How to calculate the heat of formation of benzoic acid?
Benzoic acid can be burned to produce liquid water and carbon dioxide. The molar enthalpy of combustion for benzoic acid is -3226.6 kJ/mol.
2C6H5COOH (l) -> 6H2O (l) + 14CO2 (g) ΔH= -3226.6 kJ/mol
Given the heats of formation for liquid water is -285.5kJ/mol and carbon dioxide -393.5kJ/mol, calculate the heat of formation of benzoic acid.
I need full calculations for this question, please.
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-1613.3= -3611 - ΔH(C6H5.....)
ΔH(C6H5.....)= -3611+ 1613.3 .................
first div reaction by 2
C6H5COOH (l) -> 3H2O (l) + 7CO2 (g) ΔH= -1613.3 kJ/mol
ΔH(rxn)=3ΔH(h20)+ 7ΔH(co2) - ΔH(C6H5.....)
put values and solve ull get value of ΔH(C6H5.....)
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