Difficulty: Intermediate | Prerequisites: Lewis structures, molecular geometry, polarity, moles and stoichiometry.
Intermolecular forces (IMFs) are the attractions between molecules (not within them). They determine physical properties like boiling point, melting point, and solubility. This section brings together everything from molecular geometry and polarity to explain why some liquids boil at room temperature and others need hundreds of degrees. The ideal gas law and the Clausius-Clapeyron equation then let you calculate practical quantities like moles of vapour and enthalpy of vaporisation from lab measurements. If you understand Lewis structures and molecular shape, you are ready for this material.
Stronger intermolecular forces mean higher boiling points. The three main types, from weakest to strongest, are London dispersion forces, dipole-dipole interactions, and hydrogen bonding. The ideal gas law (PV = nRT) converts between volume, pressure, temperature, and moles of gas. The Clausius-Clapeyron equation relates vapour pressure to temperature and enthalpy of vaporisation.
Intermolecular forces (IMFs)
Attractive forces between molecules (not the covalent bonds within a molecule). These are much weaker than covalent or ionic bonds but determine bulk physical properties.
London dispersion forces (LDF)
Weak, temporary attractions caused by momentary uneven distributions of electrons (temporary dipoles). Present in all molecules. Strength increases with molecular size (more electrons = more surface area = stronger LDF). Also called van der Waals forces or induced dipole forces.
Dipole-dipole interactions
Attractions between the positive end of one polar molecule and the negative end of another. Only occur in polar molecules (those with a net dipole moment). Stronger than London forces for molecules of similar size.
Hydrogen bonding
An especially strong type of dipole-dipole interaction that occurs when H is bonded to N, O, or F and attracted to a lone pair on N, O, or F of a neighbouring molecule. Think of it as the VIP tier of dipole-dipole forces.
Boiling point
The temperature at which a substance's vapour pressure equals the external atmospheric pressure (typically 1 atm for "normal" boiling point). Higher IMFs mean a higher boiling point.
Vapour pressure
The pressure exerted by a vapour in equilibrium with its liquid at a given temperature. Substances with weak IMFs have high vapour pressure (they evaporate easily).
Enthalpy of vaporisation (ΔHvap)
The energy required to convert one mole of a substance from liquid to gas at its boiling point, in kJ/mol.
Ideal gas law
PV = nRT, where P is pressure (atm), V is volume (L), n is moles, R is the gas constant (0.08206 L·atm/mol·K), and T is temperature in Kelvin.
Clausius-Clapeyron equation
Relates the vapour pressure of a substance to temperature and its enthalpy of vaporisation. The two-point form compares vapour pressures at two different temperatures to find ΔHvap (or vice versa).
From weakest to strongest:
London dispersion forces (LDF): present in all molecules. Dominant in nonpolar molecules. Increase with molecular weight and surface area. Branching reduces surface area and weakens LDF.
Dipole-dipole: present in polar molecules (those with a net dipole). Stronger than LDF for similar-size molecules.
Hydrogen bonding: requires H bonded to N, O, or F. The strongest common IMF. Responsible for water's unusually high boiling point.
The exam gives five hydrocarbons/alcohols to rank. The approach:
Identify the IMFs each molecule can form.
Among molecules with the same type of IMF, larger molecules (more electrons, more surface area) have higher boiling points.
Branching lowers the boiling point compared to a straight-chain molecule of the same formula, because it reduces surface area and therefore weakens London forces.
The five substances from Question 6:
A: pentane (C5H12, straight chain). Nonpolar, LDF only.
B: 1-butanol (C4H9OH, straight chain with OH). Capable of hydrogen bonding.
C: neopentane (C5H12, highly branched). Nonpolar, LDF only, but much less surface area than A.
D: 1-pentanol (C5H11OH, straight chain with OH). Hydrogen bonding, and larger than B.
E: 2-methylbutane / isopentane (C5H12, slightly branched). Nonpolar, LDF only, slightly less surface area than A.
Ranking (lowest to highest boiling point): C < E < A < B < D.
Reasoning: C (neopentane) is the most branched, so weakest LDF. E is slightly branched, A is straight-chain (all three are C5H12 isomers, so the difference is entirely about shape/surface area). B is smaller than D but both have hydrogen bonding, which pushes them well above the hydrocarbons. D, being a larger alcohol, has the highest boiling point.
The problem: 63.5 kJ of heat vaporises enough aniline (C6H5NH2) at its normal boiling point of 184°C to produce 45.0 L of vapour at 1 atm. Find ΔHvap in kJ/mol.
Step 1: Convert temperature to Kelvin. T = 184 + 273.15 = 457.15 K.
Step 2: Use PV = nRT to find moles of aniline vapour.
n = PV / RT = (1 atm)(45.0 L) / (0.08206 L·atm/mol·K)(457.15 K) = 45.0 / 37.51 = 1.20 mol.
Step 3: Calculate ΔHvap.
ΔHvap = q / n = 63.5 kJ / 1.20 mol = 52.9 kJ/mol.
The two-point form:
ln(P2/P1) = (ΔHvap / R) × (1/T1 - 1/T2)
where R = 8.314 J/(mol·K) and temperatures are in Kelvin.
The problem: vapour pressure increases by a factor of 2.7 from 65°C to 96°C (the normal boiling point). At the boiling point, vapour pressure equals atmospheric pressure, so you can set P2/P1 = 2.7 (you do not need the actual pressures).
Step 1: Convert temperatures. T1 = 65 + 273.15 = 338.15 K. T2 = 96 + 273.15 = 369.15 K.
Step 2: Plug in.
ln(2.7) = (ΔHvap / 8.314) × (1/338.15 - 1/369.15)
0.9933 = (ΔHvap / 8.314) × (0.002957 - 0.002709)
0.9933 = (ΔHvap / 8.314) × (0.000248)
Step 3: Solve.
ΔHvap = (0.9933 × 8.314) / 0.000248 = 8.258 / 0.000248 = 33,300 J/mol = 33.3 kJ/mol.
PV = nRTP in atm, V in litres, n in moles, R = 0.08206 L·atm/(mol·K), T in Kelvin. Always convert °C to K by adding 273.15.
\ln\left(\frac{P_2}{P_1}\right) = \frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)R = 8.314 J/(mol·K) in this equation (not the gas constant in L·atm units). Temperatures in Kelvin. P2/P1 is the ratio of vapour pressures; you do not need absolute pressures if the ratio is given.
Intermolecular forces explain why water is a liquid at room temperature while methane is a gas, even though their molecular weights are not vastly different: water has strong hydrogen bonding, methane has only weak London forces. The Clausius-Clapeyron equation is used in meteorology to predict when water vapour will condense (cloud formation) and in engineering to design distillation columns for separating liquid mixtures.
Students often confuse intermolecular forces with intramolecular bonds (covalent bonds). Boiling does not break covalent bonds; it overcomes intermolecular attractions. The O-H covalent bond in water is not what breaks when water boils.
Students often think all hydrogen bonds are the same strength. The strength depends on which atoms are involved and the geometry. H bonded to F, O, or N with a lone pair on F, O, or N nearby is the requirement.
Students often forget to convert Celsius to Kelvin before using PV = nRT or the Clausius-Clapeyron equation. This is the single most common calculation error on this material.
Students often use the wrong R value. For PV = nRT, use R = 0.08206 L·atm/(mol·K). For Clausius-Clapeyron, use R = 8.314 J/(mol·K). Mixing them up gives answers off by orders of magnitude.
⚠️ The boiling point ranking question (Q6) requires you to identify the IMF type for each molecule and then rank by strength. Branching within the same molecular formula is the tiebreaker for London forces.
⚠️ Question 7 combines the ideal gas law with enthalpy of vaporisation. The trick is recognising that you need PV = nRT to find the moles of vapour before you can calculate ΔHvap.
⚠️ For the Clausius-Clapeyron problem (Q8), you are given a vapour pressure ratio, not absolute pressures. The equation works with the ratio directly (P2/P1). Do not panic about not having actual pressure values.
⚠️ Show every step of your calculation. Partial credit is available, and a correct setup with an arithmetic slip still earns most of the marks.
True or False: London dispersion forces are only present in nonpolar molecules.
Answer: False. LDF are present in all molecules, but they are the only IMF in nonpolar molecules.
Fill in the blank: Hydrogen bonding requires H bonded to ____, ____, or ____.
Answer: N, O, or F.
True or False: At a substance's normal boiling point, its vapour pressure equals 1 atm.
Answer: True.
Fill in the blank: In the ideal gas law, temperature must be in ____.
Answer: Kelvin.
True or False: Branching increases the boiling point of a hydrocarbon.
Answer: False. Branching decreases surface area, weakens London forces, and lowers the boiling point.
Q: Rank pentane, neopentane, and isopentane by boiling point from lowest to highest. All three have the formula C5H12.
A: Neopentane < isopentane < pentane. All are nonpolar with LDF only. Neopentane is the most compact (most branched, least surface area), so it has the weakest London forces and the lowest boiling point. Pentane is the straight chain with the most surface area.
Q: Why does 1-pentanol have a higher boiling point than pentane, even though they have similar molecular weights?
A: 1-pentanol has an -OH group, enabling hydrogen bonding between molecules. Pentane is nonpolar and can only form London dispersion forces. Hydrogen bonding is much stronger than LDF, so the alcohol has a significantly higher boiling point.
Q: 63.5 kJ vaporises enough aniline at 184°C and 1 atm to produce 45.0 L of vapour. Calculate ΔHvap.
A: First find moles: n = PV/RT = (1)(45.0) / (0.08206)(457.15) = 1.20 mol. Then ΔHvap = 63.5 / 1.20 = 52.9 kJ/mol.
Q: A substance's vapour pressure increases by a factor of 2.7 as temperature rises from 65°C to its normal boiling point of 96°C. Find ΔHvap.
A: Use the two-point Clausius-Clapeyron equation. ln(2.7) = (ΔHvap / 8.314)(1/338.15 - 1/369.15). Solving gives ΔHvap = 33.3 kJ/mol.
IMFs depend on molecular polarity, which depends on molecular geometry, which depends on Lewis structures and VSEPR. This is the thread running through the entire second half of the course: structure determines shape, shape determines polarity, polarity determines physical properties.
The enthalpy of vaporisation connects back to bond energies and thermochemistry. Both are ways of quantifying energy changes, just at different scales: bond energies deal with breaking covalent bonds within molecules, while ΔHvap deals with overcoming attractions between molecules.
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