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Enthalpy of formation

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Given the following heats of combustion.

CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(l)ΔH°rxn = -726.4 kJ
C(graphite) + O2(g) → CO2(g)ΔH°rxn = -393.5 kJ
H2(g) + 1/2 O2(g) → H2O(l)ΔH°rxn = -285.8 kJ
Calculate the enthalpy of formation of methanol (CH3OH) from its elements.
C(graphite) + 2 H2(g) + 1/2 O2(g) → CH3OH(l)
Mar 19th, 2015

Hess law is employed here

i.e ΔH°rxn = Σ ΔH°comb, products minus Σ ΔH°comb, reactants

so ΔH°rxn =(726)-[(-393)+(2*-285.8)+(0.5*-285.8)]

ΔH°rxn= -381.5 kJ

Mar 19th, 2015

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