Question

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

Answer 1

Answer:

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