Review for Exam 2

Review for Exam 2* This sheet is meant as a tool for studying; do not assume that if something is not on this sheet that you do not need to know it for the exam.Chapter 10 – Liquids, Solids and Phase ChangesQuiz 3 Problems ( 10-24-05)1. Drawing molecules in 3-dimensions and determination of the net dipole moment.  Remember to include lone pairs that are relevant to the determination of the dipole moment (ie; lone pairs in water or ammonia).  You do not need to know  = Q x r2. Intermolecular forces: a. ion – dipole (between an ion and a polar solvent) b. dipole-dipole (between polar molecular) – Does the molecule have a net dipole moment? y t i v i t a g e n o r t c e Electronegativity l E c. london – dispersion (induction of a temporary dipole)  all molecules have this force, but the strength depends on the size and polarizability of the molecule. So an increase in molar mass, will contribute larger dispersion forces. d. Hydrogen bonds (between a hydrogen bonded to a very electronegative atom (O, N, F) and an unshared electron on another electronegative atom. i. Intermolecular – between two different molecules ii. Intramolecular – within the same moleculeIn general : ion-dipole > hydrogen bond > dipole-dipole > london dispersion 3. Phase ChangesHfusion water = +6.01 kJ/molHvaporization = +40.67 kJ/mol*notice the signs of the enthalpy and entropy terms of each process in the diagram.Heating curve for waterWhy is the Hvaporization usually larger than Hfusion (+40.67 versus +6.01 kJ/mol for water)? 4. Phase DiagramPhase Diagram of WaterCan you navigate around a phase diagram?What are the melting and boiling points? * assume normal melting and boiling points unless otherwise stated. 5. Vapor Pressure (Pvap) – partial pressure of a gas in equilibrium with liquid (ease with which molecules are able to escape to the vapor phase) Vapor pressure depends on the magnitude of the intermolecular forces and the temperature.  The smaller the intermolecular forces, the higher the vapor pressure because loosely held molecules escape more easily.  The higher the temperature (molecules are moving faster), the higher the vapor pressure because a larger amount of molecules will have enough kinetic energy to escape to the vapor phase. Vapor pressure (mm Hg) versus Temperature (K) is not a linear relationship. ) g H m -  Hvap m (R e p r a u v s P s e n r l p r o p a V Temperature (K) 1/T (K) ln Pvap versus 1/T is a linear relationship with a negative slope (- Hvap/R).H 1  Clausius - Clapeyron Equation: ln Pvap      C  R T  y = m x + b Remember that R needs to be in J/molK not Latm/molK Since C is a constant (same at two pressures and temperatures for the same molecule) then they can be set equal to one another.ln P  ln P R 2 1  H vap 1 1   ln P  ln P      H    2 1    vap 1 1  R T1 T2        T T    1 2   6. Solids Be able to calculate the radius of an atom  Be able to calculate density * Crystalline solids are those whose atoms, ions or molecules have an ordered arrangement. There are four types.  A crystal is made up many small repeat units called units cells stacked together. 14 different unit-cell geometries occur in crystalline solids-we looked at only those unit cell that are cubic. primitive-cubic body-centered cubic face-centered cubic Number of atoms per unit cellPrimitive (simple cubic):8 corner atoms x 1/8 = 1Total: 1atomBody-centered cubic:8 corner atoms x 1/8 = 1 1 center atomTotal: 2 atomsFace-centered cubic:8 corner atoms x 1/8 = 1 6 face atoms x 1/2 =3Total: 4 atoms Chapter 11 – Solutions and Their PropertiesQuiz 4 Quiz 5 Problems (10-31-05) Problems (11-7-05)Solute – dissolved substance Solvent – pure liquid1. Will a solution form?  “LIKE DISSOLVES LIKE” a solution will form when the three types of interactions are similar in kind and in magnitude. (solute-solute, solvent-solvent, and solute-solvent) Dissolving a solute (or liquid) into a solvent has a free energy change, G.- G (spontaneous)  substance dissolves +G (nonspontaneous)  substance does not dissolveGsoln = Hsoln - TSsoln Entropy is usually positive for a solution (more ordered to less ordered (more random) Enthalpy of the solution (Hsoln) is equal to the sum of the enthalpies from the solute-solute, solvent-solvent, and solute-solvent interactions (see Figure 11.4)2. Units of Concentration* make sure to keep track of units!(a) molarity (M) = moles of solute Liters of solution(b) mole fraction (X) = moles of component total moles making up solution(c) molality (m) = moles of solute kg of solvent(d) mass percent (%) = mass of component total mass of solution 3. Colligative Properties Colligative properties depend on the amount of dissolved solute, but not on its identity(a) vapor pressure lowering of a solution-Solutions with nonvolatile solutesRaoult’s law: Psoln = PsolvXsolvVan’t Hoff factor (i) = moles of particles in solution moles of solute dissolved-Solutions with a volatile solutePtotal = PA + PB = (PAXA) + (PBXB)(b) boiling point elevation of a solution (c) melting point depression of a solutionT = Kbmi Kb (water) = 0.51Ckg/molT = Kfmi Kf (water) = 1.86Ckg/mol-Be able to explain boiling point elevation and melting point depression with enthalpy and entropy (d) osmosis and osmotic pressureWhen a solution and a pure solvent (or two solutions of different concentration) are separated by the right kind of semipermeable membrane, solvent molecules pass through the membrane in a process called osmosis. Ions and large solute molecules can not pass through the membrane (see figure 11.15).Osmotic pressure () = iMRT R needs to be in Latm/molK Determination of molecular mass using osmotic pressure4. Uses of colligative properties -reverse osmosis -fractional distillation

Review for Exam 2

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