forces)
3
Dipole-dipole forces — polar molecules only
4
Ion-dipole forces — ions + polar molecules
5
Hydrogen bonding — strongest intermolecular force
In polar molecules like HCl, the difference in electronegativity between bonded atoms creates partial positive and partial negative charges. The positive end of one molecule attracts the negative end of another, producing dipole-dipole forces. Thermal energy prevents perfect alignment, but a net attraction persists. These forces are roughly one percent as strong as a covalent bond.
Strength depends on two factors: electronegativity difference between bonded atoms, and the distance between molecules.
In gases these forces are very weak because intermolecular distances are large; in liquids they are reasonably strong.
Greater dipole-dipole strength means higher melting points, boiling points, heats of vapourisation, and heats of sublimation.
Molecules Showing Dipole-Dipole Attractions
•
HCl — Cl(δ⁻) attracts H(δ⁺) of neighbouring molecule
•
CHCl₃ (chloroform) — three Cl(δ⁻) attract H(δ⁺) of adjacent molecule
When a polar molecule approaches a non-polar molecule, the partial charge of the polar molecule distorts the electron cloud of the non-polar one, inducing a temporary dipole. The resulting attraction is called dipole-induced dipole force (Debye force). These forces are weaker than dipole-dipole forces but important in mixtures of polar and non-polar substances.
Polar molecule acts as the source: its permanent partial charge creates an electric field that shifts the electron cloud of the nearby non-polar molecule.
Induced dipole is temporary: it exists only while the polar molecule is nearby.
Strength depends on polarizability of the non-polar molecule — larger electron clouds are more easily distorted.
London dispersion forces are the weakest intermolecular forces, yet they are present in ALL molecules — polar and non-polar alike. In a non-polar molecule like helium, the electrons are in constant motion. At any instant, the electron density may be uneven, creating an instantaneous dipole. This momentary dipole distorts the electron cloud of a neighbouring atom, inducing a second dipole. The fleeting attraction between them is the London force.
Very short-lived: dipoles vanish as quickly as they form, but new ones constantly appear in different orientations.
Dominant force for non-polar molecules like Cl₂, H₂, and noble gases (He, Ne, Ar).
Also present in polar molecules but usually overshadowed by dipole-dipole and hydrogen bonding.
Discovered by Fritz London in 1930 — sometimes called London forces or dispersion forces.
The strength of London dispersion forces depends on two key factors: the size of the electron cloud (and thus the polarizability) and the number of atoms in the molecule. Both factors explain why boiling points increase down groups in the periodic table and why larger hydrocarbons have much higher boiling points than smaller ones.
Polarizability increases with atomic/molecular size: larger electron clouds are more easily distorted, creating stronger London forces.
Noble gases (Group VIIIA): boiling points increase from He (−268.6 °C) to Rn (−61.8 °C) because outer electrons are farther from the nucleus and more easily polarized.
Halogens (Group VIIA): fluorine is a gas (B.P. −188.1 °C) while iodine is a solid (B.P. 184.4 °C) — the same trend from greater polarizability.
Number of atoms matters: C₂H₆ boils at −88.6 °C but C₆H₁₄ boils at 68.7 °C because longer chains provide more surface area for London forces to act along.
Boiling Points of Halogens and Noble Gases
•
He: −268.6 °C | F: −188.1 °C
•
Ne: −245.9 °C | Cl: −34.6 °C
•
Ar: −185.7 °C | Br: 58.8 °C
•
Kr: −152.3 °C | I: 184.4 °C
•
Xe: −107.1 °C | At: 337 °C
•
Rn: −61.8 °C
Evaporation
According to the Kinetic Molecular Theory, molecules in a liquid have a range of kinetic energies — some move slowly, others fast. If a high-energy molecule reaches the surface, it may overcome the intermolecular forces holding it and escape into the gas phase. This spontaneous conversion of liquid to vapour is called evaporation, and it occurs at all temperatures.
Evaporation is a surface phenomenon: only molecules at the surface can escape because molecules in the bulk are surrounded on all sides.
Cooling effect: when high-energy molecules leave, the average kinetic energy of the remaining liquid decreases — the liquid cools. Heat flows from the surroundings to the liquid, and the surroundings also cool.
Occurs at all temperatures: unlike boiling, evaporation does not require the vapour pressure to equal atmospheric pressure.
The rate of evaporation depends on several factors. Since evaporation occurs from the surface, increasing the surface area allows more molecules to escape per unit time. For liquids with the same surface area, the rate is controlled by temperature and the strength of intermolecular forces.
Surface area: larger surface → more molecules exposed → faster evaporation.
Temperature: higher temperature → more high-energy molecules → faster evaporation.
Intermolecular forces: weaker forces → molecules escape more easily → faster evaporation. Gasoline evaporates much faster than water because its London forces are weaker than water's hydrogen bonds.
Wind/airflow: removes vapour molecules from above the surface, shifting the equilibrium and accelerating evaporation.
Vapour Pressure
When a liquid evaporates in a closed container, vapour molecules accumulate above the surface. These molecules collide with the container walls (exerting pressure) and also with the liquid surface, where some are recaptured — a process called condensation. Eventually, the rate of evaporation equals the rate of condensation, and the system reaches dynamic equilibrium. At this point, the pressure exerted by the vapour is the equilibrium vapour pressure of the liquid.
$$\text{Rate of evaporation} = \text{Rate of condensation (at equilibrium)}$$
At dynamic equilibrium, the number of molecules leaving the liquid surface equals the number returning — the vapour pressure remains constant
Independent of amount: vapour pressure depends only on temperature and the nature of the liquid — not on the amount of liquid, the volume of the container, or the surface area.
Surface area cancels out: a larger surface area increases both the rate of evaporation AND the rate of condensation equally, so equilibrium vapour pressure is unchanged.
Nature of the liquid matters: stronger intermolecular forces mean fewer molecules escape, so lower vapour pressure. Glycerol (0.00016 torr at 20 °C) has a far lower vapour pressure than isopentane (580 torr at 20 °C).
Vapour Pressure of Water at Various Temperatures
1
0 °C → 4.579 torr
2
10 °C → 9.209 torr
3
20 °C → 17.54 torr
4
37 °C → 47.07 torr
5
50 °C → 92.51 torr
6
70 °C → 233.7 torr
7
90 °C → 527.8 torr
8
100 °C → 760.0 torr
The most important factor controlling vapour pressure is temperature. As temperature rises, the kinetic energy of molecules increases, so more molecules have sufficient energy to escape the liquid surface. The vapour pressure increases nonlinearly — the increment grows larger as the boiling point is approached.
Nonlinear increase: from 0 °C to 10 °C, water's vapour pressure rises by 4.63 torr (4.579 → 9.209), but from 90 °C to 100 °C it rises by 232.2 torr (527.8 → 760.0).
Intermolecular forces set the starting point: at 0 °C, water starts at 4.8 torr while diethyl ether starts near 200 torr because ether's weaker forces allow far more molecules to escape.
Rapid rise near boiling point: vapour pressure curves are steep near the boiling point — a small temperature increase causes a large pressure increase.
Vapour pressure can be measured experimentally using the manometric method. The liquid is frozen, the air above it is evacuated with a vacuum pump, and the process is repeated until all air is removed. The liquid is then warmed to the desired temperature in a thermostat, and the difference in mercury levels in the two limbs of the manometer gives the vapour pressure.
$$P = P_a + \Delta h$$
The vapour pressure of the liquid equals atmospheric pressure plus the mercury height difference in the manometer
$P$=Vapour pressure of the liquid(torr)
$P_a$=Atmospheric pressure(torr)
$\Delta h$=Difference in mercury column heights between the two manometer limbs(torr (mm Hg))
Freeze-evacuate-thaw cycle: repeated freezing and evacuation removes all dissolved air so that only the liquid's vapour remains above the surface.
Mercury column facing the liquid is depressed more because the total pressure on that side equals atmospheric pressure plus vapour pressure.
Thermostat maintains constant temperature during measurement — vapour pressure is highly temperature-dependent.
Boiling Point
When a liquid is heated, its vapour pressure increases. The boiling point is the temperature at which the vapour pressure of the liquid equals the external atmospheric pressure. At this temperature, bubbles of vapour form in the interior of the liquid — their internal pressure exceeds atmospheric pressure, so they rise to the surface and burst. A constant stream of bubbles characterises boiling.
$$P_{\text{vapour}} = P_{\text{external}} \quad \text{(at boiling point)}$$
Boiling occurs when the vapour pressure inside bubbles within the liquid matches the external pressure pushing down on the liquid surface
$P_{\text{vapour}}$=Vapour pressure of the liquid(torr (or atm))
$P_{\text{external}}$=External atmospheric pressure(torr (or atm))
$P_{\text{external}} = 760 \text{ torr}$
→Normal boiling point at sea level (1 atm)
Temperature stays constant during boiling: at the boiling point, added heat does not raise the temperature — it only breaks intermolecular forces and converts liquid to vapour.
Molar heat of vapourisation: the amount of heat required to vapourise one mole of a liquid at its boiling point. For water, this is 40.6 kJ mol⁻¹.
Stronger intermolecular forces → higher boiling point: water (100 °C, H-bonding) boils much higher than diethyl ether (34.6 °C, weaker forces).
Boiling Points of Common Liquids at 760 torr
•
Carbon disulphide: 46.30 °C
•
Acetone: 56.00 °C
•
Carbon tetrachloride: 76.50 °C
•
Ethanol: 78.26 °C
•
Benzene: 80.15 °C
•
Acetic acid: 118.50 °C
•
Phenol: 181.80 °C
•
Water: 100.00 °C
•
Naphthalene: 218.00 °C
Since boiling occurs when vapour pressure equals external pressure, changing the external pressure directly changes the boiling point. At higher external pressures, the liquid needs more heat to reach that pressure, so the boiling point rises. At lower pressures, the liquid boils at a lower temperature. This principle has practical applications in cooking and industrial distillation.
Higher pressure → higher boiling point: a pressure cooker traps steam, raising internal pressure and the boiling temperature of water, so food cooks faster.
Lower pressure → lower boiling point: water boils at 69 °C on Mount Everest (323 torr) and at 98 °C in Murree Hills (700 torr), compared to 100 °C at sea level (760 torr).
Vacuum distillation: by lowering the pressure, liquids boil at temperatures below their normal boiling points. Glycerin decomposes at 290 °C (its normal B.P.), but under vacuum at 50 torr it boils at just 210 °C and can be distilled without decomposition.
Energetics of Phase Changes
When matter changes state, energy is absorbed or released. At constant pressure, this heat change is called the enthalpy change ($\Delta H$). Three key enthalpy changes accompany physical transitions: molar heat of fusion, molar heat of vapourisation, and molar heat of sublimation. All three are positive (endothermic) because energy must be supplied to overcome intermolecular forces.
$$\Delta H_s > \Delta H_v > \Delta H_f$$
The enthalpy change is always largest for sublimation, intermediate for vapourisation, and smallest for fusion — reflecting the increasing molecular separation in each process
$\Delta H_s$=Molar heat of sublimation (solid → vapour)(kJ/mol)
$\Delta H_v$=Molar heat of vapourisation (liquid → vapour)(kJ/mol)
$\Delta H_f$=Molar heat of fusion (solid → liquid)(kJ/mol)
Molar heat of fusion ($\Delta H_f$): heat absorbed when one mole of solid melts to liquid at its melting point (1 atm).
Molar heat of vapourisation ($\Delta H_v$): heat absorbed when one mole of liquid vaporises at its boiling point (1 atm). Water: 40.6 kJ mol⁻¹.
Molar heat of sublimation ($\Delta H_s$): heat absorbed when one mole of solid sublimes directly to vapour.
$\Delta H_v > \Delta H_f$ because molecules separate much farther apart during vapourisation than during melting — the change in potential energy is greater.
$\Delta H_v$ values reflect intermolecular force strength: polar molecules like H₂O (40.6 kJ/mol) and NH₃ (21.7 kJ/mol) have higher $\Delta H_v$ than non-polar ones like CH₄ (8.6 kJ/mol).
Heats of Vapourisation of Selected Substances
•
H₂O: +40.6 kJ/mol
•
NH₃: +21.7 kJ/mol
•
SO₂: +24.3 kJ/mol
•
HCl: +15.6 kJ/mol
•
I₂: +22.0 kJ/mol
•
C₆H₁₄: +30.1 kJ/mol
•
C₃H₈: +16.9 kJ/mol
•
F₂: +5.9 kJ/mol
•
CH₄: +8.6 kJ/mol
Every change of state eventually reaches dynamic equilibrium — a condition where two opposing processes occur at equal rates. At 0 °C, ice and liquid water coexist in dynamic equilibrium. The concept of dynamic equilibrium extends to all reversible physical changes: solid-liquid, liquid-vapour, and solid-vapour transitions all reach this balanced state under the right conditions of temperature and pressure.
$$\text{ice} \rightleftharpoons \text{water} \rightleftharpoons \text{steam}$$
Phase changes are reversible — at equilibrium, forward and reverse processes occur at equal rates, so the amounts of each phase remain constant
Liquid-vapour equilibrium: at any temperature, a liquid in a closed container establishes dynamic equilibrium with its vapour.
Solid-liquid equilibrium: at the melting point, solid and liquid coexist in dynamic equilibrium (e.g., ice and water at 0 °C).
Dynamic equilibrium is the fate of all reversible changes — physical and chemical alike.
Hydrogen Bonding
Hydrogen bonding is the strongest type of intermolecular force. It occurs when a hydrogen atom covalently bonded to a highly electronegative atom (F, O, or N) is attracted to a lone pair on another electronegative atom in a neighbouring molecule. The small size of hydrogen allows it to approach closely, creating a strong electrostatic attraction that acts as a bridge between two electronegative atoms. Hydrogen bonds are about twenty times weaker than covalent bonds but significantly stronger than ordinary dipole-dipole forces.
Requires two conditions: a hydrogen atom bonded to F, O, or N (the hydrogen-bond donor), and a lone pair on F, O, or N on another molecule (the hydrogen-bond acceptor).
Water as the classic example: each H₂O molecule has two hydrogen atoms (donors) and two lone pairs on oxygen (acceptors), allowing it to form up to four hydrogen bonds.
Chlorine rarely forms hydrogen bonds, but in chloroform (CHCl₃), the three Cl atoms pull electron density away from C, giving the H atom sufficient partial positive charge to hydrogen-bond with the oxygen of acetone.
Strength order: H-F > H-O > H-N, reflecting electronegativity differences.
Compounds That Exhibit Hydrogen Bonding
•
H₂O — two H donors, two lone-pair acceptors on O
•
NH₃ — three H donors, one lone-pair acceptor on N
•
HF — one H donor, three lone-pair acceptors on F
•
C₂H₅OH (ethanol) — O-H group enables H-bonding
•
CH₃COOH (acetic acid) — O-H and C=O groups
•
CHCl₃ + CH₃COCH₃ (acetone) — intermolecular H-bond
Hydrogen bonding dramatically affects the boiling points of covalent hydrides. In groups VA, VIA, and VIIA, the first-row hydrides (NH₃, H₂O, HF) have anomalously high boiling points compared to the rest of their group. This anomaly is absent in Group IVA because those elements are not electronegative enough to form hydrogen bonds.
Group VIA trend: H₂O (100 °C) is liquid at room temperature, while H₂S (−61 °C), H₂Se (−42 °C), and H₂Te (−2 °C) are all gases — the gap is due entirely to hydrogen bonding in water.
Group VA trend: NH₃ (−33.4 °C) has a much higher boiling point than PH₃ (−88 °C), AsH₃ (−55 °C), and SbH₃ (−17 °C).
Group VIIA trend: HF (19.9 °C) boils far above HCl (−85 °C), HBr (−67 °C), and HI (−38 °C).
Why HF does not beat H₂O: although F is more electronegative than O, each HF molecule can form only one hydrogen bond (one H, multiple acceptors but limited by geometry). Each H₂O molecule can form two hydrogen bonds (two H atoms + two lone pairs = four attachment points), giving water a more extensive H-bond network.
NH₃ forms fewer H-bonds per molecule: it has three H atoms but only one lone pair, limiting it to roughly one effective H-bond per molecule on average.
HCl is borderline: its slightly higher boiling point than expected for pure London/dipole-dipole forces suggests very weak H-bonding, but it is not a strong hydrogen-bonded system.
Boiling Points of Hydrides
•
CH₄: −164 °C | NH₃: −33.4 °C | H₂O: 100 °C | HF: 19.9 °C
•
SiH₄: −112 °C | PH₃: −88 °C | H₂S: −61 °C | HCl: −85 °C
•
GeH₄: −88 °C | AsH₃: −55 °C | H₂Se: −42 °C | HBr: −67 °C
•
SnH₄: −52 °C | SbH₃: −17 °C | H₂Te: −2 °C | HI: −38 °C
Hydrogen bonding is the key to understanding why some substances dissolve in water and others do not. Water itself is an extensive H-bonded network. Substances that can also form hydrogen bonds with water — like ethyl alcohol and small carboxylic acids — dissolve readily. Non-polar substances like hydrocarbons cannot form H-bonds with water and are insoluble.
Like dissolves like (via H-bonding): ethanol (C₂H₅OH) is miscible with water because the −OH group forms hydrogen bonds with water molecules.
Small carboxylic acids dissolve: acetic acid (CH₃COOH) is soluble because its −OH and C=O groups can H-bond with water.
Hydrocarbons are insoluble: they are non-polar and lack −OH, −NH, or −FH groups, so no hydrogen bonding is possible with water.
Hydrogen bonding has wide-ranging applications across chemistry and biology. It explains the cleansing action of soaps, the structure and stability of proteins and DNA, the stickiness of glue and honey, the adhesive properties of paints and dyes, and the tensile strength of fibres like cotton and silk.
Soaps and detergents: the polar (hydrophilic) part of the molecule is water-soluble due to hydrogen bonding, while the non-polar (hydrophobic) alkyl/benzyl portion dissolves grease — together they emulsify dirt.
Proteins (alpha-helix): long chains of amino acids coil into helices. In the alpha-helix, >NH and >C=O groups on adjacent turns are linked by hydrogen bonds, with about 27 amino acid residues per turn of the helix.
DNA (double helix): two spiral strands coil around a common axis (18–20 Å diameter). H-bonds between complementary base pairs (A–T, G–C) link the two strands together.
Carbohydrates: glucose, fructose, and sucrose all contain −OH groups responsible for intermolecular hydrogen bonding.
Textiles: hydrogen bonding gives cotton, silk, and synthetic fibres their rigidity and tensile strength.
Paints, dyes, glue, and honey: their adhesive and sticky properties arise from hydrogen bonding between molecules.
Anomalous Behaviour of Water
Water exhibits a unique property among liquids: its density is maximum at 4 °C, not at its freezing point. When water freezes to ice, the molecules arrange into a regular tetrahedral lattice held together by an extensive hydrogen bond network. This ordered structure contains large empty spaces, causing ice to occupy about 9% more volume than liquid water. As a result, ice is less dense than liquid water and floats.
Tetrahedral structure of ice: each water molecule sits at the centre of a tetrahedron, connected to four neighbours via hydrogen bonds (analogous to the carbon lattice in diamond).
Liquid water is denser: in the liquid state, H-bonds constantly break and reform, allowing molecules to pack more closely — hence liquid water is denser than ice.
Density maximum at 4 °C: as liquid water cools from room temperature to 4 °C, it contracts and becomes denser. Below 4 °C, the expanding effect of forming more ordered H-bond clusters outweighs thermal contraction, and density decreases.
Ice floats on water: the open hexagonal structure of ice has about 9% more volume than the same mass of liquid water.
The anomalous density behaviour of water has profound ecological consequences. In winter, as atmospheric temperature drops, the surface layer of a lake or pond cools to 4 °C and sinks because it is the densest water. Below 4 °C, the cooled water becomes less dense and stays on top, eventually freezing. The layer of ice acts as an insulating blanket, preventing the water below from freezing and allowing aquatic life to survive throughout winter.
Freezing from the top down: the surface freezes first because cold water (below 4 °C) is less dense and stays on top of warmer water beneath.
Ice insulates: the layer of ice has low thermal conductivity and prevents further heat loss from the water underneath.
Survival of aquatic life: fish and plants survive under the ice blanket because liquid water persists below the frozen surface, maintaining temperatures around 4 °C.
Without H-bonding in water, ponds would freeze from the bottom up, killing all aquatic life — the pattern of life on Earth would be fundamentally different.
Application Strategy
Worked Examples
Problem Statement
A liquid has a vapour pressure of 233.7 torr at 70 °C. Identify the liquid using the vapour pressure-temperature data for water.
vapour_pressureintermolecular_forces
Strategic Toolkit
$$P = \text{vapour pressure at given temperature}$$
Step 1
Locate the temperature
The given temperature is 70 °C. We need to check the vapour pressure of water at this temperature from the reference table.
$$T = 70\degree C$$
Step 2
Read the vapour pressure from the table
From the vapour pressure table for water, at 70 °C the vapour pressure is 233.7 torr.
$$P = 233.7 \text{ torr at } 70\degree C$$
Step 3
Match the given value
The given vapour pressure of 233.7 torr matches the vapour pressure of water at 70 °C, so the liquid is water.
$$P_{\text{given}} = P_{\text{water at } 70\degree C} = 233.7 \text{ torr}$$
Problem Statement
In a manometric experiment, the atmospheric pressure is 758 torr and the difference in mercury column heights between the two limbs of the manometer is 4.2 torr. Calculate the vapour pressure of the liquid.
vapour_pressuremanometric_method
Strategic Toolkit
$$P = P_a + \Delta h$$
Step 1
Identify the given values
The atmospheric pressure is 758 torr and the mercury height difference is 4.2 torr.
$$P_a = 758 \text{ torr}, \quad \Delta h = 4.2 \text{ torr}$$
Step 2
Apply the manometric formula
Substitute the values into the manometric equation to find the vapour pressure of the liquid.
$$P = 758 + 4.2 = 762.2 \text{ torr}$$
Step 3
State the result
The vapour pressure of the liquid is 762.2 torr.
$$P = 762.2 \text{ torr}$$
Problem Statement
Explain why H₂O (molar mass 18 g/mol) has a boiling point of 100 °C while H₂S (molar mass 34 g/mol) boils at −61 °C, even though H₂S is heavier and has stronger London dispersion forces.
hydrogen_bondintermolecular_forcesboiling_point
Step 1
Note the apparent contradiction
H₂S is heavier than H₂O, so it should have stronger London forces and a higher boiling point based on mass alone. Yet the opposite is true.
Step 2
Identify the dominant intermolecular force in each
H₂O forms an extensive hydrogen-bond network because oxygen is highly electronegative (EN = 3.44) and each molecule has two H donors and two lone-pair acceptors. H₂S cannot form hydrogen bonds because sulfur is not electronegative enough (EN = 2.58) — it only has dipole-dipole and London forces.
Step 3
Conclude which force dominates
Hydrogen bonding in water is far stronger than the combined dipole-dipole and London forces in H₂S, giving water a much higher boiling point despite its lower molar mass.
Problem Statement
Water boils at 100 °C at sea level (760 torr). What is the boiling point of water at Mount Everest where the atmospheric pressure is approximately 323 torr?
boiling_pointvapour_pressureexternal_pressure
Strategic Toolkit
$$P_{\text{vapour}} = P_{\text{external}} \quad \text{(at boiling point)}$$
Step 1
Understand the boiling condition
Boiling occurs when the vapour pressure of the liquid equals the external atmospheric pressure. At Everest, the external pressure is only 323 torr.
$$P_{\text{external}} = 323 \text{ torr}$$
Step 2
Find the temperature where water's vapour pressure equals 323 torr
From the vapour pressure-temperature data for water, water reaches a vapour pressure of approximately 323 torr at 69 °C (between the 70 °C entry of 233.7 torr and 80 °C entry of 355.1 torr — interpolation gives approximately 69 °C).
$$P_{\text{water}} = 323 \text{ torr at } \approx 69\degree C$$
Step 3
State the boiling point at altitude
Water boils at approximately 69 °C on Mount Everest, well below the 100 °C sea-level boiling point.
$$\text{B.P. at Everest} \approx 69\degree C$$
Problem Statement
Use the heats of vapourisation data to rank the following substances by the strength of their intermolecular forces: CH₄ (8.6 kJ/mol), C₆H₁₄ (30.1 kJ/mol), HCl (15.6 kJ/mol), and H₂O (40.6 kJ/mol).
molar_heat_vapourisationintermolecular_forcesenthalpy_change
Strategic Toolkit
$$\Delta H_v \text{ directly reflects intermolecular force strength}$$
Step 1
List the given values
Arrange the $\Delta H_v$ values in ascending order to rank the intermolecular force strengths.
$$\text{CH}_4: 8.6 < \text{HCl}: 15.6 < \text{C}_6\text{H}_{14}: 30.1 < \text{H}_2\text{O}: 40.6 \text{ (kJ/mol)}$$
Step 2
Rank from weakest to strongest intermolecular forces
CH₄ has the weakest intermolecular forces (only London dispersion in a small molecule). HCl has moderate dipole-dipole forces. C₆H₁₄ has strong London forces due to its large size. H₂O has the strongest forces due to hydrogen bonding.
$$\text{Weakest: CH}_4 < \text{HCl} < \text{C}_6\text{H}_{14} < \text{H}_2\text{O : Strongest}$$
Step 3
Identify the dominant force for each substance
CH₄ relies on weak London forces (small molecule, low polarizability). HCl has dipole-dipole forces. C₆H₁₄ has strong London forces (large molecule, high polarizability). H₂O has hydrogen bonding — the strongest intermolecular force among these.
Problem Statement
At 20 °C, isopentane has a vapour pressure of 580 torr while glycerol has a vapour pressure of 0.00016 torr. Explain this enormous difference.
vapour_pressureintermolecular_forceshydrogen_bond
Step 1
Compare the molecular structures
Isopentane (C₅H₁₂) is a non-polar hydrocarbon with only London dispersion forces. Glycerol (C₃H₈O₃) has three −OH groups and can form extensive hydrogen bonding.
Step 2
Relate intermolecular forces to vapour pressure
Stronger intermolecular forces mean fewer molecules have enough energy to escape the liquid surface at a given temperature, resulting in lower vapour pressure. Glycerol's hydrogen bonds are far stronger than isopentane's London forces.
Step 3
State the conclusion
The vapour pressure ratio is about 3.6 million to one — glycerol's extensive hydrogen bonding network traps molecules far more effectively than isopentane's weak London forces, giving glycerol a negligible vapour pressure at room temperature.
$$\frac{580}{0.00016} \approx 3.6 \times 10^6$$
Memorize formulas
Vapour Pressure (Manometric Method)
Use when calculating vapour pressure from manometer readings. $P_a$ = atmospheric pressure, $\Delta h$ = mercury height difference between the two limbs.
Formula
$$P = P_a + \Delta h$$
Boiling Point Condition
Boiling occurs when liquid's vapour pressure equals the external atmospheric pressure. At sea level, $P_{\text{external}}$ = 760 torr.
Formula
$$P_{\text{vapour}} = P_{\text{external}}$$
Enthalpy Hierarchy of Phase Changes
Sublimation requires the most energy, vapourisation the next, and fusion the least — reflecting increasing molecular separation.
Formula
$$\Delta H_s > \Delta H_v > \Delta H_f$$
Molar Heat of Vapourisation of Water
Energy required to convert 1 mole of liquid water to steam at 100 °C and 1 atm. This value reflects the strength of water's hydrogen bonds.
Formula
$$\Delta H_v(\text{H}_2\text{O}) = 40.6 \text{ kJ mol}^{-1}$$
Maximum Density of Water
Water is densest at 4 °C (1.000 g/cm³). Below 4 °C, the expanding effect of forming ordered hydrogen-bond clusters dominates, and density decreases until ice forms at 0 °C.
Formula
$$\rho_{\text{max}}(\text{H}_2\text{O}) = 1.000 \text{ g/cm}^3 \text{ at } 4\degree C$$
Review
1
According to the kinetic molecular theory, which statement best explains why a liquid has a definite volume but takes the shape of its container?
Ans
Particles are close together but can slide past one another, allowing the liquid to flow
Review
Tap to reveal
Recall KMT for liquids
Liquid particles are in close proximity with strong enough intermolecular attractions to maintain a fixed volume, but they have enough kinetic energy to move and change positions.
Definite volume, variable shape
Compare with other states
Solids have particles in fixed positions (definite volume AND shape). Gases have particles far apart (neither definite volume nor shape). Liquids are intermediate.
Liquids flow while retaining volume
2
For a hydrogen bond to form between two molecules, hydrogen must be covalently bonded to which of the following?
Ans
Nitrogen, oxygen, or fluorine
Review
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Identify hydrogen bonding requirement
The H atom must be bonded to an atom with very high electronegativity and small atomic size so that it acquires a strong partial positive charge.
N (3.04), O (3.44), F (3.98)
Eliminate incorrect options
Carbon (EN = 2.55) does not create sufficient δ⁺ on H. Sulfur and chlorine, while electronegative, do not form hydrogen bonds of comparable strength to N, O, and F in the standard classification.
Only N, O, and F satisfy the conditions
3
Which statement correctly describes London dispersion forces?
Ans
They are present in all molecules and arise from temporary, instantaneous shifts in the electron cloud
Review
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Origin of London forces
The electron cloud around any atom or molecule is in constant motion. At any instant, the electron distribution may be uneven, creating a temporary dipole.
Instantaneous dipole → induces dipole in neighbour
Universality of London forces
Since all molecules have electron clouds, all molecules experience London dispersion forces. In nonpolar molecules, these are the only intermolecular forces.
Present in all substances
4
At what temperature does pure water reach its maximum density?
Ans
4°C
Review
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Identify the anomalous behaviour of water
Most liquids expand and become less dense as they are cooled, but water behaves differently due to hydrogen bonding forming an open cage-like structure as temperature decreases below $4\degree C$.
$$\rho_{\max}(\text{H}_2\text{O}) = 1.000 \text{ g/cm}^3 \text{ at } 4\degree C$$
Trend above and below 4°C
Above $4\degree C$: normal thermal expansion reduces density. Below $4\degree C$: hydrogen-bonded open structures increase volume and reduce density.
Peak density at $4\degree C$ only
5
Evaporation occurs because molecules at the surface of a liquid:
Ans
Gain enough kinetic energy through collisions to overcome intermolecular attractions and escape into the vapour phase
Review
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Understand the energy distribution
Molecules in a liquid have a Maxwell-Boltzmann distribution of kinetic energies. Only a fraction at the high-energy tail have enough energy to escape.
Not all molecules can evaporate — only the fastest
Surface requirement
Evaporation is limited to surface molecules because only they experience an unbalanced net attraction (molecules in the bulk are pulled equally in all directions).
Surface molecules with sufficient energy escape
6
Ethanol ($\text{CH}_3\text{CH}_2\text{OH}$) and dimethyl ether ($\text{CH}_3\text{OCH}_3$) are structural isomers. At the same temperature, which statement correctly compares their vapour pressures?
Ans
Dimethyl ether has the higher vapour pressure because ethanol forms intermolecular hydrogen bonds that hold its molecules together more strongly
Review
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Identify intermolecular forces in each substance
Ethanol: hydrogen bonding (O$-$H present) + dipole-dipole + London. Dimethyl ether: dipole-dipole + London only (no O$-$H bond).
Ethanol has stronger IM forces
Relate IM force strength to vapour pressure
Stronger intermolecular forces make it harder for molecules to escape the liquid, lowering the equilibrium vapour pressure.
$P_{\text{vapour}}$(dimethyl ether) > $P_{\text{vapour}}$(ethanol)
7
A liquid begins to boil when:
Ans
Its vapour pressure equals the external pressure acting on the surface of the liquid
Review
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State the boiling point condition
Boiling begins when the vapour pressure within the liquid equals the pressure pushing down on the liquid surface from outside.
$$P_{\text{vapour}} = P_{\text{external}}$$
Explain the physical meaning
When vapour pressure equals external pressure, bubbles can form throughout the liquid (not just at the surface), because the internal vapour pressure can push back against the external pressure.
Bubbles form throughout the bulk liquid
8
Using the manometric method, a student determines the vapour pressure of a liquid. The atmospheric pressure is $760 \text{ mmHg}$ and the mercury column in the closed limb is $28 \text{ mmHg}$ higher than in the open limb. What is the vapour pressure of the liquid?
Ans
$732 \text{ mmHg}$
Review
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Identify the manometric relationship
When the closed limb mercury is higher, the vapour pressure is less than atmospheric pressure. The difference equals the mercury height difference.
$$P_{\text{vapour}} = P_{\text{atm}} - \Delta h$$
Substitute and calculate
$P_{\text{vapour}} = 760 - 28 = 732 \text{ mmHg}$.
$$P_{\text{vapour}} = 732 \text{ mmHg}$$
9
Which of the following substances has the highest boiling point?
Ans
$\text{H}_2\text{O}$
Review
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Compare boiling points down Group 16
$\text{H}_2\text{O}$: $100\degree C$, $\text{H}_2\text{S}$: $-60\degree C$, $\text{H}_2\text{Se}$: $-41\degree C$, $\text{H}_2\text{Te}$: $-2\degree C$.
Water stands out with a dramatically higher BP
Identify the cause of the anomaly
Water forms strong hydrogen bonds (H bonded to O), while the heavier hydrides rely on weaker dipole-dipole and London forces.
Hydrogen bonding overrides the expected group trend
10
For a given substance, how do the enthalpy changes of sublimation, vapourisation, and fusion compare?
Ans
$\Delta H_s > \Delta H_v > \Delta H_f$
Review
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Sublimation ($\Delta H_s$)
Sublimation = solid → gas. All intermolecular bonds in the solid must be completely broken.
Largest enthalpy change
Vapourisation ($\Delta H_v$) and Fusion ($\Delta H_f$)
Vapourisation = liquid → gas (breaks remaining intermolecular forces). Fusion = solid → liquid (partially disrupts ordering but molecules stay close). Also, $\Delta H_s = \Delta H_f + \Delta H_v$ by Hess's Law.
$$\Delta H_s > \Delta H_v > \Delta H_f$$
11
Which of the following correctly distinguishes a hydrogen bond from a covalent bond?
Ans
A hydrogen bond is an intermolecular attractive force, whereas a covalent bond is an intramolecular sharing of electrons between atoms
Review
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Compare the nature of each bond
Covalent bond: sharing of electron pairs within a molecule (intramolecular). Hydrogen bond: electrostatic attraction between molecules (intermolecular).
Intramolecular vs intermolecular
Compare bond strengths
Covalent bonds: 150–900 kJ/mol. Hydrogen bonds: 10–40 kJ/mol. Hydrogen bonds are much weaker.
Covalent bonds are orders of magnitude stronger
12
Polarizability of an electron cloud is a key factor in determining the strength of London dispersion forces. Which statement correctly describes this relationship?
Ans
Larger electron clouds with more diffuse electrons are more polarizable and experience stronger London forces
Review
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Define polarizability
Polarizability is the ease with which an electron cloud can be distorted. It increases with atomic/molecular size because outer electrons are further from the nucleus and less tightly held.
Larger electron cloud → more polarizable
Link to London force strength
A more polarizable electron cloud can create larger instantaneous dipoles, which induce stronger dipoles in neighbouring molecules. This results in stronger London dispersion forces.
Higher polarizability → stronger London forces
13
A liquid is in dynamic equilibrium with its vapour in a sealed container at constant temperature. If the volume of the container is suddenly doubled while the temperature remains constant, what happens to the vapour pressure?
Ans
The vapour pressure temporarily decreases, then returns to its original value as more molecules evaporate
Review
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Immediate effect of volume increase
Doubling the volume halves the concentration of vapour molecules. Fewer molecules striking the walls means the measured vapour pressure drops.
Vapour pressure temporarily decreases
Restoration of equilibrium
With fewer vapour molecules, the condensation rate drops below the evaporation rate. More molecules evaporate from the liquid until the vapour pressure returns to its original equilibrium value (determined by temperature alone).
Equilibrium vapour pressure is restored
14
At an altitude where the atmospheric pressure is $530 \text{ mmHg}$, water boils at approximately $90\degree C$. If a climber ascends to a higher altitude where the atmospheric pressure drops to $470 \text{ mmHg}$, what happens to the boiling point of water?
Ans
The boiling point decreases because the external pressure is lower, so vapour pressure equals external pressure at a lower temperature
Review
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Apply the boiling point condition
Boiling requires $P_{\text{vapour}} = P_{\text{external}}$. Since $P_{\text{vapour}}$ increases with temperature, lower $P_{\text{external}}$ means the equality is met at a lower temperature.
Lower external pressure → lower boiling point
Evaluate the change
At $530 \text{ mmHg}$: water boils at $90\degree C$. At $470 \text{ mmHg}$: a lower temperature is sufficient for $P_{\text{vapour}}$ to equal the external pressure.
Boiling point decreases below $90\degree C$
15
The molar heat of vapourisation of water is $40.6 \text{ kJ mol}^{-1}$ while that of carbon tetrachloride ($\text{CCl}_4$) is $30.0 \text{ kJ mol}^{-1}$. Which statement best explains this difference?
Ans
Water molecules are held together by hydrogen bonds, which are significantly stronger than the London dispersion forces between carbon tetrachloride molecules
Review
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Identify the IM forces in each substance
Water: hydrogen bonding (H bonded to O) + dipole-dipole + London. $\text{CCl}_4$: London dispersion forces only (nonpolar molecule).
Water has stronger overall IM forces
Relate to heat of vapourisation
The molar heat of vapourisation measures the energy needed to overcome all intermolecular forces per mole. Water's $40.6 \text{ kJ mol}^{-1}$ > $\text{CCl}_4$'s $30.0 \text{ kJ mol}^{-1}$ confirms hydrogen bonding is stronger.
$$\Delta H_v(\text{H}_2\text{O}) = 40.6 \text{ kJ mol}^{-1}$$
Question 1 of 15
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1
According to the kinetic molecular theory, which statement best explains why a liquid has a definite volume but takes the shape of its container?