[MilkyCar]

I'm confused and I've been trying to solve this problem.

So it states that given the following data:
C2H4 (g) + 3O2 (g) -> 2CO2 (g) + 2H2O (l)  H= -1411.0 kJ
2C2H6 (g) + 7O2 (g) -> 4CO2 (g) + 6H2O (l) H = -3119.8 kJ
2H2 (g) + O2 (g) -> 2H2O (l)  H = -571.7 kJ

And I have to calculate H for the reaction:
C2H4 (g) + H2 (g) -> C2H6 (g)

I didn't change anything for equation one and then I did -1411.0 - -3119.8 + (-571.7) and then I divided the answer by 2 to get 568.6 kJ. I am stuck and I'm unsure how to get the right answer.

[Hunter2]

Sum of rectants minus sum of products.
Substitut C2H4 + H2 => C2H6 to the given equations.
You take the first equation how it is for C2H4.
For H2 you have to take the building of water by half and for C2H6 you also have to subtract the second equation by half.

[MilkyCar]

Okay, I tried that and I got -137.0 kJ as a result. Is this correct?

[Hunter2]

Its correct

[Borek]

Assuming I made no typos, you took first equation, subtracted second and added third, yes?

  C₂H₄ + 3O₂  2CO₂ + 2H₂O
- (2C₂H₆ + 7O₂  4CO₂ + 6H₂O)
+ (2H₂ + O₂  2H₂O)
--------------------------------------------
C₂H₄ + 3O₂ - 2C₂H₆ - 7O₂ + 2H₂ + O₂  2CO₂ + 2H₂O - 4CO₂ - 6H₂O + 2H₂O

sorting
C₂H₄ - 2C₂H₆ + 3O₂ - 7O₂ + O₂ + 2H₂  2CO₂ - 4CO₂ + 2H₂O - 6H₂O + 2H₂O

combining
C₂H₄ - 2C₂H₆ - 3O₂ + 2H₂  -2O₂ - 2H₂O

moving things with negative coefficients to other side
C₂H₄ + 2H₂ + 2CO₂ + 2H₂O  2C₂H₆ + 3O₂

Is the final equation the one you are looking for?

[Hunter2]

No
C2H4 + 3 O2 => 2 CO2 + 2 H2O -1411 kJ
Convert first
C2H6 + 3,5 O2 => 2 CO2 + 3 H2O -1559,9 kJ
C2H6 = - 3,5 O2 + 2 CO2 + 3 H2O
and H2 + 0,5 O2 => H2O  - 285,85 kJ
H2 => H2O - 0,5 O2

C2H4 + 3 O2 + H2 - 3,5 O2 + 2 CO2 + 3 H2O => 2 CO2 + 2 H2O + H2O - 0,5 O2 + C2H6
C2H4 + H2 => C2H6
-1411 kJ + (-285,85 kJ) -(-1559,9kJ) = 136,95 kJ