Create assignment, 48975, Exam 1, Feb 19 at 3:23 pm 1This print-out should have 30 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore making your selection. The due time isCentral time.PLEASE remember to bubble in yourname, student ID number, and version num-ber on the scantron!Msci 15 0906HS15:09, general, multiple choice, < 1 min, fixed.001What would be the signs of ∆H and ∆S, forthe reactionCH4(g) → CH3(g) + H(g) ?1. Both positive. correct2. ∆H = +, ∆S = −3. ∆H = −, ∆S = +4. Both negative.5. ∆H = +, ∆S = 06. ∆H = −, ∆S = 07. ∆H = 0, ∆S = +8. ∆H = 0, ∆S = −9. Both 0.10. Impossible to tell from the informationgiven.Explanation:C HHHH−−−−→CHHH+H∆H0rxn =XB.E.rct−XB.E.prod=h4 (CH)i−h3 (C H)i= (C H)= 414 kJ/mol∆H is positive (endothermic, net bond break-ing).∆S is also positive because the random-ness of the system has increased. (1 mole ofgaseous reactants forms 2 moles of gaseousproducts.)Msci 15 040215:10, general, multiple choice, > 1 min, fixed.002A 0.10 g piece of chocolate cake is combustedwith oxygen in a bomb calorimeter. The tem-perature of 4000 g of H2O in the calorimeter israised by 0.32 K. (The specific heat of the wa-ter is 1.0 cal/g·K and the heat of vaporizationof water is 540 cal/g.)What is ∆E for the combustion of chocolatecake? Assume no heat is absorbed by thecalorimeter.1. −3900 kcal/g2. −460 kcal/g3. −12.8 kcal/g correct4. −532 kcal/g5. −13.3 kcal/gExplanation:The amount of heat responsible for the in-crease in water temperature for 4000 g ofwater isq =1 calg · K· (4000 g)(0.32 K) = 1280 calThe amount of heat released by the reactionis thus 1280 cal. There were 0.10 g of cake, so−1280 cal0.10 g·kcal1000 cal= −12.8 kcal/gMlib 05 000915:06, basic, multiple choice, > 1 min, fixed.003Calculate the enthalpy change for the reaction2 SO2(g) + O2(g) → 2 SO3(g)Create assignment, 48975, Exam 1, Feb 19 at 3:23 pm 2∆Hffor SO2(g) = −16.9 kJ/mol;∆Hffor SO3(g) = −21.9 kJ/mol.1. −5.0 kJ/mol rxn2. −10.0 kJ/mol rxn correct3. +5.0 kJ/mol rxn4. −77.6 kJ/mol rxn5. +10.0 kJ/mol rxnExplanation:∆Hffor O2(g) = 0 kJ/mol∆Hrxn=Xn ∆Hf products−Xn ∆Hf reactants= (2 mol)(−21.9 kJ/mol)−(2 mol)(−16.9 kJ/mol)−(1 mol)(0 kJ/mol)= −10.0 kJ/mol rxnMlib 75 000115:99, basic, multiple choice, > 1 min, fixed.004Based on the bond energies P−H = 321kg/mol, H−F = 568 kg/mol, P−F = 490kg/mol, and H−H = 436 kg/mol, estimate∆H for the gas-phase reactionPH3+ 3 HF → PF3+ 3H2.1. −111 kJ. correct2. +111 kJ.3. −37 kJ.4. +37 kJ.Explanation:∆H =Xn BEreactants−Xn BEproducts= (3 mol)(321 kJ/mol)+ (3 mol)(568 kJ/mol)− (3 mol)(490 kJ/mol)− (3 mol)(436 kJ/mol)= −111 kJDAL 0301 0115:10, general, multiple choice, < 1 min, fixed.005For the methanol combustion reaction2 CH3OH(l)+ 3 O2(g)→ 2 CO2(g)+ 4 H2O(g)estimate the amount of P ∆V work done andtell whether the work was done on or by thesystem. Assume a temperature of 27◦C.1. 2.5 kJ, work done on the system2. 2.5 kJ, work done by the system3. 7.5 kJ, work done on the system4. 7.5 kJ, work done by the system correct5. No work is done in this reaction.Explanation:Considering only moles of gas,∆n = nf− ni= (2 + 4) − 3 = 3 .27◦C + 273 = 300 Kw = − ∆n R T= − (3 mol) (8.314 J/mol · K) (300 K)= − 7500 J = − 7.5 kJThe system expands because ∆n is positive,so the system does the work on the surround-ings. Also, when w is negative, work is doneby the system.Msci 15 143515:15, general, multiple choice, > 1 min, fixed.006Which one of the following statements is notcorrect?1. When ∆G for a reaction is negative thereaction can occur spontaneously.Create assignment, 48975, Exam 1, Feb 19 at 3:23 pm 32. When ∆G for a reaction is positive thereaction cannot occur spontaneously.3. When ∆G for a reaction is zero the systemis at equilibrium.4. When ∆H for a reaction is negative thereaction can never occur spontaneously. cor-rect5. When ∆H for a reaction is very positivethe reaction will probably not occur sponta-neously at lower temperatures.Explanation:∆G determines sponteneity. ∆G is de-pendent on ∆H, T and ∆S by the equation∆G = ∆H − T ∆S. If ∆H is negative and∆S is positive, ∆G will be negative and theprocess will occur spontaneously.DAL 12 00115:14, general, multiple choice, < 1 min, fixed.007For the four chemical reactionsI. 3 O2(g)→ 2 O3(g)II. 2 H2O(g)→ 2 H2(g)+ O2(g)III. H2O(g)→ H2O(l)IV. 2 H2O(l)+ O2(g)→ 2 H2O2(l)which ones are likely to exhibit a positive ∆S?1. II correct2. I, III and IV3. I and II4. III and IV5. all have a positive ∆SExplanation:The Third Law of Thermodynamics statesthat the entropy of a perfect pure crystal at0 K is 0. As disorder, randomness, and de-grees of freedom increase, so does S. Entropycan increase by changing phase from solid toliquid to gas, and by increasing temperature,volume, or number of particles.In reaction I, the final state has less gas par-ticles (and thus less entropy) than the initialstate. Therefore ∆S is negative.In reaction II, the final state has more gasparticles (and thus more entropy) than theinitial state. Therefore ∆S is positive.III describes a phase change. Gases havemore degrees of freedom, randomness, anddisorder (entropy) than liquids. The finalstate is a liquid and the initial state is a gas.Therefore ∆S is negative.In reaction IV, the final state has 0 gasparticles and the initial state has 1 mole ofgas particles. Therefore ∆S is negative.Msci 15 141215:15, general, multiple choice, > 1 min, fixed.008Calculate ∆G at 298 K for the reaction2 Ag2O(s) → 4 Ag(s) + O2(g) .Species ∆H◦fS◦kJ/mol J/mol ·KAg(s) 0.0 42.55Ag2O(s) −30.57 121.7O2(s) 0.0 205.01. 21.9 kJ/mol rxn correct2. 38.2 kJ/mol rxn3. 52.7 kJ/mol rxn4. −69.85 kJ/mol rxn5. 81.2 kJ/mol rxnExplanation:∆H0rxn=Xn ∆H0f prod−Xn ∆H0f rct= 0 kJ/mol− 2(−30.57 kJ/mol)= +61.14 kJ/mol∆S0rxn=Xn ∆S0f prod−Xn ∆S0f rctCreate assignment, 48975, Exam 1, Feb 19 at 3:23 pm 4=h4(42.55 J/mol · K)+ (205.0 J/mol · K)i− 2(121.7 J/mol · K)= 131.8 J/mol · K ·kJ1000 J= 0.1318 kJ/mol · K∆G = ∆H − T ∆S= (+61.14 kJ/mol)− 298 K(0.1318 kJ/mol · K)= 21.9 kJ/mol rxnMsci 15 143715:14, general, multiple choice, > 1 min, fixed.009At the normal boiling point of water,∆Hvap= 40 kJ/mole.What is the entropy change forH2O(l) → H2O(g)?1. 107 J/mol·K correct2. 40 kJ/mol·K3. −40
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