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ALEKS Chemistry O ENTROPY AND FREE ENERGY Calculating reaction free energy under nonstandard conditions A chemist fills a rea
emistry substance sº (J/mol-K) AH,° (kJ/mol) AGº (kJ/mol Aluminum -485.0 A13+ (aq) --- 0 Al(s) Al2O3 (s) Al(OH)3 (s) 0 -1675.
nistry --- D001./ Cu304/2 ) Carbon C(s) 0 0 5.7 CC14 (9) -95.7 -64.0 309.4 CCIA (1) -128.2 -68.6 214.4 CCIA (s) -136.8 CH4 (9
emistry Cl2 (9) 0 0 223.1 Cl- (aq) - 131.2 HCI (aq) - 167. HCI (9) -92.3 -95.3 186.9 Chromium Cro 2- (aq) -727.8 Fluorine F2
le/content/798394/fullscreen/4577602/View PaperCut Login fo... ALEKS Chemistry... nistry Nitrogen N2 (9) 0 0 191.6 NzH4 (9) 9
PaperCut Login fo... ALEKS Chemistry.. mistry -- Oxygen OH (aq) -157.2 H20 (1) -285.8 -237.1 70.0 H20 (9) -241.8 -228.6 188.
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solution: 2 - Norgot clcg 2 Nocle) find AGayn: AGO - E A Gif (presducts) 1xn EAG f(reactants) 2 x 66.1 kJ/ (2 2x87.6k] mol +We know that AG-AG + RT ln (2) - AG & RT ln (PNoch)? (PNO )² (BCI) AG 43 kJ mol + 0.0083145 kJ x 298.15k x malik (4.56² In (0AG = (-43k] mol 2.478968 175 kJ x x x In 20.1936 mol.lt 0-298116 x 1.48 = - 43kJ +/2.478968175 kJ x ln/20.7936 mol 0.4412 116= -33.5 kJ mol AG =-34 kJ (G round to the nearest number) mol Answer AG = - 34 kJNote:

T = 25 + 273.15 = 298.15 K

R= gas constant = 0.0083145 KJ/mol.K

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