Question

The enthalpy of combustion (∆H_c) of aluminum borohydride, Al(BH₄)₃(l), was measured to be -4138.4 kJ/mol [Rulon and Mason, J. Am. Chem. Soc., 73, 5491 (1951)]. The combustion reaction for this compound is given by

Al(BH₄)₃(l) + 6 O₂(g) → ½ Al₂O₃(s) + 3/2 B₂O₃(s) + 6 H₂O(l)

Given the following additional data, calculate the enthalpy of formation of Al(BH₄)₃(g).

Substance	Property
Al2O3(s)	∆Hf = -1669.8 kJ/mol
B2O3(s)	∆Hf = -1267.8 kJ/mol
H2O(l)	∆Hf = -285.84 kJ/mol
Al(BH4)3(l)	∆Hvap = 30.125 kJ/mol

∆H_f = ___ kJ/mol

Answer 1

Given reaction is,

Al(BH4)3(l) + 6 O2(g) → ½ Al2O3(s) + 3/2 B2O3(s) + 6 H2O(l)

Delta H reaction = 1/2 H_f Al2O3(s) + 3/2 Hf B2O3(s) + 6 x Hf H2O(l) -- ( Hf Al(BH4)3(l) + 6 x Hf O2(g) )

1 mol x -4138.4 kJ/mol = 1/2 x -1669.8 kJ/mol + 3/2 x -1267.8 kJ/mol + 6 x -285.84 kJ/mol - (Hf + 0 )

Hf Al(BH4)3(l) == -313.64 kJ/mol

we know,

delta H vap = Hf Vapor - Hf liquid

So given delta H vapor = 30.125 kJ/mol

Hf Vapor = delta H vap + Hf liquid

= 30.125 kJ/mol + (-313.64 kJ/mol)

= - 283.515 kJ/mol