Intermolecular Forces, Liquids and Solids Chap. 10

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1 Page III-10-1 / Chapter Ten Lecture Notes Intermolecular Forces, and Solids Chap. 10 States of Matter The fundamental difference between states of matter is the distance between particles. Chemistry 222 Professor Michael Russell Solids and liquids often referred to as condensed phases Intermolecular Forces The States of Matter The state a substance is in at a particular temperature and pressure depends on two antagonistic entities: the kinetic energy of the particles; the strength of the attractions between the particles. The attractions between molecules (intermolecular forces) are not nearly as strong as the intramolecular attractions that hold compounds together. Intramolecular forces: ionic, covalent, metallic Intermolecular forces are not chemical bonds! Intermolecular Forces Intermolecular Forces Kinetic Energy (Ek) vs. Attractive (IM) Force EK = lowest intermediate Intermolecular (IM) Attraction: highest intermediate highest lowest We will assume gases have no IM force in C 222 Intermolecular forces are strong enough to affect physical properties such as boiling and melting points, vapor pressures, and viscosities. See the IM Forces Guide Page III-10-1 / Chapter Ten Lecture Notes

2 -δ water dipole +δ Attraction Between Ions and Permanent Dipoles Page III-10-2 / Chapter Ten Lecture Notes water dipole +δ Water is highly polar (it Water is highly polar (it has a dipole) and can has a dipole) and can interact with positive interact with positive ions to give ions to give hydrated ions in hydrated ions in water. water. This is the Ion-Dipole IM force This is the Ion-Dipole IM force -δ Attraction Between Ions and Permanent Dipoles Attraction Between Ions and Permanent Dipoles Many metal ions are hydrated. This is the reason metal salts dissolve in water. Attraction Between Ions and Permanent Dipoles Attraction between ions and dipole depends on ion charge and ion-dipole distance. Measured by for M n > [M( 2 ) x ] n+ δ- δ+ Na + δ+ Mg 2+ Cs + δ- δ- δ kj/mol kj/mol -263 kj/mol Dipole-Dipole Forces Dipole-dipole forces bind molecules having permanent dipoles to one another. Dipole-Dipole Forces Influence of dipole-dipole forces is seen in the boiling points of simple molecules. Compd Mol. Wt. Boil Point N o C C o C Br o C ICl o C Page III-10-2 / Chapter Ten Lecture Notes

3 Page III-10-3 / Chapter Ten Lecture Notes ydrogen Bonding A special form of the dipole-dipole force which enhances dipole-dipole attractions. ydrogen Bonding in 2 Ice has open lattice-like structure. Ice density is < liquid and so solid floats on water. -bonding is strongest when X and Y are N,, or F ydrogen Bonding bonds ---> abnormally high boiling point of water. Boiling Points of Simple ydrocarbon Compounds ydrogen Bonding ydrogen bonding and base pairing in DNA. FRCES INVLVING INDUCED DIPLES ow can non-polar molecules such as 2 and I 2 dissolve in water? Page III-10-3 / Chapter Ten Lecture Notes The water dipole INDUCES a dipole in the 2 electric cloud. Dipole-induced dipole

4 Page III-10-4 / Chapter Ten Lecture Notes FRCES INVLVING INDUCED DIPLES Consider I 2 dissolving in alcohol, C 3 C 2. - δ FRCES INVLVING INDUCED DIPLES Formation of a dipole in two nonpolar I 2 molecules. Induced dipoleinduced dipole I-I - δ R + δ The alcohol temporarily creates or INDUCES a dipole in I 2. I-I + δ - δ R + δ "Induced Dipole-Induced Dipole" is also known as "London Dispersion", same thing FRCES INVLVING INDUCED DIPLES The induced forces between I 2 molecules are very weak, so solid I 2 sublimes (goes from a solid to gaseous molecules). FRCES INVLVING INDUCED DIPLES The magnitude of the induced dipole depends on the tendency to be distorted. igher molar mass ---> larger induced dipoles. Larger atoms have larger electron clouds which are easier to polarize Intermolecular Forces Strength Summary Ion-ion / metallic force strongest of all In a liquid molecules are in constant motion there are appreciable intermolecular forces molecules close together are almost incompressible do not fill the container aka ID-ID Page III-10-4 / Chapter Ten Lecture Notes

5 Page III-10-5 / Chapter Ten Lecture Notes The two key properties we need to describe are EVAPRATIN and its opposite- CNDENSATIN LIQUID evaporation---> Add energy VAPR break IM bonds make IM bonds Remove energy <---condensation To evaporate, molecules must have sufficient energy to break IM forces. Breaking IM forces requires energy. The process of evaporation is endothermic. Number of molecules 0 lower T higher T Molecular energy minimum energy needed to break IM forces and evaporate Distribution of molecular energies in a liquid. Kinetic Energy proportional to Temperature At higher T a much larger number of molecules has high enough energy to break IM forces and move from liquid to vapor state. When molecules of liquid are in the vapor state, they exert a VAPR PRESSURE EQUILIBRIUM VAPR PRESSURE is the pressure exerted by a vapor over a liquid in a closed container when the rate of evaporation = the rate of condensation. Vapor Pressure Boiling Liquid boils when its vapor pressure equals atmospheric pressure. Page III-10-5 / Chapter Ten Lecture Notes

6 Page III-10-6 / Chapter Ten Lecture Notes Boiling Point at Lower Pressure Consequences of Vapor Pressure Changes When pressure is lowered, the vapor pressure can equal the external pressure at a lower temperature. When can cools, vp of water drops. Pressure in the can is less than that of atmosphere, so can is crushed. Consequences of Vapor Pressure Changes - Whoops! Equilibrium Vapor Pressure When car cools on hot day (i.e. cleaning with cold water), vp of fumes inside drops. Pressure in the car is less than that of atmosphere, so car is crushed! The curves show all conditions of P and T where LIQ and VAP are in EQUILIBRIUM. The VP rises with T. When VP = external P, the liquid boils. This means that BPs of liquids change with altitude. If external P = 760 mm g, T of boiling is the NRMAL BILING PINT VP of a given molecule at a given T depends on IM forces. ere the VPs are in the order: Page III-10-6 / Chapter Ten Lecture Notes ether 5 C 2 C 2 5 dipoledipole alcohol 5 C 2 -bonds increasing strength of IM interactions water extensive -bonds

7 Page III-10-7 / Chapter Ten Lecture Notes Viscosity Molecules at surface behave differently than those in the interior. VISCSITY is the tendency or resistance of liquids to flow. "flow" differently due to the strength of their intermolecular bonds Glycerol Ethanol Viscosity results from several factors, including IM interactions, molecular shape and size Molecules at surface experience net INWARD force of attraction. This leads to SURFACE TENSIN - the energy required to break the surface. Surface Tension IM forces also lead to CAPILLARY action Cohesive forces: interactions between like particles. SURFACE TENSIN also leads to spherical liquid droplets. Capillary Action Adhesive forces: interactions between unlike particles. Concave Meniscus: adhesive forces cohesive forces (2 on glass) Convex Meniscus: Cohesive forces > adhesive forces (g on glass). eat Transfer with Phase Change verall patterns: solid liquid gas = endothermic reaction gas liquid solid = exothermic reaction Movement of water up a piece of paper depends on -bonds between 2 and the groups of the cellulose in the paper. Page III-10-7 / Chapter Ten Lecture Notes Chapter 5: eat Transfer with no Phase Change (q = mc T)

8 eat Transfer with Phase Change EAT F VAPRIZATIN is the heat required (at constant P) to vaporize a liquid. LIQ + heat ---> VAP Compd. vap (kj/mol) IM Force (100 o C) -bonds S (-47 o C) dipole Xe 12.6 (-107 o C) induced dipole EAT F FUSIN is the heat required (at constant P) to melt a solid. SL + heat ---> LIQ Temperature constant during phase change Page III-10-8 / Chapter Ten Lecture Notes Clausius-Clapeyron The Clausius Clapeyron Equation provides a link between vapor pressure (P), temperature (in K), and molar heat of vaporization ( vap ): ln P = - Δ vap RT + C Perform a linear regression (ln P vs. 1/T(K)) to get best values, or, with only two temps: ln P 1 P 2 = Δ! vap R R = J mol -1 K -1! " 1-1 % $ ' # T 2 T 1 & eating/cooling Curve for Water eat & Changes of State Melt ice eat water q = mc T Evaporate water Note that T is constant as as ice liquid melts water evaporates What quantity of heat is required to melt 500. g of ice at 0.0 o C and heat the water to steam at 100. o C? eat of fusion of ice = 333 J/g Specific heat of water = J/g K eat of vaporization = 2260 J/g +333 J/g J/g solid (ice) liquid (0 100 C) gas (steam) eat & Changes of State Phase Diagrams What quantity of heat is required to melt 500. g of ice at 0.0 oc and heat the water to steam at 100. o C? 1. To melt ice q = (500. g)(333 J/g) = 1.67 x 10 5 J 2. To raise water from 0.0 o C to 100. o C q = (500. g)(4.184 J/g K)( )K = 2.09 x 10 5 J 3. To evaporate water at 100. o C q = (500. g)(2260 J/g) = 1.13 x 10 6 J 4. Total heat energy = 1.51 x 10 6 J = 1510 kj Page III-10-8 / Chapter Ten Lecture Notes

9 Page III-10-9 / Chapter Ten Lecture Notes TRANSITINS BETWEEN PASES See the phase diagram for water in text Lines connect all conditions of T and P where EQUILIBRIUM exists between the phases on either side of the line. At equilibrium particles move from liquid to gas as fast as they move from gas to liquid. solid phase. equilibrium between solid and liquid phases. liquid phase. equilibrium between liquid and gas phases. gas phase. Page III-10-9 / Chapter Ten Lecture Notes

10 Page III / Chapter Ten Lecture Notes triple point. equilibrium between solid and gas phases. At the TRIPLE PINT all three phases are in equilibrium. Phases Diagrams- Important Points for Water T( C) P(mm g) Normal boil point Normal freeze point Triple point We can think of solids as falling into two groups: crystalline: particles in highly ordered arrangements amorphous: no particular order in arrangement of particles. We will focus on crystalline solids Solids Water at the Triple Point Properties of Crystalline Solids 1. Molecules, atoms or ions locked into a CRYSTAL LATTICE 2. Particles are CLSE together 3. STRNG IM forces 4. ighly ordered, rigid, incompressible Crystal Lattices Regular 3-D arrangements of equivalent LATTICE PINTS in space. Lattice points define UNIT CELLS smallest repeating internal unit that has the symmetry characteristic of the solid. ZnS, zinc(ii) sulfide Page III / Chapter Ten Lecture Notes

11 Page III / Chapter Ten Lecture Notes The Brevais Lattices 7 main types: Triclinic Monoclinic rthorhombic Tetragonal exagonal Rhombohedral Cubic (Isometric) We will use just the cubic system in C 222 Cubic Unit Cells There are 7 basic crystal systems, but we are only concerned with CUBIC (isometric). All sides equal length All angles are 90 degrees Simple cubic (SC) Cubic Unit Cells Body-centered cubic (BCC) Three types of Cubic Unit Cells: Face-centered cubic (FCC) Simple Cubic Unit Cell Simple cubic unit cell. Note that each atom is at a corner of a unit cell and is shared among 8 unit cells. See Cubic Unit Cells Guide The Simple Cubic Unit Cell Body-Centered Cubic Unit Cell Atom at each corner, nly 1 net atom per simple cubic cell Page III / Chapter Ten Lecture Notes Atom at each cube corner plus one in center Two net atoms per bcc unit cell

12 Face-Centered Cubic Unit Cell Page III / Chapter Ten Lecture Notes Unit Cells for Metals also known as cubic close packing Atom in each cube corner plus atom in each cube face, four net atoms per fcc unit cell Finding the Lattice Type To find out if a metal is SC, BCC or FCC, use the known radius and density of an atom to calc. no. of atoms per unit cell. PRBLEM Al has density = g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SLUTIN 1. Calc. unit cell edge (cm) see handout: edge = 4 * radius / 2 edge = 4 * 143 pm / 2 = 404 pm 404 pm * (10-10 cm / pm) = 4.04 * 10-8 cm Finding the Lattice Type PRBLEM Al has density = g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SLUTIN 2. Calc. unit cell volume edge = 4.04 x 10-8 cm (previous slide) V = (cell edge) 3 = (4.04 x 10-8 cm) 3 V = 6.62 x cm 3 3. Now use density to find mass mass = (6.62 x cm 3 )(2.699 g/cm 3 ) = 1.79 x g/unit cell Finding the Lattice Type PRBLEM Al has density = g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SLUTIN 4. Calculate number of Al per unit cell from mass of unit cell. Mass 1 Al atom = g mol 1 atom = x g, so 1.79 x g unit cell 1 mol x atoms 1 atom x g = 3.99 Al atoms/unit cell...more in lab and problem set Page III / Chapter Ten Lecture Notes Types of Crystalline Solids

13 Metallic Forces Metals have loosely held outer electrons - a sea of electrons over metal "ions" Easy to melt - sea of electrons flows over solid or liquid Conductors of electricity and heat - moving electrons in the "sea" Malleable - not rigid like ionics, able to reform with sea still surrounding atoms Difficult to boil - need to remove atom from "electron sea", high temperatures Page III / Chapter Ten Lecture Notes Na + - Cl - in salt - the Ionic force These are the strongest forces. Lead to solids with high melting temperatures. NaCl, mp = 800 o C Mg, mp = 2800 o C Ionic Solids Comparing NaCl and CsCl End of Chapter 10 Even though their formulas have one cation and one anion, the lattices of CsCl and NaCl are different. The different lattices arise from the fact that a Cs + ion is much larger than a Na + ion. See: Chapter Ten Study Guide Chapter Ten Concept Guide Page III / Chapter Ten Lecture Notes