Ch. 11: Liquids and Intermolecular Forces

Transcription

1 Ch. 11: Liquids and Intermolecular Forces Gases Learning goals and key skills: Identify the intermolecular attractive interactions (dispersion, dipole-dipole, hydrogen bonding, ion-dipole) that exist between molecules or ions based on their composition and molecular structure and be able to compare the relative strengths of these intermolecular forces Explain the concept of polarizability and how it relates to dispersion forces Explain the concepts of viscosity and surface tension in liquids Know the names of various changes of state for a pure substance Interpret heating curves and be able to calculate quantities related to temperature and enthalpies of phase changes Define critical pressure, critical temperature, vapor pressure, normal boiling point, normal melting point, critical point, and triple point Be able to interpret and sketch phase diagrams. Explain how water s phase diagram differs from most other substances, and why. Understand how the molecular arrangements characteristic or nematic, smectic, and choloesteric liquid crystals differ from ordinary liquids and from each other. Be able to recognize the features of molecules that favor formation of liquid crystalline phases. In gases the particles/molecules are far apart and the intermolecular forces between them are weak.... vs liquids and solids Strength of intermolecular attractions In liquids and solids this is not true. Instead we must consider the intermolecular forces, i.e., the forces that exist between atoms and molecules. States of Matter The fundamental difference between states of matter is the distance between particles. The States of Matter Because in the solid and liquid states particles are closer together, we refer to them as condensed phases. The state a substance is in at a particular temperature and pressure depends on: The kinetic energy of the particles The strength of the attractions between the particles 1

2 Intermolecular Forces Intermolecular Forces ion-dipole hydrogen bonding dipole-dipole van der Waals dispersion forces (London) forces The strength of these intermolecular forces is directly related to the melting/boiling points, enthalpy of fusion, enthalpy of vaporization, and solubility of the substances. Dispersion forces London dispersion forces Dispersion forces tend to increase with increasing molecular weight. These induced dipoles lead to intermolecular attractions. This is how nonpolar gas molecules (e.g. He, N 2, etc.) can liquefy. The ease with which the electron distribution in a molecule is distorted is called its polarizability. MW Boiling Pt. (g/mol) (K) F Cl Br I

3 Factors Affecting Dispersion Forces The shape of the molecule affects the dispersion forces: long, thin molecules (like n-pentane tend to have stronger dispersion forces than short, round ones (like neopentane). This is due to the increased surface area in n-pentane. Polar Covalent Bonds and Electronegativity Electronegativity is the ability of atoms in a molecule to attract electrons to itself. Although atoms often form compounds by sharing electrons, the electrons are not always shared equally. Dipole-Dipole Forces Dipole-Dipole Forces Influence of dipole-dipole forces is seen in the boiling points of simple molecules. MW (g/mol) Boiling Pt (K) N CO Br ICl Dipole-Dipole Forces Dipole-dipole forces increase with increasing polarity (dipole moments). MW Dipole Boiling (g/mol) Moment, µ (D) Pt (K) CH 3 CH 2 CH CH 3 OCH CH 3 CHO CH 3 CN Hydrogen Bonding A special form of dipole-dipole attraction, which enhances dipole-dipole attractions. H-bonding is strongest when X and Y are N, O, or F 3

4 Boiling Points of Simple Hydrogen-Containing Compounds Base-Pairing through H-Bonds Double helix of DNA Portion of a DNA chain Ion-dipole forces water -δ O dipole H H +δ 4

5 Intermolecular Forces Summary Increasing Strength Ionic bonding Ion-dipole Hydrogen bonding Dipole-dipole Dispersion forces (Induced dipoleinduced dipole) only Ions Ion charge, dipole moment Very polar H-X bond and lone pair of e - (X = N, O, F) Dipole moment polarizability Like dissolves like Polar molecules are more likely to dissolve in a polar solvent, and nonpolar molecules are more likely to dissolve in a nonpolar solvent. Liquids Viscosity The resistance of liquids to flow Increases with stronger intermolecular forces and decreases with increasing temperature. molecules close together appreciable intermolecular forces liquids are almost incompressible molecules are in constant motion finite volume, but indefinite shape 5

6 Surface Tension Surface tension results from the net inward force experienced by the molecules on the surface of a liquid. Cohesive and adhesive forces Cohesive forces - between the liquid molecules Adhesive forces - between the liquid molecules and another substance Properties of liquids Viscosity the resistance of liquids to flow; the thickness Surface tension energy required to break through the surface or to disrupt a liquid drop and make the drop spread out like a film. Cohesive forces forces between liquid molecules Adhesive forces forces between liquid molecules and another substance Capillary action the rise of liquids up narrow tubes and other surfaces Phase changes Changes of state involve energy (at constant T) Ice + (heat of fusion) water Water + (heat of vaporization) steam Phase changes Energy changes associated with changes of state For water H fus = 6.01 kj/mol or 334 J/g H vap = kj/mol or 2258 J/g 6

7 Heating/Cooling Curve for Water Heating/Cooling Curve for Water The total heat is the sum of each step q total = q i The ice is heated The ice melts to water The water is heated The water evaporates to steam The steam is heated q ice = s ice m T q fus = H fus m q water = s water m T q vap = H vap m q steam = s steam m T Example: Heating curve problem Determine the amount of heat (in kj) required to heat 500. g of ice from C to steam at 200. C. s ice = 2.09 J/gK H fus = 334 J/g s water = J/gK H vap = 2258 J/g s steam =1.84 J/gK q ice, to 0 C = J q fus, ice to water = J q water, 0 to 100 C = J q vap, water to steam = J q steam, 100 to 200. C = J q total = J = 1650 kj Example: Heating curve problem Determine the amount of heat (in kj) required to convert 42.0 g of ethanol from -155 C to 78 C. ethanol (C 2 H 5 OH) melts at -114 C and boils at 78 C. The enthalpy of fusion of ethanol is 5.02 kj/mol and the enthalpy of vaporization is kj/mol. The specific heats of solid and liquid ethanol are 0.97 J/gK and 2.3 J/gK, respectively. Note that most of the heat is used to convert water into steam. Critical Temperature: The highest temperature at which a distinct liquid phase can form Critical Pressure: Pressure required to bring about liquefaction at this critical temperature. Vapor Pressure Supercritical Fluid: Above this critical point, we have a supercritical fluid where the density is similar to a liquid and the viscosity is similar to a gas. 7

8 Temperature effects on the distribution of energy in a liquid Vapor pressure curves E needed to evaporate liquid Liquids that evaporate rapidly are volatile Equilibrium Vapor Pressure & the Clausius-Clapeyron Equation Clausius-Clapeyron equation used to find heat of vaporization, H vap. The logarithm of the vapor pressure P is proportional to H vap and to 1/T. ln P = ( H vap /RT) + C P H ln = P1 R 1 T2 T1 2 vap 1 Clausius-Clapeyron Equation Liquid bromine has a vapor pressure of 400. torr at 41.0 C and a normal boiling point of 58.2 C. Calculate the heat of vaporization of bromine (in kj/mol). P H ln = P1 R 1 T2 T1 2 vap 1 Phase Diagrams Phase diagrams display the state of a substance at various pressures and temperatures and the places where equilibria exist between phases. Normal Phase Diagram: CO 2 8

9 Phase Diagram of Water Note the high critical temperature and critical pressure. These are due to the strong van der Waals forces between water molecules. Info from P-T phase diagrams From a (pressure-temperature) phase diagram we can find: the normal melting point the normal boiling point the triple point whether the liquid is more or less dense than the solid the vapor pressure curve the sublimation pressure curve the critical temperature (T c ) and critical pressure (P c ) Liquid Crystals Liquid Crystals Rod-shaped and contain double or triple bond near the middle Polar groups create a dipole moment and promote alignment 9

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