## 8.57

Posts: 14
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### 8.57

We did this problem in discussion but I'm still confused why we used -286 KJ/mol of H2 instead of 0. In our reader it says that the standard enthalpy of formation of an element in its most stable form is zero so I think I just need a clarification

Mikaela Gareeb 3G
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### Re: 8.57

You're adding up the enthalpies of the reaction. You are given an equation and the enthalpies of combustion for each of the individual molecules, so when you create a combustion equation for each of the molecules (remember that combustion is a molecule + oxygen -> carbon dioxide + water) you will have a reaction enthalpy. Then you must simplify the three equations into the one that you were given; the first equation will stay the same so the delta H is still -1300 kJ, but you flipped the second one so the delta H will be 1560 kJ, and finally you have to multiply the reaction by 2 in order to have a balanced final equation, thus the last delta H is 2(-286 kJ). After solving for your reaction enthalpies for all the equations you used to attain your final equation, just add them. The 2(-285 kJ) isn't for the H2 per say, but instead for the reaction that helped you get H2 for the final equation.

Chem_Mod
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### Re: 8.57

To clarify further, the value provided is $\Delta$Hc. This indicates the change in enthalpy when the molecule is combusted (i.e. burned in the presence of oxygen). In other words, $\Delta$Hc= -286 kJ/mol is the change in heat as H2 is burned to form H2O.

Thus, this question is not asking about the Standard Enthalpy of Formation for H2 as that would be zero like you initially thought. This problem actually deals with Hess's Law. Just be mindful when looking at the subscript of $\Delta$H as c= combustion and f= formation.

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### Re: 8.57

Thank you!

604735966
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### Re: 8.57

Why are the products H2O and CO2? Why wouldn't C2H2, C2H6, and H2 be the products when finding the equations?

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