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1. Calculate*∆H for the reaction: C2H4 (g) + H2 (g) → C2H6 (g), from the following data.**
C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (l)* ∆H =*‐1411. kJ
C2H6 (g) + 3½ O2 (g) → 2 CO2 (g) + 3 H2O (l)* ∆H =*‐1560. kJ
H2 (g) + ½ O2 (g) → H2O (l)* ∆H =*‐285.8 kJ

I completely understand how to tackle the question up until this point:

I can flip equations and signs and balance equations so I have the correct enthalpies for each equation.

C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (l) ∆H1 X 1
C2H6 (g) + 3½ O2 (g) → 2 CO2 (g) + 3 H2O (l) ∆H2 X*‐1
H2 (g) + ½ O2 (g) → H2O (l) ∆H3 X 1

after this the solution to the question is

∆H =*∆H1*‐*∆H2 +*∆H3 =*‐1411 + 1560 – 285.8 =**‐*126.8 kJ

however every tutorial video I have watched tells me simply to add the enthalpies at the end to get ∆H.

and i was always taught ∆H= enthalpies of products- reactants

I have no idea why in this question the solution is to subtract a reactant from a product and a reactant.

*
2. Draw a Hess' Law triangle and write on the enthalpy changes you know first, then calculate the unknown enthalpy change from the diagram.

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Updated: August 15, 2016
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