Version PREVIEW – Exam 3 – JOHNSON – (53140) 1This print-out should have 40 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.LDE Carbon Allotropes 001001 5.0 poi ntsBased on your knowledge of carbon al-lotropes, You can say that ∆G◦fof graphite isand ∆G◦rxnof diamond → graphite is.1. zero, large2. small, negative3. large, small4. positive, positive5. zero, negative correct6. large, zeroExplanation:The standard state of carbon is graphite;consequently, graphite’s standard free energyof formation is zero. Diamond is actuallythermodynamically less stable than graphite;consequently, diamond’s reversion to graphiteis spontaneous.First Law Thermo 01002 5.0 poi ntsWhat is true about the first law of thermody-namics?1. ∆Euniv= 0 correct2. ∆Euniv> 03. ∆Euniv< 04. ∆Esys< 05. ∆Esys> 06. ∆Esys= 0Explanation:Msci 15 0505003 5.0 pointsFor a reaction in which g a ses are neither pro-duced nor consumed, ∆H is1. the same as ∆E. correct2. less than ∆E.3. unrelated to ∆E.4. greater than ∆E.Explanation:For a reaction at constant temperature andconstant pressure, ∆E = ∆H − (∆n) R T .If gases are neither produced nor consumed,∆n will be zero, so ∆E must equal ∆H.Spontaneous Endo Signs alt004 5.0 pointsWhat are the signs of ∆Ssurr, ∆Ssys, and∆Suniv, respectively, for a spontaneous en-dothermic reaction?1. −, +, + correct2. +, −, −3. +, +, +4. Cannot be predicted with certainty fromthe given informatio n5. −, +, −Explanation:Thermo Signs005 5.0 pointsWhen water condenses, what are the signs forq, w, and ∆Ssys, respectively?1. +, −, +2. −, +, +3. +, −, −Version PREVIEW – Exam 3 – JOHNSON – (53140) 24. −, +, − correct5. +, +, −6. +, +, +Explanation:LDE Definitions in Thermo 011006 5.0 poi ntsWhich of t he following state functions areextensive?I) Pressure (P)II) Temperature (T)III) Enthalpy (H)1. II, III2. None are true3. II only4. I only5. I, II6. I, III7. III only correct8. I, II, IIIExplanation:Examples of extensive state functions in-clude: mass, volume, enthalpy, entropy, inter-nal energy, free energy etc.Heat of Formation007 5.0 poi ntsWhich ofO2(g), O2(ℓ), H2(g), H2(ℓ), H2O(g), H2O(ℓ)have a heat of formation equal to zero?1. O2(g), O2(ℓ), H2(g), H2(ℓ)2. O2(g), H2(g) correct3. O2(g), H2(g), H2O(g)4. All of them, but only at absolute zero5. O2(g), O2(ℓ), H2(g), H2(ℓ), H2O(g),H2O(ℓ)Explanation:Molecules in their native state at STP havea heat of formation of zero.CIC Specific Heat 1 W008 5.0 pointsConsider the following specific heats: copper,0.384 J/g·◦C; lead, 0.159 J/g·◦C; water, 4.18J/g·◦C; glass, 0.502 J/g·◦C. Which substance,once warmed, would be more likely to main-tain its heat and keep you warm through along football game on a cold night?1. copper2. water correct3. lead4. glassExplanation:Water has the highest specific heat of thesubstances l isted, so it has the capacity toemit the largest quantity of heat with minimaltemperature loss; the emitted heat keeps youwarm. The substance continues to warm youuntil its temperature is at or below your bodytemperature.Mlib 05 0003009 5.0 points4.7 g of a hydrocarbon fuel is burned in acalorimeter that contains 26 3 grams of waterinitially at 25. 00◦C. After the combustion,the temperature is 26.55◦C. How much heatis evolved per gram of fuel burned? The heatcapacity o f the calorimeter (hardware only) is92.3 J/◦C.1. 362 J/g2. 393 J/g correctVersion PREVIEW – Exam 3 – JOHNSON – (53140) 33. 143 J/g4. 6246 J/g5. 2765 2 J/g6. 5883 J/g7. 1848 J/g8. 1705 J/gExplanation:mfuel= 4.7 g mwater= 2 63 g∆T = 26.55◦C − 26.00◦C = 1.55◦CThe amount of heat evolved by the reactionis equal to the amount of heat gained by thewater plus the amount of heat gained by thecalorimeter.The specific heat of water is 4.184 J/g·◦C,so we have to multiply by grams and temper-ature change in order to obtain Joules:∆H of water = (SH)(mwater)(∆T )= (4.184 J/g ·◦C) (263 g)× (1 .55◦C)= 1705 JThe heat capacity of the calorimeter is 92.3J/◦C. This is not per gram, so we just haveto multiply by t he temperature change to getJoules:∆H of calorimeter = (SH)(∆T )= (92.3 J/◦C)(1.55◦C)= 143 J∆H of rxn = ∆H of water+ ∆H of calorimeter= 1705 J + 143 J = 1848 JThis is the amount of heat evolved. Toget the amount per gram of fuel burned, wedivide ∆H of the reaction by the amount offuel burned (4.7 grams):1848 J4.7 g= 393 J/gLDE Bomb Calorimeter 007010 5.0 pointsIf 25.0 g of water a t 100.0◦C are mixed with15.0 g of water a t 4 0 .0◦C, what temperaturewill the 40.0 g of combined water be at oncethey reach equilibrium?1. 70.0◦C2. 62.5◦C3. 60.0◦C4. 77.5◦C correctExplanation:qcold= −qhot15 · 4.184 · (Tf− 40) = −25 · 4.184 · (Tf− 100)15 · (Tf− 40) = −25 · (Tf− 100)40 · Tf= 3100Tf= 77.5◦CChemPrin3e 06 58011 5.0 pointsCalculate the standard reaction enthalpyfor the reaction of calcite with hydrochloricacidCaCO3(s) + 2 HCl(aq) −→CaCl2(aq) + H2O(ℓ) + CO2(g)The standard enthalpies of formation are:for CaCl2(aq) : −877.1 kJ/mol;for H2O(ℓ) : −285.83 kJ/mol;for CO2(g) : −393.51 kJ/mol;for CaCO3(s) : −1206.9 kJ/mol;and for HCl(aq) : −167.16 kJ/mol.1. −215 kJ/mol2. −116 kJ/mol3. −98.8 kJ/mol4. −165 kJ/mol5. −38.2 kJ/mol6. −72.7 kJ/mol7. −15.2 kJ/mol correctVersion PREVIEW – Exam 3 – JOHNSON – (53140) 4Explanation:We use Hess’ Law:∆H◦=Xn ∆H◦j,prod−Xn ∆H◦j,reac= ∆H◦f, CaCl2(aq)+ ∆H◦f, H2O(ℓ)+ ∆H◦f, CO2(g)−h∆H◦f, CaCO3(s)+2∆H◦f, HCl(aq)i= −877.1 kJ/mol + (−285.83 kJ/mol)+ (−393.51 kJ/mol)−h−1206.9 kJ/mol+ 2 (−167.16 kJ/mol)i= −15.22 kJ/mol .ChemPrin3e 06 60012 5.0 poi ntsCalculate the reaction enthalpy for the forma-tion2 Al(s) + 3 Cl2(g) −→ 2 AlCl3(s) ,of anhydrous aluminum chlori de using thedata2 Al(s) + 6 HCl(aq ) −→2 AlCl3(aq) + 3 H2(g)∆H◦= −1049 kJHCl(g) −→ H C l(aq) ∆H◦= −74.8 kJH2(g) + Cl2(g) −→ 2 HCl(g) ∆H◦= −185 kJAlCl3(s) −→ AlCl3(aq) ∆H◦= −323 kJ1. −1826.2 kJ2. −1406.8 kJ correct3. −1450.8 5 kJ4. −1502.4 kJ5. −1883.5 kJ6. −1225.7 kJ7. −1100.3 6 kJExplanation:Using Hess’ Law:The desired reaction is obtained by addingreaction 1; 6 times reaction 2; 3 times reaction3; and 2 times the reverse of reaction 4:2 Al(s) + 6 HCl(aq) −→2 AlCl3(aq) + 3 H2(g)∆H◦= −1049 kJ6 HCl(g) −→
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