{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Ch10.ppt - CHAPTER 10 Liquids and Solids 1 THREE THREE...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHAPTER 10 Liquids and Solids 1 THREE THREE STATES STATES OF OF MATTER MATTER • Molecules in both solid and liquid states are in contact. • Molecules of gas are very far apart and completely disordered. What holds the molecules close to each other in liquids and solids? 2 3 10.1. 10.1. Intermolecular Intermolecular Forces Forces INTERmolecular forces – forces between molecules. Not the same as Chemical Bonds — the forces holding atoms together inside each molecule. 4 ATTRACTIVE FORCES Chemical Bonds These forces exist within each molecule. They influence the chemical properties of the substance. Examples: covalent bonds ionic bonds Intermolecular Forces These forces exist between molecules. They influence the physical properties of the substance. Examples: hydrogen bonds dipole-dipole forces Comparison Comparison of of magnitude magnitude of of forces forces The 3 main types of intermolecular forces: a) hydrogen bonds – relatively strong; b) dipole-dipole forces – “average”; c) dispersion (London) forces – the weakest intermolecular force. 5 6 ATTRACTIVE FORCES Chemical Bonds (within each molecule) Intermolecular forces (between molecules) Hydrogen bonds All intermolecular forces are weaker than chemical bonds. Dipole-dipole forces London forces Strong intramolecular and weak intermolecular forces. Substances with stronger intermolecular forces have higher boiling points; weaker intermolecular forces - lower boiling points 7 8 Dipole-Dipole Dipole-Dipole Forces Forces Attractive forces between polar molecules are called dipole-dipole forces. Polar molecules (molecules with dipole moments) attract each other electrostatically (positive end of one molecule attracts the negative end of another molecule). 9 Dipole-Dipole Dipole-Dipole Forces Forces solid Dipole-dipole forces exist between polar molecules. liquid 10 LONDON LONDON DISPERSION DISPERSION FORCES FORCES How can nonpolar molecules such as O2 and I2 condense to liquid? • In nonpolar molecules, instantaneous dipoles can exist temporarily. because of random motions of electrons, momentarily, most of the electrons in a molecule may happen to be at one end. • The attractive force resulting from such temporary dipole is called London dispersion force. 11 LONDON LONDON DISPERSION DISPERSION FORCES FORCES 12 LONDON LONDON DISPERSION DISPERSION FORCES FORCES Compare boiling points of several nonpolar compounds: Molecule CH4 (methane) C2H6 (ethane) Boiling Point (oC) - 161.5 - 88.6 C3H8 (propane) C4H10 (butane) - 42.1 - 0.5 What influences this trend? Boiling Points of Hydrocarbons C4H10 C3H8 C2H6 CH4 Note linear relation between boiling points and molar mass. 13 14 LONDON LONDON DISPERSION DISPERSION FORCES FORCES • Larger London forces occur in molecules with larger number of electrons. » which correlates with larger molar mass. • London forces are stronger when molar mass of a compound is higher. Hydrogen Hydrogen Bonding Bonding • Hydrogen bond (“H-bond”) is the especially strong type of intermolecular force. • Hydrogen bond occurs when all of these 3 requirements are met: 1) There must be a H atom in a molecule; 2) There must be one of the following atoms: F, O, N; 3) There must be a covalent bond between a H atom and one of the F/O/N atoms. X and Y are N, O, or F 15 16 SAMPLE PROBLEM 1 PROBLEM: PLAN: Recognizing Molecules that have Hydrogen Bonds Which of the following substances exhibits hydrogen bonding? O (a) C2H6 (b) CH3OH (d) F 2 (c) CH3C H Find molecules in which H is bonded to F, O or N. SOLUTION: (a) C2H6 doesn’t have H-bonding, because there’s no F, O, or N. (b) CH3OH has H-bonding. There’s H, there’s O, and there’s H a bond between them. H C O H O H H (c) CH3C H doesn’t have H-bonding. Even though H O C H there’s H and there’s O, there’s no bond directly H between them. (d) F2 doesn’t have H-bonding, because there’s no H. 17 H-Bonding H-Bonding Between Between Two Two Methanol Methanol Molecules Molecules - + - H-bond H-bond 18 H-Bonding H-Bonding Between Between Methanol Methanol and and Water Water - H-bond H-bond + - Hydrogen Hydrogen Bonding Bonding in in H H22O O 19 H-bonding is especially strong in water because • the O—H bond is very polar • there are 2 lone pairs on the O atom Boiling Points of Simple HydrogenContaining Compounds Figure 10.4 20 Hydrogen Hydrogen Bonding Bonding H-bonds lead to abnormally high boiling point of water. H-bonding also accounts for many other water’s unique properties. 21 Ion-Ion Ion-Ion Forces Forces for for comparison comparison of of magnitude magnitude 22 How strong are intermolecular forces in ionic compounds? Na+—Cl- forces are the same as ionic bonds. Very strong! Therefore, ionic solids have high melting/boiling NaCl, mp = 800 oC temperatures. MgO, mp = 2800 oC 23 Intermolecular Forces and Boiling Points The most important factor that influences the trend in boiling points is the type of intermolecular force. Ionic > H-bonding > dipole-dipole > dispersion 24 Among nonpolar compounds, boiling point depends on the molar mass. In polar substances, boiling point depends on dipole moment more than it does on molar mass. 25 26 27 SAMPLE PROBLEM 2 PROBLEM: Predicting the Type and Relative Strength of Intermolecular Forces For each pair of substances, identify the dominant intermolecular forces in each substance, and select the substance with the higher boiling point. (a) MgCl2 or PCl3 (b) CH3NH2 or CH3F (c) Hexane (CH3CH2CH2CH2CH2CH3) or ethane (CH3CH3) SOLUTION: (a) MgCl2 has ionic bonds. The PCl3 has covalent bonds inside the molecules, and dipole-dipole forces between the molecules. Ionic bonds are stronger than dipole-dipole forces, so MgCl2 has the higher boiling point. (b) CH3NH2 and CH3F are both covalent and both polar. The CH3NH2 can make H-bonds, while CH3F cannot. Therefore CH3NH2 has the stronger interactions and the higher boiling point. (c) Hexane and ethane are both nonpolar, and molecules interact by only dispersion forces. Hexane has the larger molar mass, thereby the greater dispersion forces and the higher boiling point. 28 SAMPLE PROBLEM 3 PROBLEM: Arrange the normal boiling points of the following substances from highest to lowest: (a) He PLAN: Arranging Substances by Order of Boiling Points (b) CH4 (c) HBr (d) Cl2 (e) HF 1) Determine which of the compounds is polar, and which is nonpolar; 2) Determine which of the compounds can form hydrogen bonds; 3) If there are several nonpolar compounds, the strength of London forces depends on molar mass. Stronger intermolecular forces -> higher boiling point SOLUTION: 1) He, CH4, and Cl2 are nonpolar; HF and HBr are polar; 2) Only HF forms hydrogen bonds; 29 SAMPLE PROBLEM 3 Arranging Substances by Order of Boiling Points continued SOLUTION: 3) Among nonpolar compounds, check for molar mass: He 4 g/mol CH4 16 g/mol Cl2 71 g/mol Highest boiling point: HF Next: HBr Next: Cl2 Next: CH4 Lowest boiling point: He Boiling points from highest to lowest: HF > HBr > Cl2 > CH4 > He 30 Changes Changes of of State State sublimation vaporization melting solid liquid freezing gas condensation requires energy (endothermic) releases energy (exothermic) 31 Sublimation Sublimation of of Iodine Iodine test tube with ice iodine solid iodine vapor iodine solid Solid I2 sublimes (goes from a solid to a gas) because the dispersion forces between I2 molecules are very weak. Heat and Changes of State 32 33 Heats of fusion and vaporization for several common substances. Heat of fusion ( Hfus) is amount of heat required to melt 1 gram of solid. Heat of vaporization ( Hvap) is amount of heat required to vaporize 1 gram of liquid. Changes of State 34 • HEAT OF VAPORIZATION is the heat required to vaporize the liquid. LIQUID + heat ---> VAPOR H = Hvap • It can be given either per gram (in kJ/g) or per mole (in kJ/mol). Compd. IM Force BP ∆Hvap (kJ/mol) H2O H-bonds 100 oC 40.7 SO2 dipole-dipole - 47 oC 26.8 Xe dispersion 12.6 -107 oC Heat Heat Transfer Transfer with with Change Change of of State State Changes of state require energy (at constant T) Ice + 333 J/g (heat of fusion) -----> Liquid water q = (heat of fusion)(mass) 35 Heating/Cooling Heating/Cooling Curve Curve for for Water Water Note that T is constant as ice melts 36 37 Quantitative Aspects of Phase Changes Within a phase, a change in heat is accompanied by a change in temperature; the specific heat formula applies: q = (mass)(specific heat capacity)(T) During a change of phase, a change in heat occurs at a constant temperature; the formula to be used is: q = (mass)(heat of phase change) either Hvap or Hfus Heat Heat & & Changes Changes of of State State What quantity of heat is required to melt 500. g of ice at 0 oC and convert the water to steam at 100 oC? Heat of fusion of ice = 333 J/g Heat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/g•K Specific heat of water = 4.2 J/g•K Heat of vaporization = 2260 J/g Heat of vaporization = 2260 J/g +333 J/g +2260 J/g heat to 100C 38 Heat Heat & & Changes Changes of of State State What quantity of heat is required to melt 500. g of ice at 0 oC and heat the water to steam at 100 oC? 1. To melt ice q = (500. g)(333 J/g) = 1.67 x 105 J 2. To raise water from 0 oC to 100 oC q = (500. g)(4.2 J/g•K)(100 ­ 0)K = 2.1 x 105 J 3. To evaporate water at 100 oC q = (500. g)(2260 J/g) = 1.13 x 106 J Total heat energy = 1.51 x 106 J = 1510 kJ 39 Vaporization of Liquids EVAPORATION EVAPORATION (liquid (liquid -> -> gas) gas) CONDENSATION CONDENSATION (gas (gas -> -> liquid) liquid) A A molecule molecule evaporates evaporates (“escapes” (“escapes” the the liquid) liquid) ifif itit can can break break the the intermolecular intermolecular forces forces that that attract attract itit to to its its neighbors. neighbors. LIQUID evaporation---> VAPOR break intermolec. forces form intermolec. forces <---condensation 40 Liquids— Evaporation To evaporate, molecules must have sufficient energy to break intermolecular forces. Breaking intermolecular forces requires energy. The process of evaporation is endothermic. 41 Distribution of Energy in a Liquid Minimum energy required to break intermolecular forces and evaporate 42 aver.KE is proportional to T. Vaporizati on Number of molecules lower T 0 higher T Molecular energy minimum energy needed to break IM forces and evaporate 43 At higher T a much larger number of molecules has high enough energy to break intermolecular forces and move from liquid to vapor state. High-energy molecules carry away Energy. You cool down when sweating or after swimming. Vapor Pressure Suppose we have a liquid in a closed container. After a while, a balance (EQUILIBRIUM) is achieved between rates of evaporation and condensation. At equilibrium, vapor pressure can be measured. 44 45 Liquid-gas equilibrium 46 Vapor Pressure • Vapor pressure depends on Temperature. • The larger the temperature, the larger the vapor pressure. – even though the relationship is nonlinear. Liquid Liquid boils boils when when its its vapor vapor pressure pressure equals equals atmospheric atmospheric pressure. pressure. FIGURE 10.42 Boiling of Liquids NORMAL NORMAL BOILING BOILING POINT POINT of of aa liquid liquid is is the the temperature temperature at at which which boiling boiling occurs occurs under under pressure pressure of of exactly exactly 1.0 1.0 atm atm (760 (760 torr). torr). What’s normal boiling point of diethyl ether? of ethanol? of water? Can water boil at T < 100ºC? 47 48 Boiling Point at Lower Pressure When pressure is lowered, the vapor pressure can equal the external pressure at a lower temperature. This means that BP’s of liquids change with altitude. 49 Vapor pressure as a function of temperature and intermolecular forces. A linear plot of vapor pressure- temperature relationship. Equilibrium Vapor Pressure & the Clausius-Clapeyron Equation • The logarithm of the vapor pressure (lnP) is proportional to ∆Hvap and to 1/T. • This is Clausius-Clapeyron equation: -Hvap 1 C ln P = R T 50 51 Graphical determination of heat of vaporization C slope = -Hvap / R ln P ln P = -Hvap/R (1/T) + C Clausius-Clapeyron Equation 52 -Hvap 1 C ln P = R T how to get rid of C? write out the equation at two temperatures, and subtract. - ln P2 = - Hvap/RT2 + C Because P ln 2 ln P2 ln P1 P1 (ln P1 = - Hvap/RT1 + C ) P2 Hvap ln ( ) = P1 R 1 T2 1 T1 53 Clausius-Clapeyron Equation The rearranged Clausius-Clapeyron Equation (more practical form): P2 Hvap ln ( ) = P1 R 1 T2 1 T1 where Hvap is heat of vaporization R = 8.31 J/mol•K both T must be in Kelvin 54 SAMPLE PROBLEM 4 PROBLEM: PLAN: Using the Clausius-Clapeyron Equation The vapor pressure of ethanol is 115 torr at 34.90C. If Hvap of ethanol is 40.5 kJ/mol, calculate the temperature (in 0C) when the vapor pressure is 760 torr. We are given 4 of the 5 variables in the Clausius-Clapeyron equation. Plug them in and solve for T2. SOLUTION: P2 Hvap ln ( ) = P1 R 1 T2 1 T1 T1 = 34.90C = 308.0K P1 = 115 torr P2 = 760 torr R = 8.31 J/mol*K Hvap = 40.5 kJ/mol = 40.5 x 103 J/mol 55 SAMPLE PROBLEM 4 Using the Clausius-Clapeyron Equation continued 760 torr ln 115 torr = = ln 6.609 1 T2 1 T2 - - 40.5 x103 J/mol 8.314 J/mol*K - 4871 x 1 308K = = 1 T2 - 1.888 - 4871 1 - 3.877 x 10-4 308K T2 = 350K = 770C 1 T2 - 1 308K 1 308K = - 3.877 x 10-4 = 2.859 x 10-3 56 SAMPLE PROBLEM 5 PROBLEM: PLAN: Using the Clausius-Clapeyron Equation In Breckenridge, Colorado, the atmospheric pressure is 520.0 torr. What is the boiling point of water (Hvap = 40.7 kJ/mol) in Breckenridge? Water (and any other liquid) always boils when its vapor pressure equals external (atmospheric) pressure. Therefore, we are looking for temperature (T1) at which vapor pressure of water is 520.0 torr (P1). We also know that at the sea level water would boil at 100.0 °C. Therefore, at temperature T2 =100.0 °C water’s vapor pressure is P2 = 760.0 torr. We can now use Clausius-Clapeyron equation to solve for T1. SOLUTION: T2 = 373 K R = 8.31 J/mol*K P2 Hvap ln ( ) = P1 R P2 = 760 torr 1 T2 1 T1 P1 = 520 torr Hvap = 40.7 kJ/mol = 40.7 x 103 J/mol 57 SAMPLE PROBLEM 5 Using the Clausius-Clapeyron Equation continued 760 torr ln 520 torr = = ln 1.462 1 T1 1 T1 - 40.7 x103 J/mol 8.314 J/mol*K - 4895 x 1 373 K = = 1 373 K 0.3798 4895 1 + 7.759 x 10-5 373K T1 = 362.5K = 89.5 0C 1 373 K 1 T1 1 T1 58 Sample Problem 6 PROBLEM: PLAN: Determining the Activation Energy A student plotted logarithm of vapor pressure vs 1/T and obtained a straight line with the slope of -4100. Find the enthalpy of vaporization. The slope of the line represents –H/R, which can be solved for H SOLUTION: ln P = - H /RT + C H 4100 R H = 4100x8.31 = 34 kJ/mol Vapor Pressure and Intermolecular Forces • At a given temperature, liquids that are volatile (close to boiling) have higher vapor pressure than liquids that are far from boiling. • Lower boiling point higher vapor pressure at given T. Vapor pressure is larger when the attractive (intermolecular) forces between molecules are weaker. 59 Vapor Pressure and Intermolecular Forces Problem: Determine from the graph, which of the substances (diethyl ether, ethanol, and water) have larger vapor pressure at 20ºC, and propose the explanation. ether O C2H5 H5C2 dipole­ dipole ethanol O H5C2 H H­bonds water O H H extensive H­bonds increasing strength of intermolecular forces 60 Using Intermolecular Forces Problem: Methanol (CH3OH) and methyl iodide (CH3I) are both polar, but methyl iodide’s vapor pressure at room temperature is almost 4 times as great as that of methanol. What is molecular basis for the difference? Problem: Which of the substances: ammonia (NH3) or phosphine (PH3) has larger vapor pressure at T = -100 °C? 61 10.9. Phase Diagrams 62 Phase Diagram for Water 63 Liquid phase Solid phase Gas phase 64 Phase Diagrams • “Phase” means state of matter. • A phase diagram is drawn in coordinates of P vs T. • Lines connect all conditions of T and P where EQUILIBRIUM exists between the phases on either side of the line. TRANSITIONS BETWEEN PHASES Melting: Solid Liquid Freezing: Liquid Solid Boiling: Liquid Gas Condensation: Gas Liquid Sublimation: Solid Gas 65 Triple Point At the TRIPLE POINT all three phases are in equilibrium. 66 Critical Point 67 As As P P and and TT increase, increase, you you finally finally reach reach the the CRITICAL CRITICAL point point • Above critical T, no liquid exists no matter how high the pressure. • The state of matter above critical T is called supercritical fluid. 68 Phase Diagram for CO2 69 Phases Diagrams— Important Points for Water T(˚C) P(atm) Triple point 0.0098 0.006 Critical point 218 atm 374 Normal boiling point 100 1.0 Normal freezing point 0 1.0 70 Normal boiling and normal freezing p Normal Boiling Point is the temperature at which liquid boils at P=1.00 atm. What is normal freezing point? How to determine these points from the phase diagram? 71 Solid-Vapor Equilibria At P < 0.006 atm and T < 0.0098 ˚C solid H2O (ice) can go directly to vapor. This process is called SUBLIMATION Solid CO2 (dry ice) can sublime even at atmospheric pressure! Phase diagrams for H2O and CO2 H2O CO2 72 Solid-Liquid Equilibria H2O Raising the pressure at constant T causes solid ice to melt. The NEGATIVE SLOPE of the S/L line is unique to H2O. Almost everything else has positive slope. 73 Hydrogen Bonding in H2O Ice has open lattice-like structure. 74 Hydrogen Hydrogen Bonding Bonding in in H H22O O Ice has open lattice-like structure. Ice density is < liquid, and so ice floats on water. One of the VERY few substances where solid is LESS DENSE than the liquid. 75 Solid-Liquid Equilibria The unique negative S/L slope of water has to do with densities of liquid and solid H2O: Liquid Liquid H H22O OSolid Solid H H22O O Density Density 11 g/cm g/cm33 0.917 0.917 g/cm g/cm33 In any system, if you increase P the DENSITY will go up. Therefore — as P goes up, the S/L equilibrium favors phase with the larger density (or smaller volume/gram). ICE ICE favored at favored at low P low P LIQUID H LIQUID H22O O favored at favored at high P high P 76 Liquid-Vapor Equilibria The equilibrium curve between liquid and vapor is explained by Clausius-Clapeyron equation. ln P = -Hvap 1 C R T 77 10.3. 10.3. celestite Metallic Metallic and and Ionic Ionic Solids Solids pyrite amethyst halite The striking beauty of crystalline solids. 78 79 Crystalline and amorphous silicon dioxide. Amorphous solids do not have a regular structure. Glass is amorphous SiO2. Crystal Lattices UNIT CELL - smallest repeating unit that has the symmetry characteristic of the solid. 80 81 The crystal lattice and the unit cell. lattice point unit cell unit cell portion of a 3-D lattice portion of a 2-D lattice Cubic Unit Cells of Metals Simple cubic (SC) Bodycentered cubic (BCC) Facecentered cubic (FCC) 82 Cubic Unit Cells of Metals 83 Unit Cells for Metals 84 85 Atom Sharing at Cube Faces and Corners Atom shared in corner --> 1/8 inside each unit cell Atom shared in face --> 1/2 inside each unit cell Simple Cubic Unit Cell • Each atom is at a corner of a unit cell and is shared among 8 unit cells. • Each edge is shared with 4 cells. • Each face is part of two cells. 86 87 Simple cubic (SC) Number of Atoms Per Unit Cell – the average number of atoms in a volume of each unit cell. 1/8 atom at 8 corners Atoms/unit cell = 1/8 * 8 = 1 88 Body-centered cubic (BCC) Number of Atoms Per Unit Cell – the average number of atoms in a volume of each unit cell. 1/8 atom at 8 corners 1 atom at center Atoms/unit cell = (1/8*8) + 1 = 2 Face-centered cubic (FCC) 89 Number of Atoms Per Unit Cell – the average number of atoms in a volume of each unit cell. 1/8 atom at 8 corners 1/2 atom at 6 faces Atoms/unit cell = (1/8*8)+(1/2*6) = 4 Number of Atoms per Unit Cell Unit Cell Type SC BCC FCC Net Number Atoms 1 2 4 90 Atom Packing in Unit Cells Packing factor (packing efficiency) – the fraction of the space occupied by atoms. 91 Packing Factors in Unit Cells • P...
View Full Document

{[ snackBarMessage ]}