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Hess's Law with Three Equations

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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.

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.

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

Its correct

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

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

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

C2H4 - 2C2H6 - 3O2 + 2H2  :rarrow: -2O2 - 2H2O

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

Is the final equation the one you are looking for?


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