63. Answer:Hf(B2H6) = 36 kJ/molStrategy and Explanation: Given a balanced thermochemical expression for a reaction and some molar enthalpies of formation, determine an unknown molar enthalpy of formation. Use the stoichiometric coefficient of the balanced equation to describe the moles of each of the reactants and products. Set up a specific version of Equation 6.11, and solve it for the unknown Hf. Look up the Hffor the rest of the reactants and products. Plug them into the equation and solve for Hf. The products are B2O3and H2O(). The reactants are B2H6and O2. O2is the elemental form for oxygen. H° = [(1 mol) Hf(B2O3) + (3 mol) Hf(H2O())] – (1 mol) Hf(B2H6) Hf(B2H6) = – H(1 mol)+ Hf(B2O3) + 3 Hf(H2O()) Look up the Hfvalue for H2O() in Table 6.2. Hf(B2H6) = – (2166 kJ)(1 mol)+ (–1273 kJ/mol) + 3 (–285.830) = 36 kJ/mol Reasonable Answer Check:The high exothermicity of the reaction can be almost completely accounted for by the formation of the products. So, it makes sense that the molar enthalpy of formation of the reactant B2H6is small. .