Answer to Question #336754 in General Chemistry for Yama

Question

The combustion of ethane proceeds according to equation 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O. The enthalpy change for combustion of 1 mol of this hydrocarbon equals –1600.0 kJ/mol. Use the following from Hess's law and calculate the enthalpy of formation of 1 mol of ethane, if ΔHf (CO₂) = −393.5 kJ/mol, ΔHf (H₂O) = −241.8 kJ/mol, ΔHf (O₂) = 0.

Answer 1

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O, ΔH = –1600.0 kJ/mol

ΔH_f(CO₂) =−393.5 kJ/mol

ΔH_f(O₂) = 0 kJ/mol

ΔH_f(H₂O) = −241.8 kJ/mol

ΔH_c = -2ΔH (C₂H₆) + 4ΔH_f(CO₂) + 6ΔH_f(H₂O) = -(–1600.0 kJ/mol) +4(−393.5 kJ/mol) + 6(−241.8 kJ/mol) = -1424.8 kJ/mol