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?)

Accepted Answer

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...

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