Intermolecular Forces- Notes

Intermolecular Forces and Physical Properties

• Physical state is a balance between two competing factors:
  • Thermal energy (favors disorder)
  • Interactions between molecules (favors order)
• Stronger attractive forces shift boiling/melting points higher because more energy is required to separate molecules

Physical Properties vs Chemical Properties

• Physical properties
  • Readily observed and do not involve transformations into different substances
  • Molecular-scale definition: properties that do not involve changing the molecules
  • Reversible changes
• Chemical properties
  • Can only be determined when a substance is changed into another substance
  • Molecular-scale definition: properties that involve making new molecules (new bonds between atoms)
  • Often involve irreversible changes

Intermolecular Forces (IM) vs Intramolecular Forces

• Intramolecular forces (within molecules):
  • Strong – hold atoms together inside a molecule
• Intermolecular forces (between molecules):
  • Weaker – determine physical properties like boiling/melting points, solubility, etc.
  • Interaction strength depends on the magnitude of electrostatic forces

Types of Intermolecular Forces

• Dispersion (London) forces
  • Present in all molecules
  • Magnitude: much less than 1 (overall small, but accumulates)
  • Depends on molecular size and surface contact
  • Origin: induced dipoles due to instantaneous fluctuations in electron distribution; polarizability plays a key role
• Dipole–Dipole interactions
  • Occur between polar molecules (electronegative differences create permanent dipoles)
  • Magnitude: less than 1; stronger than dispersion in many cases
  • Depend on the molecular dipole; the size of opposite charges matters
• Hydrogen bonds
  • Nearly as strong as covalent bonds in strength among IM forces
  • Occur when there are N–H or O–H bonds in a molecule
  • Involve a strong dipole–dipole interaction with a hydrogen atom that is covalently bonded to N or O
• Ionic bonds (between ions)
  • Magnitude ranges from 1 to 7 (very strong IM interaction in the context of condensed matter)
  • Depend on charge magnitude and ionic radii; smaller size and higher charge yield stronger interaction

Polarizability vs Polarity

• Dispersion forces arise from polarizability (temporary fluctuations in electron distribution), not from permanent polarity
  • Polarizability causes temporary charge separation; charges are very small and fleeting
• Polarity arises from permanent dipole moments due to differences in electronegativity between bonded atoms
• Polarity leads to dipole–dipole interactions; dispersion forces depend on size and shape

Dispersion Forces: Determinants

• Strength increases with molecular surface contact and surface area
• Longer and/or larger molecules have stronger dispersion forces due to greater contact and higher number of electrons contributing to polarizability

Strength of IM Forces: Comparisons

• Between two nonpolar molecules: the heavier/larger molecule generally has stronger dispersion forces
• Between a polar and a nonpolar molecule: the polar molecule usually has stronger IM forces, but dispersion forces can still be significant, especially if the nonpolar molecule is large
• In small molecules, dispersion forces are typically weak; in larger molecules they become substantial

Electronegativity and Polarity

• Electronegativity is the ability of an atom to attract electrons in a bond
• Greater electronegativity difference → more polar bond
• Bonds are polar if there is a difference in electronegativity between bonded atoms

Molecular Shape and Polarity

• A molecule is polar if it has polar bonds and the bond dipoles do not cancel due to asymmetric geometry
• If a molecule has polar bonds but a symmetric shape with canceling dipoles, it can be nonpolar

Guidelines for Polarity

• Nonpolar molecules: no polar bonds (e.g., diatomic homonuclear molecules, simple hydrocarbons)
• Polar molecules: have polar bonds and an asymmetric shape or lone pairs on the central atom (e.g., H2O, NH3)
• Nonpolar molecules with polar bonds: if all bonds are identical and there are no lone pairs on the central atom (e.g., CO2, CBr4), the molecule can still be nonpolar due to symmetry

Dipole–Dipole Interactions and Polarity

• Polar molecules have partial charges (dipoles) that attract each other via dipole–dipole interactions
• These interactions are typically stronger than dispersion forces in many polar molecules

Hydrogen Bonding: An Exceptionally Strong IM Force

• N and O are highly electronegative; hydrogen has a very small electron cloud around it when bonded to N or O
• H–bonding occurs when a hydrogen atom bonded to N or O interacts with a lone pair on N or O in another molecule
• Hydrogen bonds are strong dipole–dipole interactions and contribute significantly to physical properties like boiling points and solubility

Hydrogen Bonding in Practice

• Molecules with N–H or O–H bonds can participate in hydrogen bonding
• Example interactions: H–N–…:N or H–O–…:O (where “…” indicates lone pairs on N or O)

Identifying IM Forces in Molecules

• Start with the Lewis structure
  • Check for H-bonds (N–H or O–H bonds) that can engage with lone pairs on other molecules
  • If no H-bonds, assess polarity to determine dipole–dipole interactions
  • If molecule is nonpolar or large with many electrons, dispersion forces may dominate
• Ranking of IM forces (strongest to weakest):
  • Hydrogen bonding > Dipole–Dipole > Dispersion

Examples: Ethanol vs Dimethyl Ether

• Ethanol, CH3CH2OH
  • Contains O–H bond → Hydrogen bonding
  • Also has dipole–dipole and dispersion forces
• Dimethyl ether, CH3OCH3
  • Has polar C–O–C linkage → Dipole–dipole forces
  • No O–H or N–H bonds → No hydrogen bonding
  • Also has dispersion forces
• Ranking: Ethanol > Dimethyl ether (in terms of IM strength and associated properties like boiling point)

CH4: Strongest IM Force

• For CH4, the molecule is nonpolar and has no lone pairs or highly electronegative atoms capable of H-bonding
• The strongest IM force present is dispersion (London) forces
• Answer: Dispersion forces

Which Molecule Has Only Dispersion Forces?

• Among common options: CHCl3, H2O, CH3CN, NH2NH2, O2
• O2, being nonpolar and diatomic, has only dispersion forces
• Answer: O2

Boiling Point, IM Forces, and Phase Changes

• Stronger intermolecular forces lead to higher boiling and melting points
• More energy is required to break stronger IM forces
• The total strength of IM forces can be gauged by the energy needed to disrupt them

Surface Tension and the “Skin” of a Liquid

• Liquids minimize surface area; beading of drops into spheres and the ability to support objects denser than the liquid are due to IM forces at the surface
• Stronger IM forces yield greater surface tension
• Surface tension decreases with increasing temperature

Adhesive and Cohesive Forces; Beading and Wetting

• Beading: cohesive forces within the liquid cause surface tension at the interface
• Wetting: adhesive forces between liquid and surface
• Strong cohesive forces or weak adhesive forces lead to poor wetting (beading)
• Weak cohesive forces or strong adhesive forces lead to good wetting
• Meniscus formation:
  • Concave (adhesive > cohesive): liquid climbs up the surface (e.g., water on glass)
  • Convex (cohesive > adhesive): liquid beading on nonpolar surfaces (e.g., water on waxy surfaces)

Capillary Action

• The ability of a liquid to rise in a thin tube due to adhesive forces between the liquid and the tube walls and cohesive forces within the liquid
• Counteracted by gravity and cohesive forces within the liquid
• At a given diameter, the rise height is determined by a balance of surface tension, contact angle, density, gravity, and radius
• Height depends inversely on the tube radius and directly on surface tension
• Key relation (capillary rise):
  • $h = \frac{2 \gamma \cos\theta}{\rho g r}$
  • where $\gamma$ is surface tension, $\theta$ is the contact angle, $\rho$ is liquid density, $g$ is acceleration due to gravity, and $r$ is the tube radius

Factors Affecting Capillary Height (Quick Practice)

• Which change leads to smaller capillary rise? Increasing the diameter (radius) of the tube lowers the height
• Other options affecting height include contact angle and surface tension, not tilt alone or weaker gravity (in practical classroom contexts)

Water Transport in Plants (Capillary Action in Trees)

• Xylem vessels are highly polar and very narrow, aiding capillary rise of water from roots to leaves

Phase Changes and Energetics

• Phase changes involve absorption or release of energy and occur at constant temperature during the transition
• Endothermic phase changes require heat input
• Exothermic phase changes release heat
• Energy changes during phase transitions are governed by bond-making and bond-breaking:
  • Making bonds releases energy
  • Breaking bonds requires energy
• Within a phase, temperature changes with added heat according to:
  • $q = m \, C_s \, \Delta T$
  • where $m$ is mass, $C<em>s$ is specific heat, and $\Delta T = T</em>2 - T_1$
• Phase changes do not change the identity of the substance; only the state
• Enthalpy changes for common transitions:
  • Fusion/Melting: \Delta H_{fus} > 0
  • Freezing: $\Delta H<em>{freezing} = -\Delta H</em>{fus}$
  • Vaporization: \Delta H_{vap} > 0
  • Condensation: $\Delta H<em>{condensation} = -\Delta H</em>{vap}$
• Solid–Liquid (Melting/Fusion)
  • Temperature remains constant at the melting point during melting/freezing
  • The heat required to melt a substance: $q = n \, \Delta H_{fus}$
• Example: Ice at 0°C melting
  • Given: $\Delta H_{fus} = 6.02\ \text{kJ/mol}$ for H2O(s) at 0°C
  • Mass: 100 g, M(H2O) = 18.02 g/mol → $n = \frac{100}{18.02} = 5.55\ \text{mol}$
  • Heat to melt: $q = n \Delta H_{fus} = 5.55 \times 6.02\ \text{kJ} \approx 33.4\ \text{kJ}$
• Heating liquid water from 0°C to 25°C:
  • $q = m \; C<em>s \; \Delta T$ with $m = 100\ g,\; C</em>s = 4.18\ \text{J g}^{-1}\text{K}^{-1},\; \Delta T = 25\ \text{K}$
  • $q = 100 \times 4.18 \times 25 = 10450\ \text{J} = 10.5\ \text{kJ}$
• Combined, melting ice at 0°C and heating to 25°C liquid water: $q_{total} = 33.4\ \text{kJ} + 10.5\ \text{kJ} = 43.9\ \text{kJ}$

Phase Changes: Vaporization and Boiling

• Vaporization (evaporation vs boiling)
  • Evaporation can occur below the boiling point and occurs at a surface; no bubbles; rate increases with temperature and surface area; requires energy; happens at a fixed temperature during boiling
  • Boiling occurs at the boiling point when vapor pressure equals the external pressure; heat input continues during boiling
• Heat for vaporization: $q = n \Delta H_{vap}$
• Example: Melt 100 g ice at 0°C, heat to 100°C, and boil off all water
  • Melting: $q<em>{fus} = n \Delta H</em>{fus}$ with $n = 5.55\ \text{mol},\; \Delta H_{fus} = 6.02\ \text{kJ/mol}$ → 33.4 kJ
  • Heating liquid from 0°C to 100°C: $q = m C_s \Delta T = 100\ \text{g} \times 4.18\ \text{J g}^{-1}\text{K}^{-1} \times 100\ \text{K} = 41.8\ \text{kJ}$
  • Vaporization of the resulting water: $n = 5.55\ \text{mol},\; \Delta H_{vap} = 40.7\ \text{kJ/mol}$ → $q = 5.55 \times 40.7 \approx 226\ \text{kJ}$
  • Total: $q_{total} \approx 33.4 + 41.8 + 226 = 301.2\ \text{kJ}$

Vapor Pressure, Volatility, and Boiling Point

• Vapor pressure: pressure of evaporated gas molecules above a liquid
• Volatility: how easily a liquid evaporates
• A molecule with weaker IM forces tends to have:
  • Higher vapor pressure
  • Higher volatility
  • Lower boiling point
  • Smaller \Delta H_vap

Phase Changes: Evaporation vs Boiling and Vapor Pressure

• Evaporation occurs below the boiling point and is surface-based
• Boiling occurs at the boiling point when the vapor pressure equals the external pressure
• Vapor pressure increases with temperature and decreases with stronger IM forces

Clausius–Clapeyron Equation (Vapor Pressure vs Temperature)

• Linearized form for a liquid:
  • $\ln P<em>v = -\frac{\Delta H</em>{vap}}{R} \left( \frac{1}{T} \right) + C$
• Two-point form (given Pvap1 at T1, Pvap2 at T2):
  • $\ln\left( \frac{P<em>2}{P</em>1} \right) = -\frac{\Delta H<em>{vap}}{R} \left( \frac{1}{T</em>2} - \frac{1}{T_1} \right)$
• Using two known points allows solving for Pvap at a new temperature or for an unknown \Delta H_{vap}

Example: Vapor Pressure of Water at 25°C

• Given: normal boiling point T1 = 100°C = 373 K; Pvap1 = 1 atm; \Delta H_{vap} = 40.7 kJ/mol; R = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K)
• Set T2 = 25°C = 298 K
• Compute:
  • \ln\left( \frac{P_2}{1 \text{ atm}} \right) = -\frac{40.7}{0.008314} \left( \frac{1}{298} - \frac{1}{373} \right)
  • Numerically: \left( \frac{1}{298} - \frac{1}{373} \right) ≈ 0.003356 - 0.002681 = 0.000675\,\text{K}^{-1}
  • Factor: \frac{40.7}{0.008314} ≈ 4897.7
  • Thus: \ln P_2 ≈ -4897.7 × 0.000675 ≈ -3.305
  • P_2 ≈ e^{-3.305} ≈ 0.0366\ \text{atm}
• Result: P2 ≈ 0.0366–0.0369 atm at 25°C (approximately 0.037 atm)
• This demonstrates how vapor pressure rises with temperature and depends on the enthalpy of vaporization

Phase Diagrams: Dependence of Phase on Pressure and Temperature

• Phase of a substance is a function of pressure (P) and temperature (T)
• Boundaries on a P–T diagram:
  • Solid–liquid boundary (melting/freezing curve)
  • Liquid–gas boundary (vapor-pressure curve, i.e., boiling point line)
  • Solid–gas boundary (sublimation curve)
• Key features:
  • At the CO2 triple point: solid, liquid, and gas are in equilibrium
  • At the CO2 critical point: beyond this point, liquid and gas phases are indistinguishable (supercritical fluid)
• Important CO2 data (as example from the slides):
  • Triple point: approximately attained at approximately \(-56.4^{\circ}\text{C}, 5.11\ \text{atm})\
  • Critical point: approximately \(31.1^{\circ}\text{C}, 73.0\ \text{atm})\
• For water, melting point curve slopes downward on a pressure scale because ice is less dense than liquid water

Supercritical Fluids

• Definition: The region above the critical point (high P and high T) where distinct liquid and gas phases do not exist
• Properties:
  • Dense like a liquid
  • Fluid like a gas; can fill containers completely and dissolve substances
  • Higher density and higher pressure → more liquid-like behavior; higher temperature and lower pressure → more gas-like behavior

CO2 Phase-Diagram Practice Question (Summary)

• Determine the state of CO2 under these conditions using the CO2 phase diagram:
  1) 1.0 atm and 25°C → likely gas
  2) 25°C and 2500 psi (≈ 170 atm) → likely liquid
  3) -100°C and 1.0 atm → below triple point temperature, at low pressure → solid (dry ice)