Calculate the standard enthalpy of formation of the fictitious compound, gaseous dideborane (D2H6) using the following thermochemical information: (in description)

Question

4D (s) + 3O2 (g) 2D2O3 (s) Delta H -2345.1 kJ

2H2 (g) + O2 (g) 2 H2O (l) Delta H - 571.7 kJ

D2H6 (g) + 3O2 (g) D2O3 (s) + 3H2O (l) Delta H -2134.5 kJ

Answer 1

This is a problem using Hess' law

TARGET EQUATION: 2D + 3H₂ ==> D₂H₆

1). 4D (s) + 3O₂ (g) ==> 2D₂O₃ (s) Delta H -2345.1 kJ

2). 2H₂ (g) + O₂ (g) ==> 2 H₂O (l) Delta H - 571.7 kJ

3). D₂H₆ (g) + 3O₂ (g) ==> D₂O₃ (s) + 3H₂O (l) Delta H -2134.5 kJ

Procedure:

reverse eq 3: 2D₂O₃ + 3H₂O ==> D₂H₆ + 3O₂ ∆H = +2134.5 kJ

eq.1÷2: 2D + 3/2 O₂ ==> D₂O₃ ∆H = -1173 kJ

eq 2 x 3/2: 3H₂ + 3/2 O₂ ==> 3H₂O ∆H = -858 kJ

__________________________________________ Add them together and cancel/combine like terms..

2D₂O₃ + 3H₂O + 2D + 3/2 O₂ + 3H₂ + 3/2 O₂ ==> D₂H₆ + 3O₂ + D₂O₃ + 3H₂O

You are left with the TARGET EQUATION: 2D + 3H₂ ==> D₂H₆

∆H = 2135 kJ - 1173 kJ - 858 kJ

∆H = 104 kJ