CBSE Class 11-science Answered
bondhaber cycle of AlCL3
Asked by Kashif | 19 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 | 09 Nov, 2012, 09:20: AM
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