Question

Calculate ΔH∘f (in kilojoules per mole) for benzene, C6H6, from the following data:
2C6H6(l)+15O2(g)→12CO2(g)+6H2O(l) ΔH∘ = -6534kJ
ΔH∘f (CO2) = -393.5kJ/mol
ΔH∘f (H2O) = - 285.8kJ/mol

I thought I was supposed solve for benzene by doing the products - reactants thing, so I did: { 12(-393.5) + 6(-285.8)} - x = -6534 but the answer I got (97.2) was incorrect.

Answer 1

delta H for the reaction = [sum of heat of formation for products] - [sum of heat of formation for reactants]

(-6534) = [(12)(-393.5) + (6)(-285.8)] - (2X)

(-6534) = -6436.8 - 2X

2X = 97.2

X = +48.6 kJ, which is the heat of formation for benzene.

Answer 2

Thank you so much!