Use values of standard enthalpies of formation to calculate the enthalpy change for the reaction below:

Question

Use values of standard enthalpies of formation to calculate the enthalpy change for the reaction below:

C2H5OH (l)+3O2(g)——>2CO(g)+3H2O(l)

give the quantity and unit (kJ) and don’t forget the + or - sign.

Answer 1

∑∆H_{formation products} - ∑∆H_{formation reactants} = ∆H_reaction

(note: I believe you have written an incorrect equation in the question since it isn't balanced. I have written the correctly balanced equation for the complete combustion of C₂H₅OH below)

C₂H₅OH(l) + 3O₂(g) ===> 2CO₂(g) + 3H₂O(l) ... balanced equation

-277.6.............0.....................-393.5..........-285.8........values are in kJ/mole

∑∆H products = 2x-393.5 + 3x-285.8 = -787 + -857.4 = -1644.4 kJ

∑∆H reactants = -277.6 kJ

∆Hreaction = -1644.4 kJ - (-277.6 kJ) = -1366.8 kJ (reaction is exothermic)