Pentaborane B₅H₉(s) burns vigorously in O₂ to give B₂O₃(s) and H₂O(l) . Calculate $\Delta$ H_rxn for the combustion of 5.00 mol of B₅H₉.
$\Delta$ H^$\circ$_f[B₂O₃(s) ] = -1,273.5 kJ/mol
$\Delta$ H^$\circ$_f[B₅H₉(s) ] = 73.2 kJ/mol
$\Delta$ H^$\circ$_f[H₂O(l) ] = -285.8 kJ/mol

Question

Pentaborane B₅H₉(s) burns vigorously in O₂ to give B₂O₃(s) and H₂O(l) . Calculate

\Delta

H_rxn for the combustion of 5.00 mol of B₅H₉.

\Delta

H^$\circ$_f[B₂O₃(s) ] = -1,273.5 kJ/mol

\Delta

H^$\circ$_f[B₅H₉(s) ] = 73.2 kJ/mol

\Delta

H^$\circ$_f[H₂O(l) ] = -285.8 kJ/mol

Quizplus · Accepted Answer

Using the given enthalpies of formation, we can calculate the delta H for the reaction:
ΔH = (5 mol)(-1273.5 kJ/mol) - (19 mol)(73.2 kJ/mol) - (1 mol)(285.8 kJ/mol) = -22,698 kJ
The negative sign indicates that the reaction is exothermic (i.e. releases energy). The answer is C, which is the closest value to -22,698 kJ.

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Pentaborane B₅H₉(s) Burns Vigorously in O₂ to Give B₂O₃(s) and H₂O(l)