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H14- Thermo 3-solutions.pdf

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Department
Chemistry
Course Code
CH 301
Professor
Sutcliffe

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ngac (ttn793) – H14: Thermo 3 – mccord – (51600) 1 This print-out should have 22 questions. For the CO gas volume expanding 10▯ 2 Multiple-choice questions may continue on isothermally, 2 =10 V1and the next column or page – find all choices ▯ ▯ before answering. V2 ▯S = nR ln This is your LAST homework on Quest for V1 CH301. =(1 .00 mol)(8.314 J · mol1 · K1 ) ▯ln(10) 001 10.0 points ▯1 Consider the following processes. (Treat all =+38 .29 J · K gases as ideal.) We expect a positive answer since volume I) The pressure of one mole of oxygen gas increased. is allowed to double isothermally. 1 For the nitrogen gas compressed to origi- II) Carbon dioxide is allowed to expand 2 isothermally to 10 times its original vol-nal volume isothermally, 1 =2 V2and ume. ▯ ▯ V2 III) The temperatu▯e of one mole of helium ▯S = nR ln V is increased 25 Catcotpressure. 1 ▯ ▯ IV) Nitrogen gas is compressed isothermally ▯1 ▯1 1 =(1mol)(8 .314 J · mol · K )ln 2 to one half its original volume. V) A glass of water loses 100 J of energy = ▯5.76 J · K▯1 reversibly at 30 C. Which of these processes leads to an increase We expect a negative answer since volume in entropy? decreased. For the cooling glass of water, ▯ 1. V T =30 C+273 .15 = 303.15 K 2. IandII q ▯200 J ▯1 ▯S = T = 303.15 K = ▯0.6597 J · K 3. IandIV The last situation (heating the 1 mol of He) 4. II and III correct does not give enough data to calculate an answer but from the formula 5. III and V ▯ ▯ T2 Explanation: ▯S = nC p, mln ▯1 ▯1 T1 R =8 .314 J · mol · K Assume 1 mol in each case. Entropy de- n =1m lndfraoamials Cp, mapetfil an.iF creases if ▯S is negative. ▯ ▯ For the oxygen gas pressure doubling increases this means 2 >T 1 so ln T2 will isothermally, 2 =2 P 1nd T1 be positive. ▯ ▯ P1 We expect a positive answer since tempera- ▯S = nR ln P ture increased. 2 ▯ ▯ ▯1 ▯1 1 =(1mol)(8 .314 J · mol · K )ln 002 10.0 points 2 For the four chemical reactions = ▯5.76 J · K▯1 I) 3O (g) ▯ 2O (g) 2 3 II) 2H2O(g) ▯ 2H (2) + O (2) We expect a negative answer since pressure III) H2O(g) ▯ H 2(▯) increased. IV) 2H O(▯)+O (g) ▯ 2H O (▯) 2 2 2 2 ngac (ttn793) – H14: Thermo 3 – mccord – (51600) 2 which one(s) is/are likely to exhibit a positive 5. positive, negative, positivcorrect ▯S? Explanation: The process is spontaneous, which means 1. IandIIonly ▯S > 0acdioteSodLaw universe 2. II only correct of Thermodynamics. Entropy (S)siihfrsmswihih degrees of freedom, disorder or randomness 3. I, III and IV only and low for systems with low degrees of free- dom, disorder or randomness. 4. All have a positive ▯S. S(g) >S (▯) >S (s). 5. III and IV only The system is Explanation: The Third Law of Thermodynamics states H2O(▯) ▯ H O2g) that the entropy of a perfect pure crystal at so ▯S system> 0. 0Ksi. Asid,nome,ndd- grees of freedom increase, so does Stpy The entropy of the surroundings must be can increase by changing phase from solid to negative. Energy is removed from the sur- roundings to get the water to evaporate. liquid to gas, and by increasing temperature, volume, or number of particles. In reaction I, the final state has less gas par- 004 (part 1 of 5) 10.0 points ticles (and thus less entropy) than the initial Asamplosigf2 .5mlfaprcy ideal diatomic gas is expanded isothermally state. Therefore ▯S is negative. at 265.5Kfrom56Lto113 .8L.Calculate w. In reaction II, the final state has more gas particles (and thus more entropy) than the initial state. Therefore ▯S is positive. Correct answer: ▯3913.06 J. III describes a phase change. Gases have Explanation: more degrees of freedom, randomness, and n =2 .5mol T =265 .5K disorder (entropy) than liquids. The final V i56L V =113 .8L state is a liquid and the initial state is a gas. f Therefore ▯S is negative. Vf In reaction IV, the final state has 0 gas w = ▯nRT ln particles and the initial state has 1 mole of V i = ▯(2.5mol)(8 .314 J/mol · K)(265.5K) gas particles. Therefore ▯S is negative. ▯ ln113.8L 003 10.0 points 56 L For the evaporation of water from an open = ▯3913.06 J ▯ pan at 25 C, the values of ▯S for the water, the surroundings, and the universe must be, 005 (part 2 of 5) 10.0 points respectively, Calculate q. 1. None of these is correct. Correct answer: 3913.06 J. 2. positive, positive, positive. Explanation: q = ▯w =3913 .06 J 3. positive, negative, zero. 4. negative, negative, negative. 006 (part 3 of 5) 10.0 points Calculate ▯U. ngac (ttn793) – H14: Thermo 3 – mccord – (51600) 3 Consider the unbalanced reaction Correct answer: 0 J. Al 2 3s) + CO(g) ▯ Al(s) + CO 2g) Explanation: By definition ▯U =0J. Species ▯H 0 S 0 f 007 (part 4 of 5) 10.0 points kJ/mol J/mol · K Calculate ▯H. Al 2 3 ▯16076.0 5 .92 CO 7 ▯1190.5 1 .6 Correct answer: 0 J. Al 8 0.0 2 .3 Explanation: CO 2 3 ▯3913.5 2 .6 By definition ▯H =0J. Estimate ▯G at 298 K if 6.7m efo 008 (part 5 of 5) 10.0 points Al2O3(s) is reacted. Calculate ▯S. Correct answer: 5433.72 kJ. Correct answer: 14.7385 J/K. Explanation: Explanation: T =298K nAl O =1mol 2 3 ▯S = q = 3913.06 J =14 .7385 J/K Al2O 3s) + 3CO(g) ▯ 2Al(s)+3CO 2(g) T 265.5K 0 ▯ 0 ▯ 0 009 10.0 points ▯H rxn= n▯H fprod▯ n▯H frct Calculate the change in entropy when 3.7mol =3( ▯▯93.5kJ /mol) of monatomic perfect gas is heated from 265 Kto62 .5Kandilylpdd ▯ 3(▯110.5kJ /mol) ▯ from 39 L to 70.2L. +(▯1676 kJ/mol) Correct answer: 60.3615 J/K. =827kJ /mol Explanation: Do this in 2 steps that are easily calculated. ▯ ▯ ▯S 0 = n▯S 0 ▯ n▯S 0 Step 1: Isothermal expansion from 39 to 70.2. rxn ▯ fprod frct Use Vf = 3(213.6J /mol · K) ▯S 1 nRln ▯ V i +2(28 .3J /mol · K) ▯S =3 .7(8.314)ln 70.2 =18 .0814 J/K ▯ 1 39 Step 2: Constant volume temperature in- ▯ 3(197.6J /mol · K) crease. C for monatomic ideal gas is 3/2R. ▯ v +(50 .92 J/mol · K) ▯S = nC ln Tf 2 v Ti =53 .68 J/mol · K 662.5 ▯S 23 .7(1.5 ▯ 8.314)ln =4 2 .2801 =0 .05368 kJ/mol · K 265 J/K Now combine the steps: ▯S =▯ S +▯ S ▯G =▯ H ▯ T ▯S 1 2 =827kJ /mol ▯S =18 .0814 + 42.2801 = 60.3615 J/K ▯ (298 K)(0.05368 kJ/mol · K) 010 10.0 points =811 .003 kJ/molrxn ngac (ttn793) – H14: Thermo 3 – mccord – (51600) 4 5433.72 kJ 1molrxn · · 6.7molAl 2O3 mol rxn 1molAl 2O 3 013 (part 1 of 3) 10.0 points =5433 .72 kJ An electric heater is rated at 1.8kW. At what rate does it generate entropy in a room maintained at 25.9 C? 1 W = 1 J/s 011 10.0 points For the vaporization reaction Correct answer: 6.02208 J . Br2(▯) ▯▯Br (2) K · s at a certain pressure, ▯H =33kJ /mol and Explanation: 0 rate of
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