Question

Deteriminati entalpia de formare standard a enterolului,C2H5OH cunoscand reactia termochimica care are loc: C2H5OH+3O2g->2CO2g+3H2O1
^1H=-1365kj si entaliile
de formare standard.
^f H°CO2(g)=-393,5KJ/mol
^f H°H2O=-285,KJ/mol
.

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Answer 1

C2H5OH(I) + 3O2(g) --> 2CO2(g) + 3H2O(I)

ΔHr° = ∑ΔHr°produsi - ∑ΔHr°reactanti

ΔHr° = [3ΔHf°H2O(l) + 2ΔHf°CO2] - [3ΔHf°O2 + ΔHf°C2H5OH]

-1365 = [3(-285,5) + 2(-393,5)] - [0 + ΔHf°C2H5OH]

=> ΔHf°C2H5OH = -278,5 kJ/mol