Cal. enthalpy of formation of acetic acid CH 3 COOH ,if its enthalpy of combustion is -867kl/mol . The enthalpies of formation of CO2 and H2O are -393.5 and -285.9 kj/mol respectively?
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CH3COOH+2O2=2CO2+2H2O
Given:
H2O=-285.9 kj/mole
CO2=-393.5 KJ/MOLE
CH3COOH=-867 KJ/MOLE
[2(-393.5)+2(-285.9)]-[2(0)+(-867)]
=[-787-571.8]-[-867]
=-787-571.8+867
=-491.8 KJ/Mole
Given:
H2O=-285.9 kj/mole
CO2=-393.5 KJ/MOLE
CH3COOH=-867 KJ/MOLE
[2(-393.5)+2(-285.9)]-[2(0)+(-867)]
=[-787-571.8]-[-867]
=-787-571.8+867
=-491.8 KJ/Mole
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