8.57

[Madison_Moore_1I]

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]

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]

To clarify further, the value provided is H_c. This indicates the change in enthalpy when the molecule is combusted (i.e. burned in the presence of oxygen). In other words, H_c= -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 H as c= combustion and f= formation.

[Madison_Moore_1I]

Thank you!

[604735966]

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