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°_f[B₂O₃(s)] = -1,273.5 kJ/mol $\Delta$H°_f[B₅H₉(s)] = 73.2 kJ/mol $\Delta$H°_f[H₂O(l)] = -285.8 kJ/mol A)- 45,400 kJ B)45,400 kJ C)- 22,700 kJ D)- 9,090 kJ E)- 8,790 kJ

Question

Quizplus · Accepted Answer

The reaction equation is: 5B₅H₉(s) + 34O₂ -> 15B₂O₃(s) + 45H₂O(l) ΔH°rxn = ΣnΔH°f(products) - ΣnΔH°f(reactants) ΔH°rxn = [15(-1273.5) + 45(-285.8)] - [5(73.2)] ΔH°rxn = -22,700 kJ The negative sign indicates that the reaction is exothermic, and the value of 22,700 kJ represents the amount of heat released when 5.00 mol of pentaborane B₅H₉ is combusted. Therefore, the correct answer is C.

Pentaborane B5H9(s)burns Vigorously in O2 to Give B2O3(s)and H2O(l) Δ\DeltaΔ Hrxn for the Combustion of 5

Pentaborane B₅H₉(s)burns Vigorously in O₂ to Give B₂O₃(s)and H₂O(l) $\Delta$ H_rxn for the Combustion of 5