bondhaber cycle of AlCL3

Asked by Kashif | 19th Oct, 2012, 11:31: AM

Expert Answer:

AlCl3(s)  ?  Al(s)  +  3/2 Cl2(g)    ?Hf° = 704.2 kJ/mol 
 
Al(s)  ?    Al(g)          ?Hf° = 326 kJ/mol 
 
Al(g)  ?  Al+3(g)*  +  3e-      ?Hf° = 5138 kJ/mol *Note: the first three ionization energies of aluminum are added up here! 
 
 
3/2 Cl2(g)  ?  3Cl(g)     3?Hf° = 3(121.68 kJ/mol)
3Cl(g)  +  3e-  ?  3Cl-(g)    3?Hf° = 3(-349 kJ/mol)  
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?HL° = 5486 kJ/mol

Answered by  | 9th Nov, 2012, 09:20: AM

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