Question

Chapter 1 Thermodynamics Page 16 16. The standard enthalpy of combustion of solid phenol (CH3OH) is -3054 kJ moli' at 298K and its standard molar entropy is 144,0 J K'moll. Calculate the standard Gibbs free energy of formation at 298K. (81,98 kJ mol) 17. Calculate the standard Helmholtz energy of formation of CH2OH(l) at 298K from the standard Gibbs energy of formation, on the assumption that He and O2 are perfect gases. (-160 kJ mol') 18. a. Calculate the heat generated when 50 cm 1 mol/dm’ aqueous HCI is neutralised by 1 mol/dm NaOH. (2,79 kJ) b. 50 cm 1 mol/dm' aqueous acetic acid is neutralised by 1 mol/dm NaOH. (2,87 kJ) 50 cm 1 mol/dm' aqueous ammonia is neutralised by 1 mol/dm HCI. (2,59 kJ) d. 50 cm 1 mol/dm'aqueous ammonia is neutralised by 1 mol/dm'acetic acid. (2,67 kJ) c.

Answer 1

D AH=1072 KJ/mol AH = 468 kJ/mol For of solid x-D. lose to solid 19 X-D-glucose (aq) B-D glucose (aq) ,DH² 116 kJ / mal **) d-D-glucose (s) —>-D-glucose (aq) sin $-D-gluwels) — B-D- B-D-glucose lag) the muto rotation -D-glucose B-D-glucose. Reverse the egn lii) and add all three egns. d-D-gluwte lag) -> B-D-glutse tag) e AH²=-116 kJ l mal -D-glucose (9) »d-D-glucose (aq) B -D-glucone Tony B-D-glucose (3) B-D-glucose (s) 14° = 1072 KJ imal 64° =-468 kJ lonel -De glucose (s) DH6116). + (1072)+(-468) = 488 kJ/mel Ans

13 elgraphite) - (1g) 14° = 716.7k 5 lmol DH° = 43660 kJ/mol Hg (g) - 9H1) formation of propane : 3 cl graphite ) + ungly) - 4glg) into are ous atoms DHE = {OH to turn elements into gareous state - EDH to turn compounds gaseous 30lgraphite) + 4H-H (g) – H H -4 (9) H H H Bond enthalpy (ko mel C-C 348 Any = (3.3u [C] 4 ux & 14 )]- [ex[=c7+ 8x8((-43] 9 * = (3x1116.7) 44(436) - (0 x 348) + (8x413) C-H 413 4-H 436 11 3894.1 4000 =-105.9 kJ/mel, Ans.

14. a) Hydrolysis of wiea : H₂NCONH₃ + H₂O CO2 + 2NH3 Entropy change = positive (because number of molecules are in creasing) b) H+ + oH HO Entropy Chonge = positive (electoic field prenwie de preses) c) CH₂COOH CH₃COO- +ht Entropy chonge = negative (eltebuie field pemune; electrostarction increases) S, CM₂ ( 50 ) COOCH₃ + Br entropy change = positive (cle dripstorets on deceases in seaction) d) CHg Br COOCH3 +

substance OG °(KJ/mey 94, (x) + 120,43) ► 1200, (9)+ 114,0) AG= {AG (Po ducts) - Esej (upact onts) G. Had 99, (1) 1544.3 $C = [12564(0,3+0 (C0g)+ 110c (4,0)-[cj(s.14.9, 1826199 O, 19) = [136-394.3 + (-237177771544.3+0] 401) O -237.14 CO(g) -394. 39 0 = -4732.68 - 2608.54 + 1544.3 sa = -5797 kJ lmol = 5800 kJ/mol AG =-5800k 5 lmel Ans.