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 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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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...

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?)

Expert Answers

jerichorayel

| Certified Educator

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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