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...

Question

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?)

Answer 1 · 2012-10-27T10:16:21+00:00

jerichorayel | College Teacher | (Level 2) Senior Educator

Posted on October 27, 2012 at 10:16 AM

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 rewrite the heat of combustion of benzoic acid

2C6H5COOH (l) + 15O2 -> 6H2O (l) + 14CO2 (g)
ΔH= -3226.6 kJ/mol

6H2O (l) +14CO2 ---> 2C6H5COOH + 15O2 (g) 3226.6kJ/mol

14C(s) + 14O2 (g) ---> 14CO2 (g) -5509 kJ/mol

2H2(g) + O2 (s) ----> 2H2O (l) -571 kJ/mol

----------------------------------------------------------

4H2O + 14C + 2H2 ---> 2C6H5COOH -2853.4 kJ/mol

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