5.76 4 B + 3 O2  2 B2O3 O2 + 2 H2  2 H2O 3O2 + B2H6  B2O3 + 3 H2O THERE ARE ONLY 2 B IN THE NET, YOU MUST MULTIPLY THIS REACTION BY ½. THERE ARE 3 H IN THE NET, YOU MUST MULTIPLY THIS REACTION BY 3/2. 4 B + 3 O2  2 B2O3 ∆H1=(-2509.1kJ) O2 + 2 H2  2 H2O ∆H2=(-571.7 kJ) 3O2 + B2H6  B2O3 + 3 H2O ∆H3=-2147.5 kJ) REVERSE THIS REACTION TO GET B2H6 AS A REACTANT. AND TO CANCEL B2O3 2B + 3H2  B2H6 This is the formation, ∆Hf of B2H6, NET Rx

This is the formation, ∆Hf of B2H6, NET Rx 2 B + 3/2 O2  B2O3 ∆H1=1/2(-2509.1kJ) 3/2O2 + 3 H2  3 H2O ∆H2=3/2(-571.7 kJ) B2O3 + 3 H2O  B2H6 + 3O2 ∆H3=-(-2147.5 kJ) 2B + 3H2  B2H6 This is the formation, ∆Hf of B2H6, NET Rx ∆Hf=+35.4 kJ