CHEMISTRY - TRO 4E CH.11 - LIQUIDS, SOLIDS & INTERMOLECULAR FORCES

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CONCEPT: INTERMOLECULAR FORCES When looking at a molecular substance such as H 2 O you will discover two types of electrostatic forces at work: forces exist within a molecule and influences the properties of the substance. forces exist between molecules and influence the properties of the substance. is the force that exists between an ion and a polar compound. (Strongest) Ex: is the force that exists when H is directly connected F, O, N. (2 nd Strongest) Ex: is the force that exists when two polar covalent compounds interact. (3 rd Strongest) Ex: is the force that exists when a nonpolar covalent compound interacts with a polar covalent compound. (4 th Strongest) Ex: is the force that exists when two nonpolar covalent compounds interact. (Weakest) Ex: Page 2

PRACTICE: INTERMOLECULAR FORCES EXAMPLE: Based on the given compounds, answer each of the following questions: a. CH3CH3 b. KBr c. C6H5OH d. CaS e. Ne a) Which compound will have the lowest boiling point? b) Which compound will have the highest surface tension. c) Which compound will have the highest vapor pressure. PRACTICE 1: The predominant intermolecular force in C6H5NH2 is: a. London Dispersion b. Hydrogen Bonding c. Ion-Dipole d. Dipole-Dipole e. Dipole-induced Dipole PRACTICE 2: The predominant intermolecular force in HBr is: a. London Dispersion b. Hydrogen Bonding c. Ion-Dipole d. Dipole-Dipole e. Dipole-induced Dipole PRACTICE 3: The predominant intermolecular force in ZnBr2 with H2O is: a. London Dispersion b. Hydrogen Bonding c. Ion-Dipole d. Dipole-Dipole e. Dipole-induced Dipole PRACTICE 4: The predominant intermolecular force in Ne with H2O is: a. London Dispersion b. Hydrogen Bonding c. Ion-Dipole d. Dipole-Dipole e. Dipole-induced Dipole Page 3

CONCEPT: SOLUBILITY According to the theory of dissolves compounds with the same intermolecular force or polarity will dissolve into each other. EXAMPLE: Identify the intermolecular forces present in both the solute and the solvent, and predict whether a solution will form between the two. a. CCl4 and P4 b. CH3OH and C6H6 c. C6H5CH2NH2 and HF d. IF4 and NH3 PRACTICE: Which of the following statements is/are true? a. Methane will dissolve completely in acetone, CH3COCH3. b. Hydrofluoric acid (HF) will form a heterogeneous mixture with tetrachloride, CCl4. c. Pentane will form a homogeneous mixture with CBr4. d. Methanethiol (CH3SH) is miscible in fluoromethane (CH3F). Page 4

CONCEPT: PHASE DIAGRAMS Under appropriate conditions of pressure and temperature, most substances can exist in 3 states of matter:, and. Microscopic Explanation for the Behavior of Gases, Liquids and Solids Gas Liquid Solid Assumes the and of its container. Assumes the of the portion of its container it occupies, but not the. Maintains a fixed and compressible compressible compressible Viscosity Viscosity Viscosity Viscous Viscous Viscous Now, a convenient way to show the effect that temperature and pressure has on a pure substance in a closed system without any air is to use a phase diagram. Page 5

PRACTICE: PHASE DIAGRAMS a) At what temperature can we no longer tell the difference between the liquid and the gas? b) Which point represents an equilibrium between the solid, liquid and gas phase? c) Which line segment represents an equilibrium between fusion and freezing? d) Which line segment represents an equilibrium between sublimation and deposition? e) Which line segment represents an equilibrium between condensation and vaporization? f) What is the normal freezing point of this unknown substance? g) What is the normal boiling point of this unknown substance? Page 6

CONCEPT: HEATING & COOLING CURVES In heating and cooling curves we have the representation of the amount of heat absorbed or released during phase changes. Heating Curve Temperature ( o C) Time Cooling Curve Specific Heat of Ice 2.09 ΔH Fusion 334 g Specific Heat of Water 4.184 ΔH Vaporization 2260 g Specific Heat of Steam 1.84 Temperature ( o C) Time Page 7

PRACTICE: HEATING & COOLING CURVE CALCULATIONS 1 EXAMPLE: How much energy (k) is required to convert a 76.4 g acetone (molar mass = 58.08 g/mol) as a liquid at -30 o C to a solid at -115.0 o C? Specific Heat of Solid 1.65 ΔH Fusion 7.27 k mol Specific Heat of Liquid 2.16 Specific Heat of Gas 1.29 T Melting 95.0 o C Page 8

PRACTICE: HEATING & COOLING CURVE CALCULATIONS 2 PRACTICE: If 53.2k of heat are added to a 15.5g ice cube at - 5.00 o C, what will be the resulting state and temperature of the substance? Specific Heat of Ice 2.09 ΔH Fusion 334 g Specific Heat of Water 4.184 ΔH Vaporization 2260 g Specific Heat of Steam 1.84 Page 9

CONCEPT: CLASIUS-CLAPEYRON EQUATION By using the Clasius-Clapeyron equation a quantitative relationship between and can be established. ln P 2 = ΔH vap P 1 R # 1 1 & % ( $ T 2 T 1 ' ` EXAMPLE : The heat of vaporization ( Hvap) of water is 40.3 k/mol at its normal boiling point at 100 o C. What is the vapor pressure (in mmhg) of water at 60 o C? Page 10