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

ΔH 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 = [(-4722) + (-1714.8)] - 2X

-6534 = -6436.8 - 2X

-6534 + 6436.8 = -2X

-97.2 = -2X
 
then, divide each side by two to cancel out the numerical coefficient.

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

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