What is the heat of formation of Benzoic acid? I have asked this question before but made a mistake in my question. The molar enthalpy of combustion for benzoic acid is -3226.6 kJ/mol. The thermochemicasl equation is: 2C6H5COOH (l) + 15O2 -> 6H2O (l) + 14CO2 (g)ΔH= -3226.6 kJ/mol Given the heats of formation for liquid water is -285.5 kJ/mol and carbon dioxide is -393.5 kJ/mol, calculate the heat of formation of benzoic acid.Please show all proper units because that is where I got mixed up (how do I change -3226.6 kJ/mol into kJ to be used in the standard heats of formation to find x?)

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First we try to arrange the given value for the enthalpy of formation of water and Carbon dioxide.

H2 (g) + 1/2 O2 (g)  --> H2O (l)      ΔH = -285.5

C(s) + O2 (g) --> CO2 (g)              ΔH = -393.5

and we...

(The entire section contains 114 words.)

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