Complete Combustion of Hydrocarbons or Compounds with Only C, H $\Delta$

Complete combustion of hydrocarbons or compounds with only C, H, and O gives CO₂ and H₂O as the only products. If carried out under standard conditions, the CO₂ is a gas and the H₂O is a liquid. Given these standard enthalpies of combustion: C₆H₁₂(l) = -3919.86 kJ mol^-1, C₆H₆(l) = -3267.80 kJ mol^-1, H₂(g) = -285.90 kJ mol^-1, C(s) = -393.50 kJ mol^-1, calculate $\Delta$ H°_reaction for the reaction,C₆H₆(l) + 3 H₂(g) $\rarr$ C₆H₁₂(l) Hint: Remember it is always products minus reactants when performing enthalpy calculations.

A) -205.64 kJ
B) +366.16 kJ
C) +759.66 kJ
D) +2155.36 kJ
E) +5684.36 kJ