McCord Thermo Test 2008

McCord Thermo Test 2008 - 4\1 Version 229 — Exam 4 v This...

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Unformatted text preview: \ 4; \1 Version 229 — Exam 4 v This print-out should have 26 questions. Multiple-choice qumtions may continue on the next column or page — find all choices before answering. 1 cal = 4.184 J 1 L atm = 101.325 .1 R = 8.314 J/mol K our 10.0 points At the normal boiling point of water, AHnP = 40 kJ/mole. What is the entropy change for 1120(2) —r 1120(g) 7 1. 40 kJ/mol -K 2. —40 lrJ/mol ~K 3. 400 J/mol -K 4. -107 J/mol ‘ K 5.107 kJ/mol -l( 6. 107 J/mol -1( correct Explanation- ‘ 002 ‘10. rats " _ The standard molar enthalpy of formation of hydrazine, Ngimtj', is +5053 ch/mol. What is the standard molar internal energy of for— mation of hydrazine? 1. +5311 kJ/rnol 2. $43.19 kJ/mol ,. 3; +5559 kJ/mol 4“. +4815 kJ/mol 5. +£8.07 kJ/mol correct . / ./ s. +671T55'ri7inoi 7. +5063 kJ/mol Version 229 — Exam 4 ~ The specific heat of water is 4.184 .l/g-“C, so we have to multiply by grams and temper- ature change in order to obtain Joules: AH of water = (SHmenm) (AT) : (4.134 J/g -“ c) (238 g) x (155°C) = 1543 J The heat capacity of the calorimeter is 92.3 .1/“C. This is not per gram, so we just have to multiply by the temperature change to get Joules: AH of calorimeter = (SH)(AT) = (92.3 J/°C)(l.55°C) = 143 .1 AH of Dm : AH of water + AH of calorimeter =1543J+143J=1686J This is the amount of heat evolved. To get the amount per gram of fuel burned, we divide AH of the reaction by the amount of fuel burned (3.4 grams): 1685 J ——:' .1 3.4g "195“ 007 10.0 points Calculate the standard reaction enthalpy for the reaction N2H4(€) + P1205) —' 2 NHule‘) given NzHAU) + Ozlgl —’ N2lg) + lezolg) AH“ = 7543 kJ - moi“ 2H2(g) + 02(3) p 211200;) AH” = —434 k.) - moi“ Nzlg) +3Ii2(g) - ZNHAg) AH” = 79221:.1 mol“ 1. V243 k.) moi“ 2. 4335 k] ~ mol" McCord — (53745) 1 s. +45.s7 kJ/mol Explanation: AH = 50.63 kJ/mol T = 298.15 K (standard temperature) N2(g) + 2 H2(g) a N2H4(1) n;=(1+2)mol=3mol n; = 0 mol An. = (0 — 3) mol = —3 mol AE=q+w, q=AHand —PAV = —AnRT — -(—3 mol)(8.314 J/mol A K) 1:1 x(298.15 K) (m) = 7.43646 14.] Work is per 1 mole OlN2H4 formed, so AE = AH +111 = 50.63 kJ/mol + 7.43646 kJ/mol = 58.0665 kJ/mol 003 10.0 points Given the standard heats of reaction Reaction AH 0 Ms) + 2 X204) —> Mug) —73.7 moi X2(s) —* 2 X(g) +2823 kJ/mol We) -’ M(5) —15.1 kJ/mol calculate the average bond energy for a single M — X bond. /a 191.85 x 203.85 ‘r 200.85 175.85 184.35 163.35 178.85 206.35 . 194.35 10. 173.85 wwflpwewwr Correct answer: 163.35 kJ/mol. Ebcplanation: McCord — (53745) 3 3. ~59 k1 marl 4. —1119 kJ ‘ moi-1 5. —151 kJ - 11ml“l correct Explanation: We need to reverse the second reaction and add them; N21{4(l)+-92(g) —> w +enyeeg+ AH = —543 kJ/rnol wet ~ 2 Hug) + 92%} AH = +434 kJ/mol w + 3H2(g) —' 2 NHalg) AH = —92.2 kl/mol NZHilE) + 3 Halt) -’ 2 H2(g) + 2 Mine) Nam“) + Hz(g) —’ 2 NH3(E) AH = —151.2 kJ/mol 008 10.0 points For the reaction 2 (3(3) + 2 H200 a Czfldg) on: = +523 kimol“ and AS,“ = —53.07 J -K“‘ -mol‘1 at 298 K. This reac- tion will be spontaneous at 1. no temperature. correct 2. temperatura below 1015 K. 3. temperatures below 985 K. 4. temperatures above 985 K. 5. all temperatures. Explanation: AG : AH e TAS is used to predict sponv taneiry. (AG is negative for a spontaneous reaction.) T is always positive. We thus have AG = (—1—) » T<~), For AG to be neg— ative. T would have to be negative. which is not physically possible, so the reaction is nouspontaneous at any temperature. Version 229 - Exam 4 — McCord - (53745) 2 The correct reaction will he i [mo —» Me) Mme] so you must flip the first and third reactions and double the second reaction. AH for this reaction is —135.28 kcah 11811.7 kcal is released, how many molx of CD must have reacted? a: '1. 5.41 mol l 2. 16.3 mol 3. 40.5 mol 4. 6.00 mol 5. 270.5 moi 3: 6. 1.35 mol ‘7. 12.01130] correct Explanation: , Ali‘: —l35.28 kczll 77cc 12.0 mol (—135.28 heal) (1 mol rxn ,7; 1 moern 2 11101 CO "*-«~_x(12mol C0) = —811.68 kcal so 811.68 kcal were released. W1 3 005 j10.0 points If you have endothermic proc in which the diange in fifiopy is positive, how can you y’make it spontaneous? l 1. Reducing the entropy change 2. Decreasing the volume 3. Increasing the temperature correct 4. Increasing the pressure Version 229 — Exam 4 — 009 10.0 points Calculate the average H —S bond enthalpy in HgS(g) given the standard enthalpies of for mation for H»;S(g), H(g). and 5(3) as —20.1, 218, and 223 kJ - Incl—x, respectively. 1. 340 kJ ~ inol‘1 correct 2. 679 lol Imol 1 10.1 k1 -mor‘ 4. 461 kJ - moi“ 5. 231 kJ ‘ moi“ Explanation: Write an equation which represents the bond energy (or a. multiple of it)- H2$(g) H 2Hng + 3(a) AH. = 2 (31311-3) Reactants: . AH; 325m) = —20.15 kJ/mnl Products: AH, ME, = 218 kJ/mol AHPS z 223 kJ/mol 1mg“ = Z n AHSFrad — Z n AHPm = [2 (218 kJ/rnol) + 223 kJ/mol] 7 (-201 kJ/mol) z 579.] kJ/mol rxn, so Hgm = 2 x BE 679.1 M . BE = ~—=“°'m‘ : 339.55 1‘7 . 2 mo] rxn 010 10.0 points Which of the following is not a definition of internal energy or change in internal energy? 1. The total energy content of a system. 2. A measure of the change in heat of a system at constant volume. 3. A measure 01. the spontaneity of a reac— tion. correct 5. Decreasing the temperature Explanation: AG AH—TAS AH > 0 for endothermic processes. AG < 0 for spontaneous processes. T is always positive, so AG AH—TAS =(+)—TAS AG is negative if T is very large, so in- creasing the temperature malces the process endothermic. 006 10.0 points 3.4 g of a hydrocarbon Fuel is burned in a calorimeter that contains 238 grams of water initially at 25.00°C. After the combustion, the temperature is 26.55°C. How much heat is evolved per gram of fuel burned? The heat capacity of the calorimeter (hardware only) is 92.3 J/°C. 1. 1686 J/g 2. 7363 J/g 3. 25037 J/g 4. 143 .1 /g 5. 1543 J/g 6. 495 .l/g correct 7. 453 J/g s. 7818 J/g Explanation: mmd = 3.4 g mm“ = 238 g AT = 26.55°C — 26.00°C = 155°C The amount of heat evolved by the reaction is equal to the amount of heat gained by the water plus the amount of heat gained by the calorimeter. McCord ' (53745) 4 4. The sum of heat (q) and work (w). Explanation: Gibbs free energy is a measure of the spon- taneity of a reaction. All of the other state- ments are mathematical identities describing internal energy or prose restatements of the same. .1 '\ 011 '_ 10.0 points Phosphine (thé'hommon name for P113, :1 highly mic gas used for fumigation), has a. AHJM = 14.6kJ~m01 1 and a 53",, = 78.83J-mol“ -l(“l. What is the normal boiling point of phosphine expressed in centi— grade? 1. —0.2 DC 2. 273 QC 3. —87.8 “C correct 4. 185.2 “C Explanation: Because boiling is an equilibrium process, A05",| n AHfiu, — TA55u1,. And so TASS“ = AHfiup and T = AHSBI, A53”, = 14,600.1- marl/78.33 J .mor‘ -K“ :2 185.2 K = 437.8 “C 012 10.0 points Calculate AG for the process He(g. 1 atm, 298 K) a He(g, 10 Mm. 298 K) 1. 2.10 kJ/rnol 2. 4.80 kJ/mOl 3. 3.501(rl/rnol 41. ~2.1O kl/mol 5. —19.1k.l/mol l? :7 Version 229 — Exam 4 — McCord — (53745) 5 6. 5.70 k.l/ruol correct 7. seem/mar .. ~ " / s. —4.80 k'J/mol kylanation: The temperature is the same before and afler the change so the change is isothermal (AH=0)a.nd ' AG=0—TAS=—TAS. Foii an isothermal change AS = 1112 ln and All of the answers are intensive so 11 drops out of the equation: AG = —(8.314 .l/mol ~K) (293 K) ln = «5704.82 .l/mol = —5.70482 kJ/mol 013 10.0 points The definition of internal energy is 1 AU = q + w . Which of these three values are state func- tions? 1. q and w only 2. q and 111, but only at constant volume 3. AU , but only at constant volume 4. AU, q, and w 5. AU only correct Explanation: 014 10.0 points Which of the statements below concerning thermodynamic sign convention is NOT true: 1. Work is done on the system when AV is negative. 2. AS is positive whai there is increasing disorder. 3. w is positive When work is done by the i, system. correct , 4. AH is negative when heat is released to 5 the surroundings. 5. AG is negative when a reaction is spon- tmieous. Explanation: w is positive only when work is done ON the system, not BY the system. When the system does work, volume increasa or the number of moles increasm (AV > 0, An > O). u: = —P AV = ~‘Anl-ZT and will therefore be negative when AV or A11 is positive. #— «,- 015 2.10.0 points The temperature of a. container of BF; (g) goa from 29W 373 K at a constant pra— surc of 100 atm when 6.24 k] is added as heat. How many moles of BF3 does the container hold? You should treat BF3 ideally. 1. 2.86 mol 2. 1.50 mol 3. 3.33 mol 4. 6.67 mol 5. 2.50 mol correct 6. 4.00 mol Explanation: Cp for a non~linear ideal gas is 4R. The temperature increase is 75 K. q=nCpAT Version 229 — Exam 4 — McCord — (53745) 7 5. +2339 .l/K correct Explanation: T1 =10"C = 283 K T2 = 150 C = 423 K n=2mol R=8.314m For a linear ideal gas at constant pressure CW" = g R. so T2 AS = nCmm ln more =,23.3912 J/K 020 10.0 points Calculate the change in entropy when 2.00 moles of ‘metbane are compremed isother— mally to one fifth ofiLs original volume. Treat methane as an ideal gas. 1. +131 J-K“ 2. — 13.4 .l-K“ 3. —1,39 J-K“ 4. +208 J-K" 5. — 26.8 .I~l(_1 correct Explanation: If compressed to ou‘e/ fifth the original vol— . 1 V 1 urne this means V2 2 m- i . X. 115:7”? ln = (2.00 mol)(8.3ltl.l~mol‘1‘K‘ll = —25.m17.1 x“ lVe expect a negative :msn'er since the .volume decreased. 10.0 points f the following statements is the best fiatemétimfithe Second Law of Thermo- dynamics? 1. For any spontaneous process. the change in entropy of the universe must be negative. 2. The absolute entropy of a. perfect crystal at 0 K would be 0. 3. For any spontaneous process, the total entropy of the universe must increase. cor- reel: 4. For any reversible procms, the entropy change of the universe must be positive. 5. For any spontaneous process, the entropy of the system must increase. Explanation; The Second Law of Thermo dynamics states that in spontaneous changes, the universe tends toward a state of greater disorder. 022 10.0 points For the methanol combustion reaction 2 ongonuiisfiug) e 2 C02(gl + 4 H»;O(g) estimate the amount of PAV work done and tell whether the work was done on or by the system. ASSILTDE a temperature of 27°C. 1. 7.5 kl, work done by the system correct 2. 2.5 k1. work done on the system 3. 2.5 in]. work done by the system 4. 7.5 kl. work done on the system 5. No work is done in this reaction. Explanation: T:27~C+273= 300K Considering only moles ofg'us, i1.=n._n,:(2+1)‘3= “\l Version 229 — Exam 4 ~ 6240 = n(4 - s.314)(75) n = 2.5018 mol 016 10.0 points 3.5 g of a hydrocarbon fuel is burned in n. ves— sel that contains 250. grams of water initially at 25.00 C. After the combustion, the temper« nture of the water is 26.55 G. How much heat is evolved per gram of fuel burned? 1. 26555 J/g 2.1764 .l/g 3. 1105 J/g 4. 143 .l/g 5. 1263 J/g 6. 511 .l/g 7. 73.5 J/g 8. 463 1/; correct Explanation: (1 m Cp AT (1 = 250 (4.184) 1.55 = 1621.3 .1 pergram will he 1621.3 .l / 3.5 g = 463 J/g 017 10.0 points The internal energy of a. substance can be Jfirought to zero by cooling the substance all j! the way down to absolute zero ( T = 0 K). i 1. False correct 2. True Explanation: The very [act that you have any matter at all means you have some internal energy - even at absolute zero. 018 10.0 points When a. sugar cube dissolves in a. cup of notice (an endothermic process), entropy changes of the sugar plus water. the surroundings. and Version 229 ‘ Exam 4 — «— AnRT — (3 mol) (8.314.1/mol -K) (300 K) = —7500J = —7.5kJ The system expands because An is positive. so the system does the work on the surround— ings. Also, when w is negative, work is done by the system ,. . 023 For which reactions 1) 020:) +‘Hz<g) —> H2020) n) C(s. diamond) + 02(g) —» C02(g) m) 112(4) + 3 F2(g) _. 2 NF3(L7) 3 IV) C(s. graphite) + 5 Mg) +H2(gl —» 001(g)+ H20(g] V) 2 Fe(s) + g 02(g) —. F9203(s) would AH," = AHA-“‘2 1. 10.0 points 1. IV and V only 2. [and 11 only 3.1.11] and V only 4. I. III. IV and V only 5. ll, Ill and TV only 5. H and 1V only 7.1.11. Ill IV and V 8. land V only correct Explanation: 024 10.0 poinm How is ASHngv related to AG5\.5'? 1. AG”. T A5...» , AS -. 2. no... : 3. [AGsys = —AS..,.2\- McCord — (53745) 6 the universe are, respectively, 1. None of these is correct. 2. positive, negative. positive. correct 3. negative, positive, positive. 4. positive, positive, positive. 5. negative. negative, negative. Explanation: Let’s consider each of the entropy changes individually. A55”: The entropy of any substance in— creases as the substance goes from solid to liquid to gas. Solutions are inherently more disordered than pure substances. so in our system (the process inside the coffee cup) eu- tropy increasa when we dissolve the sugar cube in water; i.e., ASS},s is positive. A591": Here we know that the disolution procem is endothermic. The system draws heat from the surroundings7 causing molec- ular motion immediately around the coli‘ee cup to slow down. in turn, as molecula slow down, less random motion occurs, and entropy is decreased, causing AS...“ to be negative. Asmfivz The process of dissolving sugar in Wat-er is spoutiuieuus, so ASWv must be pos- itive by the Second Law of Thermodyainics. (In spontaneous changes, the universe tends toward a. state of greater disorder.) 019 10.0 points The temperature of 2.00 mol 002(5) is in- creased firom 10°C to 150”C at constant pres- sure. Calculate the change in the entropy of carbon dioxide. Assume ideal behavior. 1. -1170 J/K 2. +1711 J/K 3. +1170 J /K 4. —23.39 J/K McCord — (53745) 8 5. AG... = A5... 6. AGsys = ‘TASuniv correct Explanation: 025 10.0 points The entropy of fusion of asubstance is always smaller than its entropy of vaporization. 1. False 2. True correct Explanation: The entropy effusion is the entropy change upon melting a solid into a liquid. The en- tropy change of vaporization is the entropy change when a liquid becomes a gas. The en- tropy of a. gas is significantly greater than that of a liquid, which is a little more than that of a solid. This is because in a gas the particles are able to move around much more. 026 10.0 points A friend says that as a plant grows and takes an unordered set of molecules and puts them in a highly organized system the second law of thermodynamics is violated. Your best response is that 1. while the entropy involved in the plant growth decreased, the entropy in the universe increased as a result of this process. correct 2. the plant is not at equilibrium and hence the argument is invalid. 3. you must be thinking of the third law. 4. the second law applies only to chemical systems and not biological systems. Explanation: The plant became more organized. but the universe became less organized. Chemical I’euvinns Within who plan: are ezrprgnnir (—AG) and drive the plant growth forward. ...
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McCord Thermo Test 2008 - 4\1 Version 229 — Exam 4 v This...

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