Class Note for CHEM 112 at UMass(3)
Class Note for CHEM 112 at UMass(3)
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This 18 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Massachusetts taught by a professor in Fall. Since its upload, it has received 17 views.
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Date Created: 02/06/15
Emmi pr ade Tremwmu e 4 I 47quot quot W tvf iv I39i 7 39 sihis ty wish T Eva pavatv39uu mm 5 increases a large amount with phase changes mm 594 Heating Liquid Mem39ng mum Tempevalme K Evaluating the entropy and the Gibbs function 151 Physrcal Chemistry P W Atkins melt b0 2nd Ed W H Freeman 8t Co San Francrsco CA 1982 extrapolate TI Th Fig 56 The calculation of entropy from heat capacity data Example Objective 19 The Third Law entropy of nitrogen gas at 29815 K and 1 atm has been calculated from the following data ASmJ K 1 mol 1 T3 extrapolation 0 10 K 192 graphical integration eqn 521 2525 phase transition at 3561 K 643 graphical integration eqn 521 2338 phase transition fusion at 6314 K 1142 graphical integration eqn 521 1141 phase transition vaporization at 7732 K 7213 perfect gas behaviour eqn 522 3920 from 773229815 K correction for non ideality 092 Total entropy change 19206 By setting the entropy at absolute zero to zero we get the Third Law standard entropy see below of nitrogen gas as S2 29815 K 19206 I K 1 mol l 9 a we l I 5 It Move Como cplE A HU 4 le gem luvx fl l39wrf IL En39hro y cL uj vcv VIM luvjg from And quotMilly Ag h Af d riu ILquot LaJ Exam 9013 F 30 m AVH 4M 3 T 773ISH l39 AS Ltmu4acs 2r quot T 739 3731551 05 tom SK 3 TABLE 191 Standard Molar Entropy Values at 298 K Substance Entropy JKmol Cgraphite 56 Cdiamond 2377 Cvapor 1581 H2g 1307 029 2051 H20g 18884 H20 6995 m m 6 l S randar39d Molar En rr opies 14134 dudd rH Some Standam Molar Entropy Values 31298 K Enlropy 5 Enlropvy sa Element lK mol Compound JK mnl Cgravhite 55 Guy 4 1353 Cdiamund 2377 21159 If 2232 e Cvapuv 1531 canxm K 2703 amp Cas 1159 CHXOHU 1272 Avg 15m 79 1977 My 1307 020 2137 023 2051 H20g 13334 N2g 1313 H200 5995 My 1 2023 Hum 1302 c149 K 2231 NaCLs 7211 3121 1522 94905 2035 125 1151 CaCOs 917 Calculating ASOfor39 a Reaction A80 Z S0 products Z S0 reactants S 1 Consider 2 H2g 029 2 HZOqu AS 2 S HZO 2 8 H2 8 02 A80 2 mol 699 JK Xmol 2 mol 1307 J Kmol 1 mol 2053 JKmo A80 3269 JK Note that there is a decrease in S because 3 mol of gas give 2 mol of liquid 2nd Law of Thermodynamics A reaction is spontaneous if A8 for the universe is positive AS A8 AS universe system surroundings Asumverse gt O for spontaneous process Calculate the entropy created by energy dispersal in the system and surroundings 9 113 019 F JHAOKJ 4351 959 TK AS39 Sun J15 BA AMY 2 46 716 3 71 3 1 AS I a quot quot339 17sz K Hl39 TK Souhum I ASIAMus an Slit 39337 SK 445 quot 39SK Spontaneous or39 Not TABLE 192 Predicting Whether a Reaction Will Be Spontaneous Under Standard Conditions Reaction Type AHquot system AS system Spontaneous Process Standard Conditions Positive gt 0 1 AMxothermic lt 0 Exothermic lt O Negative lt 0 Endothermic gt 0 Positive gt 0 4 Endothermic gt 0 Negative lt 0 Spontaneous at alt temperatures AS universe gt O Depends on relative magnitudes of AH and ASquot Spontaneous at Lower temperatures Depends on relative magnitudes of AH and A5 Spontaneous at higher temperatures Not spontaneous at any tem eraturet AS universe lt 0 21 EmoksJCaie Cengane Learning Remember that AH sys is proportional to AS An exothermic process has AS Surr gt O SUI I Gibbs Free Energy 6 ASuniv ASsurr ASSYS AH sys ASuniv T ASSVS Mumply through by 39T Jrilulard Gibbs rAsumV AHSYS TASsys 18391903 TASuniV change in Gibbs free energy for the system AG system Under standard conditions AGOSys AHOSYS TASOsys 10 AGO AHO TASO Gibbs free energy change total energy change for system energy lost in energy dispersal If reaction is exothermic negative AHO and entropy increases positive A80 then AGO must be NEGATIVE reaction is spontaneous and productfavored 11 AGO AHO TASO Gibbs free energy change total energy change for system energy lost in energy dispersal If reaction is endothermic positive AHO and entropy decreases negative ASO then AGO must be POSITIVE reaction is not spontaneous and is reactant favored 12 AF l TA Aquot 39 AS 15 rx n 10 s 7 9quot 3 and 39CAMH o39ech J 2 mel L T 7 Rm L9 7 61 3 cut 3 menu 4 euJo Junu regLA Cmch 13 Gibbs Free Energy G AGO AHO TASO Two methods of calculating AGO a Determine AH and A5quot and use Gibbs equation b Use tabulatgd value of free energies of formation AfGO AFGO 2 AG products 2 AG reactant5 14 Free Energies of Formation Standard Molar Free Energies of Formation of Some Substances at 298 K ElementCompound AGkJmol ElementCompound AG kJmol Mo 0 C02g 3944 029 0 CHAN 508 0 2286 Cgraphite O H20t 2372 Cdiamond 2900 NH3g 164 COg 1372 F9203s 7422 Note that AfG for an element O Am Ho 399 am etcwn E 0 09 we I l 15 Calculating Ar6 Combustion of acetylene 02H2g 52 029 gt 2 0029 H20g Use enthalpies of formation to calculate ArHO 12556 kJ Use standard molar entropies to calculate ArSO 973 JK or 00973 kJK ArGO 12555 kJ 29815 K00973 kJK 42255 kJ Reaction is productfavored in spite of negative ArSO Reaction is enthalpy driven 16 Calculating ArG it E NH4NO3s heat NH4NO3aq Is the dissolution of ammonium nitrate product favored If so is it enthalpy or entropydriven 17 Calculating Ar6 NH4NO3s heat NH4NO3aq From tables of thermodynamic data we find ArHO 257 kJ ArSO 1087 JK or 01087 kJK ArGO 257 kJ 29815 KO1087 JK 67 kJ Reaction is productfavored in spite of negative ArHO Reaction is entropy driven 18
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