Create assignment, 54705, Quiz 6, Nov 27 at 9:12 am 1This print-out should have 8 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering. The due time is Centraltime.Msci 15 010819:03, general, multiple choice, > 1 min, .001If a system absorbs heat and also does workon its surroundings, its energy1. must increase.2. must decrease.3. must not change.4. may either increase or decrease, depend-ing on the relative amounts of heat absorbedand work done. correctExplanation:∆E = q + wq > 0 because heat is absorbed and w < 0 be-cause the system does work on its surround-ings. Therefore ∆E = (+) + (−). ∆E canbe positive only if q > w, and negative only ifw > q.ChemPrin3e T06 1419:03, general, multiple choice, < 1 min, .002A system had 150 kJ of work done on it andits internal energy increased by 60 kJ. Howmuch energy did the system gain or lose asheat?1. The system lost 90 kJ of energy as heat.correct2. The system lost 210 kJ of energy as heat.3. The system gained 60 kJ of energy asheat.4. The system gained 90 kJ of energy asheat.5. The system gained 210 kJ of energy asheat.Explanation:DAL Thermo Instability20:06, general, multiple choice, > 1 min, .003Consider the following compounds andtheir thermodynamic data:Compound ∆H◦fS◦∆G◦f¡kJmol¢ ¡Jmol·K¢ ¡kJmol¢CH4−75 186 −50CH2O −108 218 −102C6H5NH287 166 319C2H452 68 219Using this data, which of the following an-swers includes the compounds that are ther-modynamically unstable?1. CH4, CH2O, C2H42. CH2O, C6H5NH23. CH4, C2H44. C6H5NH2, C2H4correct5. Cannot be determined from the data pro-vided.6. All of the compounds are thermodynami-cally stable.Explanation:ChemPrin3e T07 1520:04, general, multiple choice, < 1 min, .004The enthalpy of fusion of H2O(s) at its normalmelting point is 6.01 kJ · mol−1. What is theentropy change for freezing 1 mole of water atthis temperature?1. +20.2 J · K−1· mol−12. 0 J · K−1· mol−13. − 20.2 J · K−1· mol−1Create assignment, 54705, Quiz 6, Nov 27 at 9:12 am 24. +22.0 J · K−1· mol−15. − 22.0 J · K−1· mol−1correctExplanation:ChemPrin3e T07 04a20:04, general, multiple choice, < 1 min, .005The temperature of 2.00 mol Ne(g) is in-creased from 25◦C to 200◦C at constant pres-sure. Assume the heat capacity of Ne is 20.8J/K-mol. Calculate the change in the entropyof neon. Assume ideal behavior.1. +7.68 J·K−12. +19.2 J·K−1correct3. − 7.68 J·K−14. − 19.2 J·K−15. +9.60 J·K−1Explanation:Msci 15 150920:06, general, multiple choice, > 1 min, .006If you have an endothermic process in whichthe change in entropy is positive, how can youmake it spontaneous?1. Increasing the pressure2. Decreasing the volume3. Increasing the temperature correct4. Decreasing the temperature5. Reducing the entropy changeExplanation:∆G = ∆H − T ∆S∆H > 0 for endothermic processes.∆G < 0 for spontaneous processes.T is always positive, so∆G = ∆H − T ∆S= (+) − T ∆S∆G is negative if T is very large, so in-creasing the temperature makes the processendothermic.Msci 15 141220:05, general, multiple choice, > 1 min, .007Calculate ∆G at 298 K for the reaction2 Ag2O(s) → 4 Ag(s) + O2(g) .Species ∆H0fS0kJ/mol J/mol ·KAg(s) 0.0 42.55Ag2O(s) −30.57 121.7O2(g) 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 rct=h4(42.55 J/mol · K)+ (205.0 J/mol · K)iCreate assignment, 54705, Quiz 6, Nov 27 at 9:12 am 3− 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)= 21.8636 kJ/mol rxnChemPrin3e T07 4220:04, general, multiple choice, < 1 min, .008The entropy of fusion of water is +22.0J·K−1·mol−1and the enthalpy of fusion of wa-ter is +6.01 kJ·mol−1at 0◦C. What is ∆Stotalfor the melting of ice at 0◦C?1. −6010 J ·K−1·mol−12. 0 correct3. −22.0 J· K−1·mol−14. +6010 J· K−1·mol−15. +22.0 J·
View Full Document
Unlocking...