HW-Solns-Chap-6-1413

# 63 answer h f b 2 h 6 36 kjmol strategy and

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63 . Answer: H f (B 2 H 6 ) = 36 kJ/mol Strategy 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 H f . Look up the H f for the rest of the reactants and products. Plug them into the equation and solve for H f . The products are B 2 O 3 and H 2 O( ). The reactants are B 2 H 6 and O 2 . O 2 is the elemental form for oxygen. H° = [(1 mol) H f (B 2 O 3 ) + (3 mol) H f (H 2 O( ))] – (1 mol) H f (B 2 H 6 ) H f (B 2 H 6 ) = – H (1 mol) + H f (B 2 O 3 ) + 3 H f (H 2 O( )) Look up the H f value for H 2 O( ) in Table 6.2. H f (B 2 H 6 ) = – ( 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 B 2 H 6 is small. .

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