Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Enthalpy Change using Enthalpies of Formation?
Don't know how to tackle this problem, has been puzzling me for about 2 hours and I just don't understand!!!
I'm not interested in the final answer, I'm hoping to find out HOW to solve the equation.
Use the enthalpies of formation below to calculate the enthalpy change for the following reaction:
3 Fe(s) + 4 H2O(g) → 4 H2(g) + Fe3O4(s)
ΔH f: H2O(g) -242; Fe3O4(s) -1117 kJ mol-1
All help is highly appreciated!!!!!
Still struggling on an understanding of this. Can anyone explain why the process taken is taken? And explain specifically how to calculate it? Sorry for being such an eediot!
3 Answers
- Caroline MillerLv 79 years agoFavorite Answer
∆Hrxn = ∆Hformation all products - ∆Hformation all reactants
the thing is this: elements don't have ∆Hformation values since they are elements
plus, using the balanced equation, you need to take kJ/mole and multiply by the number of moles of the compounds, for Fe3O4 = 1 and H2O = 4
∆Hrxn = ∆Hf Fe3O4 - ∆Hf H2O
∆Hrxn = -1117kJ - (4)(-242kJ) = -149kJ
- ChemTeamLv 79 years ago
Here are some solved example problems using standard enthalpies of formation:
- Anonymous9 years ago
it is the delta h of the products - the delta h of the reactants. you have to multiply the enthalpy of formation of each compound by their stoichiometric coefficients.