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This lecture covers the different states of matter, intermolecular forces, and their effects on physical properties such as boiling and melting points. It also explains dispersion forces, dipole-dipole interactions, hydrogen bonding, and their relation to molecular size and shape.
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Lecture Presentation Chapter 11Liquids and Intermolecular Forces Dr. Subhash Goel South GA College Douglas, GA 31533
States of Matter The fundamental difference between states of matter is the distance between particles.
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 two factors: • The kinetic energy of the particles. • The strength of the attractions between the particles.
Intermolecular Forces The attractions between molecules are not nearly as strong as the intramolecular attractions that hold compounds together.
Intermolecular Forces These intermolecular attractions are, however, strong enough to control physical properties, such as boiling and melting points, vapor pressures, and viscosities.
Intermolecular Forces These intermolecular forces as a group are referred to as van der Waals forces.
Kinds of Attractive Forces • Temporary polarity in the molecules due to unequal electron distribution leads to attractions called dispersion forces. • Permanent polarity in the molecules due to their structure leads to attractive forces called dipole–dipole attractions. • An especially strong dipole–dipole attraction results when H is attached to an extremely electronegative atom. These are called hydrogen bonds.
Dispersion Forces • Fluctuations in the electron distribution in atoms and molecules result in a temporary dipole. • A region with excess electron density has partial (–) charge. • A region with depleted electron density has partial (+) charge. • The attractive forces caused by these temporary dipoles are called dispersion forces. • aka London forces • All molecules and atoms will have them. • As a temporary dipole is established in one molecule, it induces a dipole in all the surrounding molecules.
+ + + + + + + + + + + + + + + + - - - + + + + - - + + + + + + - - - - − − − − - − − − − − − − - - − − Size of the Induced Dipole • The magnitude of the induced dipole depends on several factors. • polarizability of the electrons • volume of the electron cloud • larger molar mass = more electrons = larger electron cloud = increased polarizability = stronger attractions • shape of the molecule • more surface-to-surface contact = larger induced dipole = stronger attraction Molecules that are flat have more surface interaction than spherical ones. Larger molecules have more electrons, leading to increased polarizability.
Effect of Molecular Sizeon Size of Dispersion Force As the molar mass increases, the number of electrons increases. Therefore, the strength of the dispersion forces increases. The noble gases are all nonpolar atomic elements. The stronger the attractive forces between the molecules, the higher the boiling point will be.
Dipole–Dipole Interactions • Molecules that have permanent dipoles are attracted to each other. • The positive end of one is attracted to the negative end of the other, and vice versa. • These forces are only important when the molecules are close to each other.
Dipole–Dipole Interactions The more polar the molecule, the higher its boiling point.
Which Have a Greater Effect?Dipole–Dipole Interactions or Dispersion Forces • If two molecules are of comparable size and shape, dipole–dipole interactions will likely be the dominating force. • If one molecule is much larger than another, dispersion forces will likely determine its physical properties.
Hydrogen Bonding • Hydrogen bonding arises in part from the high electronegativity of nitrogen, oxygen, and fluorine.
Hydrogen Bonding • The dipole–dipole interactions experienced when H is bonded to N, O, or F are unusually strong. • We call these interactions hydrogen bonds. • The energies of H-bond varies from 5kJ/mol to 25kJ/mol • H-bond weaker then covalent bond (150-1100kJ/mol)
Due to H-bonding ice has lower density (0.917g/mL) then liquid water. In ice, the water molecules assume the ordered, open arrangement as shown in figure
Sample Exercise 11.1Identifying Substances That Can Form Hydrogen Bonds In which of these substances is hydrogen bonding likely to play an important role in determining physical properties: methane (CH4), hydrazine (H2NNH2), methyl fluoride (CH3F), hydrogen sulfide (H2S)? The foregoing criteria eliminate CH4 and H2S, which do not contain H bonded to N, O, or F. They also eliminate CH3F, whose Lewis structure shows a central C atom surrounded by three H atoms and an F atom. (Carbon always forms four bonds, whereas hydrogen and fluorine form one each.) Because the molecule contains a C F bond and not a H F bond, it does not form hydrogen bonds. In H2NNH2, however, we find N H bonds, and the Lewis structure shows a nonbonding pair of electrons on each N atom, telling us hydrogen bonds can exist between the molecules:
Ion–Dipole Interactions • Ion–dipole interactions (a fourth type of force) are important in solutions of ions. • The strength of these forces is what makes it possible for ionic substances to dissolve in polar solvents.
Summary • Dispersion forces are the weakest of the intermolecular attractions. • Dispersion forces are present in all molecules and atoms. • The magnitude of the dispersion forces increases with molar mass. • Polar molecules also have dipole–dipole attractive forces.
Summary (cont’d) • Hydrogen bonds are the strongest of the intermolecular attractive forces. • a pure substance can have • Hydrogen bonds will be present when a molecule has H directly bonded to either O , N, or F atoms. • only example of H bonded to F is HF • Ion–dipole attractions are present in mixtures of ionic compounds with polar molecules. • Ion–dipole attractions are the strongest intermolecular attraction. • Ion–dipole attractions are especially important in aqueous solutions of ionic compounds.
Intermolecular Forces Affect Many Physical Properties The strength of the attractions between particles can greatly affect the properties of a substance or solution.
Viscosity • Resistance of a liquid to flow is called viscosity. • It is related to the ease with which molecules can move past each other. • Viscosity increases with stronger intermolecular forces and decreases with higher temperature.
Surface Tension • The layer of molecules on the surface behaves differently than the interior. • because the cohesive forces on the surface molecules have a net pull into the liquid interior • The surface layer acts like an elastic skin. • allowing you to “float” a fishing lure even though steel is denser than water
Factors Affecting Surface Tension • The stronger the intermolecular attractive forces, the higher the surface tension will be. • Raising the temperature of a liquid reduces its surface tension. • Raising the temperature of the liquid increases the average kinetic energy of the molecules. • The increased molecular motion makes it easier to stretch the surface.
Capillary Action • Capillary action is the ability of a liquid to flow up a thin tube against the influence of gravity. • The narrower the tube, the higher the liquid rises. • Capillary action is the result of the two forces working in conjunction—the cohesive and adhesive forces. • Cohesive forces hold the liquid molecules together. • Adhesive forces attract the outer liquid molecules to the tube’s surface.
Capillary Action • The adhesive forces pull the surface liquid up the side of the tube, while the cohesive forces pull the interior liquid with it. • The liquid rises up the tube until the force of gravity counteracts the capillary action forces. • The narrower the tube diameter, the higher the liquid will rise up the tube.
Meniscus • The curving of the liquid surface in a thin tube is due to the competition between adhesive and cohesive forces. • The meniscus of water is concave in a glass tube because its adhesion to the glass is stronger than its cohesion for itself. • The meniscus of mercury is convex in a glass tube because its cohesion for itself is stronger than its adhesion for the glass. • Metallic bonds are stronger than intermolecular attractions.
Energy Changes Associated with Changes of State The heat of fusion is the energy required to change a solid at its melting point to a liquid.
Energy Changes Associated with Changes of State The heat of vaporization is defined as the energy required to change a liquid at its boiling point to a gas.
Energy Changes Associated with Changes of State The heat of sublimation is defined as the energy required to change a solid directly to a gas.
Energy Changes Associated with Changes of State • The heat added to the system at the melting and boiling points goes into pulling the molecules farther apart from each other. • The temperature of the substance does not rise during a phase change.
Sample Exercise 11.3Calculating H for Temperature and Phase Changes Calculate the enthalpy change upon converting 1.00 mol of ice at 25 C to steam at 125 C under a constant pressure of 1 atm. The specific heats of ice, liquid water, and steam are 2.03 J/g-K, 4.18 J/g-K, and 1.84 J/g-K, respectively. For H2O, Hfus = 6.01 kJ/mol and Hvap = 40.67 kJ/mol. Solve For segment AB in Figure 11.22, we are adding enough heat to ice to increase its temperature by 25 C. A temperature change of 25 C is the same as a temperature change of 25 K, so we can use the specific heat of ice to calculate the enthalpy change during this process: AB: H = (1.00 mol)(18.0 g/mol)(2.03 J/g-K)(25 K) = 914 J = 0.91 kJ
Sample Exercise 11.3Calculating H for Temperature and Phase Changes Continued For segment BC in Figure 11.22, in which we convert ice to water at 0 C, we can use the molar enthalpy of fusion directly: The enthalpy changes for segments CD, DE, and EF can be calculated in similar fashion: The total enthalpy change is the sum of the changes of the individual steps: BC: H = (1.00 mol)(6.01 kJ/mol) = 6.01 kJ CD: H = (1.00 mol)(18.0 g/mol)(4.18 J/g-K)(100 K) = 7520 J = 7.52 kJ DE: H = (1.00 mol)(40.67 kJ/mol) = 40.7 kJ EF: H = (1.00 mol)(18.0 g/mol)(1.84 J/g-K)(25 K) = 830 J = 0.83 kJ H = 0.91 kJ + 6.01 kJ + 7.52 kJ + 40.7 kJ + 0.83 kJ = 56.0 kJ
Vapor Pressure • At any temperature some molecules in a liquid have enough energy to break free. • As the temperature rises, the fraction of molecules that have enough energy to break free increases.
Vapor Pressure As more molecules escape the liquid, the pressure they exert increases.
Vapor Pressure The liquid and vapor reach a state of dynamic equilibrium: liquid molecules evaporate and vapor molecules condense at the same rate.
Vapor Pressure • The boiling point of a liquid is the temperature at which its vapor pressure equals atmospheric pressure. • The normal boiling point is the temperature at which its vapor pressure is 760 torr.
Phase Diagrams Phase diagrams display the state of a substance at various pressures and temperatures, and the places where equilibria exist between phases.
Phase Diagrams The liquid–vapor interface starts at the triple point (T), at which all three states are in equilibrium, and ends at the critical point (C), above which the liquid and vapor are indistinguishable from each other.
Phase Diagrams Each point along this line is the boiling point of the substance at that pressure.
Phase Diagrams The interface between liquid and solid marks the melting point of a substance at each pressure.
Phase Diagrams • Below the triple point the substance cannot exist in the liquid state. • Along the solid–gas line those two phases are in equilibrium; the sublimation point at each pressure is along this line.
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.
Phase Diagram of Water • The slope of the solid– liquid line is negative. • This means that as the pressure is increased at a temperature just below the melting point, water goes from a solid to a liquid.
