Lecture 28 - 3/22/10 Determine the bond energy of a F-F...

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3/22/10 1 45 Determine the bond energy of a F-F bond, by considering the reaction H 2 (g) + F 2 (g) ! 2 HF (g). A) -542.2 kJ/mol B) -413.2 kJ/mol C) 151.8 kJ/mol D) 158.0 kJ/mol E) Cannot be determined ! H rxn = [1 mol " D(H-H) + 1mol " D(F-F)] # 2 mol " D(H-F) D(H-H) = 436 D(H-F) = 565 kJ/mol ! H rxn = 2mol "! H f °(HF) # [1mol "! H f °(H 2 ) + 1mol "! H f °(F 2 )] = 2mol( # 271.1 kJ/mol) # [ 0 kJ+0 kJ] = # 542.2 kJ D(F-F) = [ ! H rxn # 1 mol " D(H-H) + 2 mol " D(H-F)]/1 mol = [( # 542.2kJ) # 1mol " D(436 kJ/mol)+2mol " (565 kJ/mol)]/1 mol = 151.8 kJ/mol Alternate solution using different rxn (online notes) 46 Alternate solution if reaction H 2 (g) + F 2 (g) ! 2 HF (g) not specified. ! H rxn = 0 # 1 mol " D(F-F) ! H rxn = 1mol "! H f °(F 2 ) # 2mol "! H f °(F) = (0kJ) # 2mol(78.99kJ/mol) = # 158.0 kJ D(F-F) = #! H rxn = 158.0 kJ/mol Consider 2 F (g) ! F 2 (g) 47 ! G and Spontaneity Summary Summary : (see Table 20.1) ! H ! S ! G Result negative positive ! G = ! H # T ! S negative Always spontaneous positive positive positive Low T (non-spontaneous) negative High T (spontaneous) negative positive
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Lecture 28 - 3/22/10 Determine the bond energy of a F-F...

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