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Gen Chem

by: Dr. Drew Flatley

Gen Chem CHEM 112

Dr. Drew Flatley
GPA 3.96

Edward Voigtman

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Edward Voigtman
Class Notes
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This 101 page Class Notes was uploaded by Dr. Drew Flatley on Friday October 30, 2015. The Class Notes belongs to CHEM 112 at University of Massachusetts taught by Edward Voigtman in Fall. Since its upload, it has received 32 views. For similar materials see /class/232345/chem-112-university-of-massachusetts in Chemistry at University of Massachusetts.


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Date Created: 10/30/15
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rmquot MIion nt lc1u3 w I39 Ll wakv diamfr 19 Q Q 3971 quotquot37 F Hxig gtuImlt4hm eu uln f Lyn453 ouuIH 39an mch olublnquot gt 1439 MIA Water surroundmg 7 7 Chil ConfounJ I a cation a Emckycowe Cequot aaaaaaaaa n9 9 a Water surrounding an anion monkskm c eeeeeeeeeeee 19 A n Anion inHrc lhs wH k l n 44 clip I TM anLW M 7m4 ru39l39h3 Jaw1 s causal HY NRAhug like 419uJS Alhaqu k c e I Hi17 a go WAham ion H f3 X nine4 whoIt A is Sllvmx L t Ankh E H lfy 39 Ivdb f u39f t I A139 4 6 ItoJ ALA44fo vrf39 3914 NS IIlull K r 133 pm AH 321kJmol eeeee rankCD Thurman Emma de Tmmqmu e T v39f39v39 Entro yv 5 Evaporation Heating liquid f mm 5 inCreases a39 arge amount with phase changes MeL ng mum Temperature 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 ea wol l5 n T 4 H19 Move Como HL W extrapolate Fig 56 The calculation of entropy from heat capacity data Q m l calculated from the following data Example Objective 19 The Third Law entropy of nitrogen gas at 29815 K and 1 atm has been 0 i C e l t 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 3 29815 K 19206 I K 1 mol l En39hro y cL uj vcv VIM luvjg AS fruitng e Ans vlu quotmuff 39 hukarrd riu ILquot cLaJz aamfk HgoA Hm 33 Far leIx 90quot 7 quot 7735 K a L ac 2r AS T 739 373651 05 tom SK jmkk TABLE 191 Standard Molar Entropy Values at gk Only 2 Substance Entropy 5 JKmol N 7394quot Cgraphite 56 quott 7 Cdiamond 2377 suLstu Cvapor 1581 0 mm H2g 1307 Ru Zero 029 2051 mm H20g 18884 H20 6995 Emomrcowe Cenuaqe L ggggg ng hud as Standard Molar En rr opies m 141 dyadd I Some Standam Molar Enuopy Values 31298 K 5 Entropy s Entropy sa Element lK mol Compound JK mnl Cgraphite 55 mg 4 135 3 Cdiamund 2377 term If 229 2 e Cvapuv 1531 51mg K 2703 amp Cas 1159 CHXOHU 1272 Arg 1549 mg 1977 1129 1307 mm 2137 029 2051 H20g 13334 Nzg 1916 Hzom 5935 93 1 21123 HClg 1332 c149 K 2231 NaCLs 7211 Bu 1522 MgOs 2535 115 1151 megs 917 Calcula ring ASOfor a Reac rion A80 Z S0 products Z S0 reactants X 1 Consider 2 H2g 029 2 HZOqu A80 2 so H20 2 8 H2 8 02 A80 2 mol 699 JKmol 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 9 113 013 g 3amp061 ARM 126 9 5 K Agil uhas f39 quot quot AH3 4M T 2 46m S7LC 3 5771 3 AS39a W mm TK ll BA BMW quwhus I AS HITS1r 3n SK 15 1 Spontaneous or39 Not TABLE 192 Predicting Whether a Reaction Will Be Spontaneous Under Standard Conditions Reaction Type AH system AS system Spontaneous Process Standard Conditions 1 AMxothermic lt O Exothermic lt O Endothermic gt 0 4 Endothermic gt 0 Positive gt 0 Negative lt 0 Positive gt O Negative lt 0 Spontaneous at all temperatures AS universe gt 0 Depends on relative magnitudes of AH and A5 Spontaneous at lower temperatures Depends on relative magnitudes of AH and AS Spontaneous at higher temperatures Not spontaneous at any tem eraturei A5 universe lt 0 21 e EmoksJCalz Cengage Lemurquot Remember that AH sys is proportional to AS 7 An exothermic process has AS Surr gt O SUI I Gibbs Free Energy 6 ASuniv ASsurr sys AH sys ASuniv T ASSYS Multiply through by T J39ijard Gibbs rAsumV AHSYS TASsys 18391903 TASuniV change in Gibbs free energy for the system AGSystem Under standard conditions o o 0 AG sys AH sys TAS sys 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 am M Aquot a a TX I Mn 3 andL 3M J39 u J T Qua 4 L T 44 Lgk T m4quot CAIquotJ 10 5 61 3 4 3 mmu A eulg A Jurnrg 6 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 reactants 14 Free Energies of Formation Standard Molar Free Energies of Formation of Some Substances at 298 K ElementCompound AGHkJmol ElementCompound AGkJmol H29 0 C02g 3944 029 0 CH4g sos 0 2286 Cgraphite 0 H20 2372 Cdiamond 2900 NH3g 164 C0g 1372 F9203s 7422 Note that AfG for an element O Avu AN 399 au etcw E 0 09 we 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 42555 kJ 29815 K00973 kJK 12266 kJ Reaction is productfavored in spite of negative ArSO Reaction is enthalpy driven 16 Calculating AFG i I z k 1 3977 l 7i quot 739 rquot 39 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 3mm 3 m su gtol an rnrnil DLKIIK 7 50 P13 0 aviatisv E F2 FNsh Pwan nrir 5 g my Emu wg 52 T 753 sftELLK BF n5 T QQE EVENT mm N NIL N 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