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?)
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
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