Version PREVIEW – Exam 1 – VANDEN BOUT – (53585) 1This print-out should have 30 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.Mlib 05 3027001 10.0 po intsAt a certain elevated temperature and pres-sure, diamond and graphite are in equilib-rium. When graphite changes t o diamondunder these conditions1. the change in standard molar Gibbs freeenergy is zero.2. the change in molar Gibbs free energy isa minimum.3. the molar Gibbs free energy for diamondis zero and so is that of graphite.4. the change in molar Gibbs free energy i sa maximum.5. the change in molar Gibbs free energy i szero. correctExplanation:At equili brium, ∆G for the process is zero.Msci 14 0212002 10.0 po intsConsider a series of chloride salts (MCl2).As the charge-to-size ratio of M2+(decreases,increases) the hydration energy of M2+(de-creases, increases, does not change) in magni-tude and the crystal lattice energy of MCl2(decreases, increases, may increase or de-crease) in magnitude.1. increases; decreases; decreases2. decreases; increases; increases3. increases; does not change; may increaseor decrease4. increases; increases; decreases5. increases; increases; increases correct6. decreases; decreases; increases7. increases; decreases; increasesExplanation:As charge-to-size ratio increases, hydrationenergy and crystal latice energy a lso increase.Sparks vp 010003 10.0 po intsConsider two closed containers. ContainerX is a 2 L container that contains 0.5 L o facetone. Container Y is a 3 L container thatcontains 1.8 L of acetone. Both containers andcontents are at 28◦C. Which of the followingis true?1. You would need information about theshape of the containers to be able to answerthis question.2. The vap or pressure in container X isgreater.3. The vap or pressure in container Y isgreater.4. The vapor pressures in both containersare equal. correctExplanation:Msci 13 1304004 10.0 po intsConsider the phase dia gram for water (notto scale)Version PREVIEW – Exam 1 – VANDEN BOUT – (53585) 2SolidLiquidGasAB CD2.14.6355218atm0.01 374Temp erat ure (◦C)Pressureand for carbon dioxide (not to scale)SolidLiquidGasAB C151073atm−78 −57 31Temp erat ure (◦C)PressureWhich of the following statements is NOTtrue?1. Liquid water is more dense than ice.2. Carbon dioxide cannot exist as a liquid attemperatures below −57◦C.3. Water cannot exist as a liquid at −5◦C.correct4. Water cannot exist a s a liquid at pressuresbelow 4.6 torr.5. We could cause gaseous carbon dioxide tosolidify at −78◦C by increasing the pressureto greater than 1 atm.Explanation:Starting in the solid phase of water and in-creasing the pressure (i.e., i ncreasing density)it becomes liquid and vice versa. The liq-uid section for carbon dioxide is to the ri ghtof −57◦. By increasing pressure water canremain liquid at temperatures well below itsstandard freezing poi nt. The liquid sectionfor water is completely above 4.6 torr. Anypoint for carbon dioxide that is below −78◦and above 1 atm is in the solid section.ChemPrin3e T08 35005 10.0 po intsThe phase diagram for CO2is given bel ow.SolidLiquidVaporTemp erat ure, KPressure, atmThe triple point is at 5.1 atm and 217 K. Whathapp ens if carbon dioxide at −50◦C and 25atm is suddenly brought to 1 atm?1. The solid vaporizes. correct2. The solid remains stable.3. The liquid and solid are in equilibrium.4. The solid and vapor are in equil ibrium.5. The solid melts.Explanation:Sparks phase change calc 001006 10.0 po intsHow much energy is released when 150 g waterat 52◦C freezes and forms ice with a temper-ature of −14◦C? The specific heat of waterin the liquid state is 4.18 J/g◦C, in the solidstate is 2.09 J/g◦C, and in the gaseous stateVersion PREVIEW – Exam 1 – VANDEN BOUT – (53585) 3is 2.03 J/g◦C. The heat of fusion is 334 J/gand the heat of vaporization is 2260 J/g.1. 93 kJ2. 45 kJ3. 22 kJ4. 102 kJ5. 37 kJ6. 87 kJ correctExplanation:Msci 14 0707007 10.0 po intsThe solubility of a gas in water increases with1. increase of pressure or decrease of tem-perature. correct2. decrease of pressure or decrease of tem-perature.3. the effect of temperature and pressuredepend on the identity of the gas.4. decrease of pressure or increase of tem-perature.5. increase of pressure or increase of temper-ature.Explanation:Henry’s Law states that as the pressure ofthe gas above a sol ution surface increases, theconcentration of the gas increases. In otherwords, it becomes more soluble. Conversely,solubility decreases with pressure.Solubility of a g as also increases when tem-perature is decreased. Gas dissolvi ng inwater is exothermic, therefore according toLeChatelier’s principle, if you add more heat(increase temperature), the gas is going tobubble out (be less sol uble). Inversely, if youdecrease temperature, the g as is going to bemore soluble.Mlib 04 4055008 10.0 po intsWhich of the following alcohols would be theleast mi scible with water?1. pentanol (CH3CH2CH2CH2CH2OH)2. hexanol (CH3CH2CH2CH2CH2CH2OH)correct3. propanol (CH3CH2CH2OH)4. methanol (CH3OH)5. ethanol (CH3CH2OH)Explanation:The polar OH gro up i s miscible with wa-ter but as the nonpolar hydrocarbon chainlengthens, solubility decreases.Msci 14 0904009 10.0 po intsThe vap or pressure of pure CH2Cl2(molecu-lar weight = 85 g /mol) is 133 torr at 0◦C andthe vapor pressure o f pure CH2Br2(molecu-lar weight 174 g/mol) is 11 torr at the sametemperature. What is the total vapo r pres-sure of a solution prepared from equal massesof these two substances?1. vapor pressure = 93 torr correct2. vapor pressure = 124 torr3. vapor pressure = 72 torr4. vapor pressure = 89 torr5. vapor pressure = 144 torr6. vapor pressure = 105 torrExplanation:For CH2Cl2,P0= 133 torr MW = 85 g/molFor CH2Br2,P0= 11 torr MW = 174 g/molThis is a combination of Raoult’s Law andDalton’s Law of Partial Pressures. The an-Version PREVIEW – Exam 1 – VANDEN BOUT – (53585) 4swer does not depend on what t he masses are,as long as they are equal. Yo u can choose anymass you like, but to speed up calculations, itis convenient to choose the mass the same asone of the molecular weights given, so that thenumber of mol es for o ne of the components i sexactly ONE.So, for a rgument’s sake, choose 85 g to be themass of each of the components. That wayyou have:(85 g CH2Cl2)1 mol CH2Cl285 g CH2Cl2= 1.0 mol CH2Cl2Now calculate the moles of the other compo-nent.(85 g CH2Br2)1
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