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

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.