Propane cap c sub 3 cap h sub 8 g , ΔH_f=−103.8kJ/mol reacts with oxygen to produce carbon dioxide CO sub 2 , ΔH_f=−393.5kJ/mol and water cap h sub 2 cap o , ΔH_f=−241.82kJ/mol according to the equation. cap c sub 3 cap h sub 8 g plus 5 cap o sub 2 g rightarrow 3 CO sub 2 g plus 4 cap h sub 2 cap o g What is the enthalpy of combustion (per mole) of cap c sub 3 cap h sub 8 g ? Use ΔH_rxn=∑(ΔH_f_1products)−∑(ΔH_f_1reactants)
Answer
A
531 point 5 kJ over mol
B
negative 531 point 5 kJ over mol
C
2 comma 044 point 0 kJ over mol
D
negative 2 comma 044 point 0 kJ over mol
Explanation
Use the formation-enthalpy equation: ΔH_rxn = ΣΔH_f(products) − ΣΔH_f(reactants). For the products, 3(−393.5) + 4(−241.82) = −2147.78 kJ/mol. For the reactants, propane contributes −103.8 kJ/mol and O₂ contributes 0, so ΔH_rxn = −2147.78 − (−103.8) = −2043.98 kJ/mol, or about −2,044.0 kJ per mole of propane. The correct choice is D, negative 2,044.0 kJ/mol.
AI-written and checked against the verified answer.