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CHEM 1112, Week 1 and 2

by: Hayley Seal

CHEM 1112, Week 1 and 2 CHEM 1112

Hayley Seal
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Class notes for January 12-21.
General Chemistry 2
Martin Zysmilich
Class Notes




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This 6 page Class Notes was uploaded by Hayley Seal on Friday January 22, 2016. The Class Notes belongs to CHEM 1112 at George Washington University taught by Martin Zysmilich in Spring 2016. Since its upload, it has received 82 views. For similar materials see General Chemistry 2 in Chemistry at George Washington University.


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Date Created: 01/22/16
CHEM 1112 Dr. Martín Zysmilich Class Notes for January 12-21 (including answers to sample problems) Chapter 12 (January 12-14)  Relationships between kinetic energy of molecules and intermolecular interactions in different states of matter: o Gas: KE >> IMI o Liquid: KE ≈IMI o Solid: KE <<IMI  Intermolecular forces (van der Waals forces) o London dispersion forces are the weakest  Polarization = fluctuations in electron density distribution in an atom or molecule  Linear molecules have more points of interaction with other molecules (more neighbors) than spherical molecules which causes more of an effect on other molecules  linear molecules create stronger LD forces than spherical molecules o Dipole-dipole forces exist in polar covalent bonds  Molecules have permanent dipole moments (bond dipoles)  Larger dipole moment (larger electronegativity difference and smaller distance between the atoms) causes stronger dipole-dipole forces o Hydrogen bonds (H-N, H-O, H-F)  Strong dipole-dipole bonds between molecules  Isomers with hydrogen bonds have much higher boiling points than isomers without them  Strength of H-N < H-O < H-F o Ion-dipole interactions are the strongest o Stronger IMF increase surface tension and viscosity  Vapor pressure of liquids o Rate of evaporation (proportional to surface of liquid) remains constant o Rate of condensation increases as # of molecules in gas phase increases until dynamic equilibrium is reached o Vapor pressure of liquid = when rate of evaporation equals rate of condensation  Dependent on temperature only o IMF and kinetic energy of molecules are in the same order of magnitude, so increasing temperature allows more molecules to escape to the gas phase  As temperature increases, molecules gain kinetic energy and more of them escape into the gas phase  vapor pressure increases o Boiling point = when vapor pressure equals atmospheric pressure and all molecules have enough kinetic energy to escape into the gas phase  Normal boiling point = when atmospheric pressure is 760 torr = 1 atm o Clausius-Clapyron equation ????????????????????  ???????????? = − ???????? + ????  R = 0.082 L∙atm/K∙mol = 8.314 J/K∙mol  Inverse relationship between vapor pressure and ΔH vap(lower vapor pressure when IMF are stronger, so more energy required to vaporize liquid)  Direct relationship between temperature and pressure (both increase together)  Energy Changes Associated with Changes of State o Enthalpy of fusion (ΔH fusf a substance is much lower than its enthalpy of vaporization (ΔHvap  In fusion, IMF are still strong and still hold molecules together; they are just “interrupted”  In vaporization, IMF must be completely broken o Heat required to change temperature of a substance without phase change:  ???? = ????????∆???? (calorimetry throwback)  SAMPLE PROBLEM: The normal melting point of water is 0°C. Its normal boiling point is 100°C. The heat capacities of water solid, liquid, and gas are 38 J/K∙mol, 75 J/K∙mol, and 33 J/K∙mol, respectively. The enthalpy of fusion of water is 6.0 kJ/mol, and its enthalpy of vaporization is 41 kJ/mol. a) What is the difference between molar heat capacity and specific heat? Molar heat capacity is heat required to raise 1 mole of substance by 1°C while specific heat is the heat required to raise 1 gram of substance by 1°C. b) Calculate the enthalpy change upon converting 1.00 mole of ice at -25°C to steam at 125°C under a constant pressure of 1.00 atm. Tm= 0°C T b 100°C csolid 38 J/K∙mol cliquid 75 J/K∙mol cgas= 33 J/K∙mol 3 4 ΔH fus= 6.0 kJ/mol = 6.0 x 10 J/mol ΔH vap= 41 kJ/mol = 4.1 x 10 J/mol m = 1.00 mol ΔT solid0°C – (-25°C) = 25°C = 25 K ΔT liquid100°C – 0°C = 100°C = 100 K ΔT gas 125°C – 100°C = 25°C = 25 K Because the question and the values are in terms of moles, we can substitute moles for mass in this equation. total enthalpy change = (heat required to change temperature of solid to melting point) + (heat of fusion) + (heat required to change temperature of water to boiling point) + (heat of vaporization) + (heat required to change temperature of gas to 125°C) ΔH total mc solidsolid mΔH +fus liquidliquid m ΔH vap+ mc gasΔT gas 3 = (1.00 mol)(38 J/K∙mol)(25 K) + (1.00 mol)(6.0 x 10 J/mol) + (1.00 mol)(75 J/K∙mol)(100 K) + (1.00 mol)(4.1 x 10 J/mol) + (1.00 mol)(33 J/K∙mol)(25 K) 4 = 5.6 x 10 J = 5.6 kJ Chapter 13 (January 19)  Solution = homogenous mixture of 2 or more substances o Solvent is usually (but not always) the substance in the largest amount; it determines the physical properties of the solution o Ex. In a solution of a liquid and a gas, if the solution itself is a liquid, the liquid is the solvent and the gas is the solute o Solubility = concentration of a saturated solution at a given temperature  Solvation = when molecules of solute are separated from each other and surrounded by solvent molecules o Requires 1) disruption of IMF in solute, 2) disruption of IMF in solvent, and 3) interaction of molecules in solute and solvent o Induced dipole on nonpolar molecules or atoms depends on their polarizability o Ex. Ionic compound in water (NaCl in H2O)  Dipole in H2O interacts with ions in NaCl  water molecules surround ions and separate them from each other (ions become hydrated)  Partial negative charge on oxygen atom attracts positive charge of sodium ion  Partial positive charge on hydrogen atoms attract negative charge of chlorine ion  Requires 1) breaking of ion-ion bonds in NaCl, 2) breaking of hydrogen bonds in water, 3) interaction of molecules in NaCl and water together  Energy involved in solvation: o ΔH 1 separation of molecules in solute; endothermic o ΔH 2 separation of molecules in solvent; endothermic o ΔH 3 solvation of solute molecules (surrounded by solvent); exothermic o ΔH solutionΔH1+ ΔH 2 ΔH 3 o Whether solution is exothermic or endothermic depends on strength of (ΔH 1 ΔH )2 compared to (ΔH )3 if the interactions within the solute and solvent are stronger than the interaction between the molecules of the solute and solvent, the reaction may be endothermic (and vice versa)  2 law of thermodynamics: the universe moves in order of increasing entropy (chaos); there is a natural tendency for energy to disperse o Entropy = measure of how dispersed or spread out energy is in a system  Like dissolves like: substances with similar type and magnitude of IMF are soluble in each other o Ethanol forms hydrogen bonds, therefore it is soluble in any solvent that can also form hydrogen bonds (like water) o Oil is non-polar (only London dispersion forces present); does not dissolve in water, which forms much stronger H-bonds o Vitamin A – mostly nonpolar carbon chains, soluble in fats (not so much in water) o Vitamin C – smaller and more H-bonds, therefore it is more soluble in water  Concentration = amount of solute relative to amount of solvent or volume mol o Molarity (M) = L ; common and easy to make but slightly dependent on temp. o Mole fraction (χ ) = # ???????? ???????????????????? ???????? ???? ; equal to partial pressure fraction and A ???????????????????? # ???????? ???????????????????? ???????? ???????????????????????????????? useful for gases o Molality (m) = # ???????? ???????????????????? ???????? ????????????????????; easy to make and independent of temperature 1 ???????? ???????? ???????????????????????????????? ???????????????? ???????? ???????????????????????? o % by mass = ???? 100% ???????????????? ???????? ????????????????????????????????  SAMPLE PROBLEM: 7.5 g of CH 3H is dissolved in 245 g of H 2. Find the: a) mole fraction of CH 3H molar mass of CH O3 = 32 g/mol molar mass of H O2= 18 g/mol ???????????????????? ???????? ????????3???????? χCH3OH= ???????????????????? ???????? ????????3????????+???????????????????? ???????? ????2???? 32 ???????? (7.5 ???? ÷ ????????????) = 32 ???? 18 ???? = 0.017 ( 7.5 ???? ÷ ????????????))+( 245 ???? ÷(????????????)) b) mass percent of CH O3 % = ???????????????? ???????? ????????3???????? ???? 100% ???????????????? ???????? ????????3????????+???????????????? ???????? ????2???? 7.5 ???? = 7.5 ????+245 ???????? 100% = 3.0% c) molality 245 gH2O = 0.245 kg H2O moles of CH 3H = 7.5 ???? ÷ ( 32 ) = 0.234 mol ???????????? ???????????????????? ???????? ???????????????????????? m = 1 ???????? ???????? ???????????????????????????????? 0.234 ???????????? ???? = 0.245 ???????? 1 ???????? X = 0.96 m d) molarity Assume that the density of CH O3 = density of solution = density of water = 1 g/mL 1 ???? -3 7.5 gCH3OH÷ ( ????????) = 7.5 mL CH3OH= 7.5 x 10 L 1 ???? 245 gH2O ÷ ( ) = 245 mL H2O = 0.245 L ???????? moles of CH 3H = 7.5 ???? ÷ ( 32 ????) = 0.234 mol ???????????? ???????????????????? ???????? ???????????????????????? M = ???? ???????? ???????????????????????????????? 0.234 ???????????? = ( −3 ) = 0.93 M 7.5 ???? 10 ???? +(0.245 ????) Chapter 13 continued (January 21) Factors that affect solubility  Temperature: solubility generally increases with temperature o In some cases this is not true: NaCl appears temperature independent and the solubility of C2 (SO4 3decreases with increasing temperature o If the reaction is exothermic, adding heat decreases enthalpy o Solutions with positive enthalpy = endothermic = temperature increases solubility  Solute + solvent + heat  solution o Solutions with negative enthalpy = exothermic = temperature decreases solubility (heat is a product, so adding heat pushes the system back to its original state)  Solute + solvent  heat + solution  Example: when gas is a solute in a liquid solvent, increase in temperature decreases solubility because of kinetic molecular theory (gas particles move faster, less are captured by liquid)  Pressure only affects a solution when at least one of the components is not in the condensed phase/is a gas o Solubility of a gas increases as pressure increases o Henry’s Law: c = kP  c = concentration of gas (mol/L)  P = pressure of gas (atm)  k = constant  Colligative properties depend on number of solute particles only, not the nature of them  Vapor-pressure lowering: o Raoult’s Law: P 1 χ 1 1 o Vapor pressure of solvent over solution (P 1 is equal to the mole fraction of the solvent (χ1) multiplied by the vapor pressure of the pure solvent (P1°) o ΔP = P 1 χ2where χ 2 mole fraction of the solute o Entropy of a solution is higher than that of pure solvent, so there is less of a difference between the solution and a gas than the pure solvent and its gas phase  Vapor pressure = tendency of molecules to escape into gas phase  Therefore vapor pressure is lower for solution than pure solvent (entropy is already higher so there is less of a tendency for molecules to escape to a higher entropy) o For multiple volatile compounds in the solution, the total pressure of the solution is the sum of the partial pressures of the volatile substances  P totalχ1P 1 + χ2P2°  Volatile = has measurable vapor pressure o Solutions that follow Raoult’s law = ideal solutions  Lower vapor pressure of solutions accounts for boiling point elevation and freezing/melting point depression of solutions o ΔT b K b  ΔT b change in boiling point temperature (°C)  K b boiling-point elevation constant (°C/m)  m = molality o ΔT f K f  ΔT f change in freezing point temperature (°C)  K f freezing-point depression constant (°C/m)  m = molality  SAMPLE PROBLEM: How much C H 2 8us2 be dissolved in 10.0 mL of water to lower its freezing point to -25°C? The value for Kfof water is 1.86 °C/m. 1 ???? 10.0 mL x ???????? = 10.0 gH2O ΔT = 25°C f ΔT = K m  m = ∆???????? f f ???????? 25°???? = = 13.44 m 1.86℃/???? 13.44 m = 13.44 ???????????????????? ???????? ????2????8????2= ???? ???????????????????? ???????? ????2????8????2 1000 ???? ????2???? 10.0 ???? ????2???? 64 ???????????? X = 0.1344 mol ÷ ???? = 8.6 g  Osmotic pressure (π) is also a colligative property; increases for more concentrated solutions and decreases for more dilute solutions o π = MRT o Example of practical application: seawater becomes freshwater by process of reverse osmosis as the osmotic pressure of seawater is applied  SAMPLE PROBLEM: 35.0 g of hemoglobin (Hb) is dissolved in water to make 1.0 L of solution. The osmotic pressure of the solution at 25°C is 10 mmHg. What is the molar mass of Hb? 1 ???????????? 10 mmHg x 760 ???????????????? = 0.0132 atm 25°C + 273 = 298 K ???????????? πL π = MRT  π = ( ???? ) RT  mol = RT (0.0132 ????????????)(1.0 ????) mol Hb (0.082 ????∙????????????/????∙????????????)(298 ????) 0.0005395 mol 35.0 ???? = 6.5 x 10 g/mol 0.0005395 ????????????


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