Calculate the enthalpy of the following reaction: 4 b (s) + 3 o2 (g) → 2 b2o3 (s) given the following pertinent information: (a) b2o3 (s) + 3 h2o (g) → 3 o2 (g) + b2h6 (g), δhoa = +2035 kj (b) 2 b (s) + 3 h2 (g) → b2h6 (g), δhob = +36 kj (c) h2 (g) + latex: \frac{1}{2} 1 2 o2 (g) → h2o (l), δhoc = −285 kj (d) h2o (l) → h2o (g), δhod = +44 kj

4B + 3O₂ => 2B₂O₃; ΔH° = -3673Kj

Explanation:

Work these type problems in pairs of rxns… That is, add Rxn-1 & Rxn-2 => Rxn-1,2; then add Rxn-3 to Rxn-1,2 => Rxn- 1,2,3. Rxn-4 is not needed to obtain target rxn.  

Target Rxn => 4B + 3O₂ => 2B₂O₃

Given …

(1)B₂O₃ + 2H₂O => 3O₂ + B₂H₆

       => reverse and double

       => 2B₂H₆ + 6O₂ => 2B₂O₃ + 6H₂O

(2)2B + 3H₂ => B₂H₆ => double and add to Rxn-1 => 4B + 6H₂ => 2B₂H₆

         2B₂H₆ + 6O₂ => 2B₂O₃ + 6H₂O

              4B + 6H₂ => 2B₂H₆

∑(1,2)    4B + 6O₂ + 6H₂ => 2B₂O₃ + 6H₂O; ΔH°₁₂ = -2035Kj + (+72Kj) = -1963Kj

(3)=> H₂ + ½O₂ => H₂O

       => reverse and multiply by 6, then add to (1,2) => 6H₂O => 6H₂ + 3O₂

                           6H₂O => 6H₂ + 3O₂

           4B + 6O₂ + 6H₂ => 2B₂O₃ + 6H₂O  

∑[(1,2,3) 4B + 3O₂ => 2B₂O₃; ΔH°₁₂₃ = -1963Kj + 6(-285Kj) = -3673Kj