CHEM 1442 1st Edition Lecture 1Outline of Last Lecture Outline of Current Lecture I. ExamplesII. Physical Transformations of Pure SubstancesIII. Phase DiagramsIV. BondsV. Intermolecular ForcesVI. Comparing Intermolecular ForcesCurrent LectureExample: Calculate the enthalpy change upon converting 15.00 g of water at 25.0°C to 75.0°C under a constant pressure of 1 atm.Answer: *Specific Heat Capacity* -- energy required to change the temperature of 1 g of a substance by 1oC q = mc∆Tq = (15 g)(4.184 J/g∙oC)(75oC – 25oC)q = + 3.14kJExample: Convert 4.184 J/g∙oC as a molar heat capacity.Answer: *Molar heat capacity* -- energy required to change the temperature of 1 mol of a substance by 1oC4.184 J/g∙oC x 18.02 g / 1 mol = 75.40 J/mol∙oCPhysical Transformations of Pure SubstancesPhase = a form of matter that is uniform throughout in chemical composition and physical state.Misconception: A Phase = A Physical State (WRONG)Solid to Liquid: Melting (Fusion)These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Liquid to Solid: FreezingLiquid to Gas: VaporizationGas to Liquid: CondensationGas to Solid: DepositionSolid to Gas: SublimationMisconception: Sublimation of water is a process in which you heat ice tomelt, then heat water to boil to evaporate it.(This is WRONG!)What (state variable) determines a phase?- Temperature & PressurePhase DiagramsPhase diagrams tell us which phase will exist at the given temperature and pressure.- The line between solid and liquid is known as the Fusion Curve, where solids and liquids coexist.- The line between liquid and gas is known as the Vaporization Curve, where gas and liquid coexist.- The line between gas and solid is known as the Sublimation Curve, where solids and gases coexist.- Critical Point: at and above the critical temperature and pressure, where liquid and gas states coalesce into supercritical fluid- Triple Point: the solid, liquid, and gas coexistExample: Predict the sign of the enthalpy of reaction.H2O(l) H2O(g) Heat (Enthalpy) of VaporizationΔHovap = + 40.7kJ (at 100oC)*Vaporization: heat is absorbed (endothermic) – positive sign*Condensation: heat is released (exothermic) – negative signMelting point and boiling point are “pressure”-dependentP = 1 atm: normal melting point and normal boiling point*At higher altitudes, the pressure is smaller than 1 atm.*Example: Calculate the enthalpy change upon converting 15.00 g of water to vapor at 100.0°C under a constant pressure of 1 atm.Answer: ΔHovap = + 40.7kJ (at 100oC)15.00 g x 1 mol/18.02 g x 40.7 kJ/1 mol = +33.9 kJBondsExample: Select all ionic compoundsA. CH4B. CaCl2C. Cl2OD. HClE. PCl3Answer:B. CaCl2Example: Which compound has the stronger bond? Why?Melting Point oC Boiling Point oCKCl 771 1500MgO 2852 3600Answer: MgO because it takes more energy to break the bond. It has a higher boiling point and melting point.*The higher the mp (or the bp) of an ionic compound, the more energy is required to “break” ionic bonding.**The higher the mp (or the bp) of an ionic compound, the stronger the ionic bonding between ions operating in an ionic compound is.*Example: Which bond is stronger? Why?H–F or H–O Melting Point oC Boiling Point oC Bond Energy(kJ/mol)HF -83.6 19.5 565H2O 0 100 467Answer: H—F because of its polarity. This question is only comparing one hydrogen bond in the water, not both.H–F has a stronger bond energy.Covalent bonds: They are attractive forces that holds atoms together in a molecule. They make covalent bonds. Its strength is measures through bond energy.Intermolecular forces: They are attractive forces that hold molecules together in a covalent compound. They do not make covalent bonds. Its strength is measured by the melting point or the boiling point.Intermolecular ForcesThese are attractions that are not present between ions.Type 1: Dipole—Dipole ForcesExample: Which of the following molecules (SO2 and CO2) would have dipole-dipole attractions between themselves?Answer:SO2 (polar) and CO2 (nonpolar)So the answer is sulfur dioxide*Dipole—Dipole forces do not exist in non-polar; they are only found in polar molecules*The strength correlates to the magnitude of dipole moment:Dipole moment = charge x distance (D = Q x I)Unit for dipole moment: debye (D)*More polar = stronger dipole moment*Type 2: Hydrogen Bonding- It is a special case of dipole—dipole interactions- It is an electrostatic attraction between:o The partially-positive (δ+) hydrogen atom in a polar bond (H—N, H—O, or H—F)o A lone pair of negative electrons on an electronegative atom (N, O, or F)Type 3: London Dispersion Forces- They are found in all substances- Dispersion forces increase as molecular weight increases- Dispersion forces between spherically-shaped molecules are smaller than those between more cylindrically-shaped moleculesIntermolecular Force = Hydrogen Bond + Dispersion Force*Larger molecule = stronger dispersion force*Example: Which is correct about London Dispersion Forces?Answer:B. The larger the molecule is, the stronger the dispersion force is.C. The strength of London Dispersion forces depend on the shape of the molecule (surface contact)Example: Identify Intermolecular forcesAnswer:1. BaCl2 = none; ionic bonding2. H2 = London Dispersion; not dipole—dipole because nonpolar3. CO = dipole—dipole because it is polar and London Dispersion forces4. HF = hydrogen bond; London Dispersion forces; dipole—dipole 5. Ne2 = London Dispersion forces onlyType 4: Ion—Dipole forces- Important attractive forces particularly in solutions that contain ions (charged particles) as solute- Another kind of attractive force is important particularly in solutions that contain ions (charged particles) as solute- It exists between an ion and the partial charge on the end of a polar molecule*Water molecule: as a whole molecule, the negative center and the positive center separate: total dipole moment*Comparing Intermolecular ForcesAlthough you need to consider many factors such as molecular weight and molecular shape, in general the order of strengths of intermolecular forces is:Dispersion Forces < Dipole—Dipole Forces < Hydrogen Bonding < Ion—Dipole Forces*These are all intermolecular forces**There is an assumption when using the Intermolecular Comparison rule: Molecules that are being compared have similar size and shape.*Intermolecular