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LIQUIDS, SOLIDS, and INTERMOLECULAR FORCES

LIQUIDS, SOLIDS, and INTERMOLECULAR FORCES. Intermolecular Forces and Some Proper- ties of Liquids Vaporization of Liquids: Vapor Pressure Some Properties of Solids Phase Diagrams Van der Waals Forces Hydrogen Bonding. Chemical Bonds as Intermolecular Forces

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LIQUIDS, SOLIDS, and INTERMOLECULAR FORCES

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  1. LIQUIDS, SOLIDS, and INTERMOLECULAR FORCES Intermolecular Forces and Some Proper- ties of Liquids Vaporization of Liquids: Vapor Pressure Some Properties of Solids Phase Diagrams Van der Waals Forces Hydrogen Bonding

  2. Chemical Bonds as Intermolecular Forces Crystal Structures Energy Changes in the Formation of Ionic Crystals I. Intermolecular Forces and some Properties of Liquids At high pressures and low temp, inter- molecular forces cause gas behavior to deviate from ideality. When these forces

  3. are sufficiently strong, a gas condenses to a liquid. Intermolecular Forces : a.) Cohesive forces – intermolecular forces between like molecules; Ex. H2O, oil. b.) Adhesive forces – between unlike molecules.Ex. H2O- wood and Hg-glass Properties of Liquids: a.) Surface Tension – the energy required

  4. 0 to increase the surface area of a liquid by a unit of area. - It results from an imbalance of intermole - cular forces, the cohesive forces between molecules. - represented by gamma (γ ) ; has units energy/area , joules/m2. - it decreases with increased temperature.

  5. The molecules at the surface of this sample of liquid water are not surrounded by other water molecules. The molecules inside the sample are surrounded by other molecules.

  6. The unbalanced attraction of molecules at the surface of a liquid tends to pull the molecules back into the bulk liquid leaving the minimum number of molecules on the surface. It required energy to increase the surface area of a liquid because a larger surface area contains more molecules in the unbalanced situation.

  7. b.) The difference in strength between cohesive forces and adhesive forces determine the behavior of a liquid in contact with a solid surface. • Water does not wet waxed surfaces because the cohesive forces within the drops are stronger than the adhesive forces between the drops and the wax. • Water wets glass and spreads out on it because the adhesive forces between the liquid and the glass are stronger than the cohesive forces within the water.

  8. c.) Formation of meniscus Mercury confined in a tube has its surface (meniscus) convex shaped because the cohesive forces in liquid mercury tend to draw it into a drop. Liquid water confined in a tube has its surface (menis- cus) concave shaped because it wets the surface and creeps up the side.

  9. d.) Capillary Action -rise of a liquid that wets a tube up the inside of a small diameter tube (i.e., a capillary) immersed in the liquid. • The liquid creeps up the inside of the tube ( result of adhesive forces between the liquid and the inner walls of the tube) until the adhesive and cohesive forces of the liquid are balanced by the weight of the liquid. • The smaller the diameter of the tube, the higher the liquid rises.

  10. e.)Viscosity - the liquid’s resistance to flow; •Stronger intermolecular attraction, greater viscosity. Cohesive forces within liquid creates “internal friction” which reduces liquid flow. Ex: honey & heavy motor oil. • Weaker cohesive forces, low viscosity. Ex: water & ethyl alcohol.

  11. II. VAPORIZATION OF LIQUIDS : VAPOR PRESSURE • Vaporization ( Evaporation ) - passage of molecules from the surface of the liquid into the gaseous or vapor state; molecules have sufficiently high energies to overcome intermolecular forces and escape from a liquid; increases with tempe – rature and weaker intermolecular forces.

  12. a.) Enthalpy of Vaporization – quantity of heat to be absorbed if certain quantity of liquid is vaporized at constant temp. Stated mathematically as, ΔHvaporization = Hvapor – Hliquid - ΔHvap is measurable (table 13.1 p.423); positive since vaporization is endothermic. Ex. Determining an Enthalpy of Vapor. To vaporize 1.75 g of acetone , (CH3)2CO , at 298 K,

  13. 767 J of heat is required. What is the enthalpy of vaporization of acetone , in kJ/ mole ? Sol: Determine the no. of moles of acetone in 1.75 g. Convert 767 J to kJ and divide this quantity of heat by the moles of acetone . 1 kJ 767 J --------- 1000 J ΔHvap = ------------------------------------ = 25.5 kJ/mol 1 mol 1.75 g (CH3)2 CO ----------- 58.08 g

  14. Ex 2. How much heat is reqd to vaporize a 2.35 g sample of diethyl ether at 298 K ? Using the data in table of enthalpies p 423 , and if heat reqd ΔHvap = ---------- , then mole heat reqd = ΔHvap ( mol ) . So (29.1 kJ / mol) ( 2.35 g ) = ------------------------------ = 0.923 kJ 74.12 g / mol

  15. • condensation- conversion of gas to liquid .It is the reverse of vaporization. It is opposite in sign but equal in magnitude to ΔHvap thus , ΔHcon is always negative. Condensation is exothermic. b.) Vapor Pressure – the pressure exerted by a vapor in dynamic equilibrium with its liquid. Dynamic equilibrium – a condition at which two opposing processes occur simultaneously and at equal rates. ( Refer to fig 13.6 , p. 424 of the book )

  16. - dependent only on the particular liquid and its temperature , not on the amount of the liquid. - Liquids with high vapor pressures at room temp are said to be volatile , whose intermolecular forces are weaker. Liquids with low vapor pressure are nonvolatile , intermolecular forces great. c.) Boiling and the Boiling Point Boiling – process at which the pressure exerted by the escaping molecules of a liquid equals that of the

  17. The normal boiling point is the temperature at which its vapor pressure is 760 torr. • The boiling point of a liquid is the temperature at which its vapor pressure equals atmospheric pressure.

  18. - happens when a liquid is heated in a container open to the atmosphere. Vaporization occurs throughout the liquid. - In boiling , temp is constant until all the liquid has boiled away. - pressure exerted bymolecules of the atmosphere; • Normal boiling point – temp at which vapor pressure of a liquid equal to the atmospheric pressure. The normal b.p. of several liquids can be determined from fig 13.8 ( vapor pressure curve of liquid.)

  19. d.) Critical Point – the point at which the liquid and vapor pressure of a pure stable substance have the same density, surface tension zero and meniscus between vapor and liquid disappears Happens if a liquid is heated in a sealed container where boiling does not occur. Instead , temp and vapor pressure rise continuously. The temp at critical point , critical temp Tc and the pressuure is critical pressure , Pc . ( Table 13.3)

  20. e.) Using Vapor Pressure Data 1.) Section 6-6 concerns collecting gases over liquids, water. 2.) Predicting whether a substance exists solely as as a vapor or as a liquid and vapor in equilibrium.( Refer fig 13-11, p.427 ) Ex 13.2. As a result of a chem rxn , 0.132 g H2O is produced and maintained at a temp of 50.0oC in closed flask of 525 - ml volume. Will the water be present as vapor only or liquid & vapor in equilibrium?

  21. Sol : First use the ideal gas eqtn to calculate the pressure that would exist if the water were present as vapor only. P = nRT ÷ V 1 mol 0.08206 L atm = 0.132 g x -------- x ---------------- x 323.3 K 18.02 g mol K 0.525 L 760 mm Hg = 0.370 atm x = 281 mm Hg 1 atm

  22. Now obtain the vapor pressure of water at 50.0oC from table 13.2. Because the calculated pressure exceeds the data from table, some of the vapor must condense to liquid. The water is present as liquid and vapor in equi- librium at 92.5 mm Hg. f.) An Equation for Expressing Vapor Pressure Data A particularly common form of vapor pressure eqtn is shown below, which expresses the natural loga rithm(ln) of vapor pressure as a function of the reci – procal of the Kelvin temp (1/T). The relationship is

  23. that of a straight line and the plots are drawn in fig 13.13, p 429 . ln P = - A ( 1/ T ) + B eqnt of straight line y = m x + b The constant A is related to the enthalpy of vapo - rization of liquid : A = ΔHvap/ R , where ΔHvap is in J/mol and R is 8.3145 J mol-1 K-1. It is customary to eliminate B ( Refer to Appendix A – 4) by rewriting the eqtn in the form called Clausius – Clapeyron eqtn.

  24. P2 ΔHvap 1 1 ln = --- - --- P1 R T1 T2 Ex. Applying the Clausius – Clapeyron Eqtn. Calculate the vapor pressure of water at 35.00C with data from Table 13.1 and 13.2 Sol: Let P1 = unknown vapor pressure T1 = 35.00C + 273.2 = 308.2 K From table 13.2, choose for P2 and T2 known data at temp close to 35.00C. At T2 = 40.00C = 313.2 K, P2 = 55.3 mm Hg. Assuming ΔHvap is independent of temp, its value from table 13.2 is 44.4 kJ/mol, which we convert to J/mol. Using the Clausius – Clapeyron equation ,

  25. 55.3 mm Hg 44.0 x 103 J mol-1 1 1 Ln = - K-1 P1 8.3145 J mol-1 K-1 308.2 313.2 = 5.29 x 103 ( 0.003245 – 0.003193 ) = 0.28 Determine that e0.28 = 1.32 ( Appendix A) . Thus, 55.3 mm Hg = e0.28 = 1.32 P1 P1 = 55.3 mm Hg / 1.32 = 41.9 mm Hg

  26. III. Some Properties of Solids a.) Melting, Melting Point and Heat of Fusion Melting(Fusion) – solid to liquid Freezing(solidification) – liquid to solid Melting pt of solid and freezing pt of liquid are iden- tical. Solid and liquid coexist in equilibrium . Enthalpy of fusion (ΔHfus )) – heat required to melt a solid. Expressed in kJ/mol , table 13.4. Supercooling – the condition at which the tempe - rature may drop below the freezing point without any solid appearing.

  27. Supercooling – the condition at which the tempe - rature may drop below the freezing point without any solid appearing.Liquid must contain some small particles ( suspended dust particles ) on which crystals can form.; the lesser the number of particles the crystals can grow, it may super cool for a time before freezing. (Refer to fig 13-14 for diagram.)

  28. a.) Sublimation – direct passage of molecules from solid to the vapor state. Reverse process is called Deposition. When both processes occur, dynamic equilibrium exists between solid and its vapor. The vapor exerts a characteristic pressure called sublimation pressure. Enthalpy of Sublimation – quantity of heat needed to convert a solid to vapor. At melting point , sublimation is equivalent to melting followed by vaporization.Thus , ΔHsub = ΔHfus + ΔHvap

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