# Delta H formation calculation.

Discussion in 'MCAT Discussions' started by EECStoMed, 08.04.07.

1. ### EECStoMedPersistence > Intel

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I've always thought that it's always products minus reactants. However, it seems that in order to calculate the heat of formation of something you do the opposite:

Delta H formation = Energy input of reactants - energy output of products.

Is this right?
2. ### gymgirl

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Yeap, because Enthalpy is the ( Energy of breaking bonds - Energy of Making Bonds).
The way to remember this is, you are breaking the bonds of the reactants first to "form" the products, therefore giving you the heat of "formation".
3. ### EECStoMedPersistence > Intel

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What does this mean?

The combustion of methane gas is represented by the reaction: CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
The calculation for the heat of combustion for methane is then:
heat of combustion =

[heat of formation of CO2 (g) + 2 x heat of formation of H2O (g)] -
[heat of formation of CH4 (g) + 2 x heat of formation of O2 (g)] = [ -393.5 kJ + 2x(-241.8 kJ)] - [-74.8 kJ + 2x(0 kJ)]
= - 802.3 kJ
4. ### bigman225

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Are you talking about when using bond-energies? Then its bonds broken - bonds formed. I remember this because bond breaking is endothermic while bond breaking is exothermic..
5. ### greg1184

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The easiesst way to do BDE questions is: DeltaH = (SUM)BDE reactants - (SUM)BDE products (Which is the oppisite when you do it with enthalpies.)
6. ### axp107UCLA 09': Italian Pryde

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It's just energy of bonds broken + bonds formed right?
7. ### EECStoMedPersistence > Intel

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Whoa, I'm hella confused now. Could someone please explain?
8. ### cheezerGuest

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there are three ways (that i know of) to solve for delta h. one-the one you mentioned- involves looking up the heat of formation for each molecule and doing a "products-reactants" taking into account the stoichometry of the reaction. another scenario is if you are given a set of lesser reactions, each with their own delta h, and manipulation is needed to solve for the delta h of the main reaction. yet another way is the method others are mentioning, which is to draw the structures out, count up the different types of bonds, look up the bond enthalpies of each, and then do "bonds broken-bonds formed," again taking stoichiometry into account
9. ### Zoom-Zoom

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Delta H = Bonds broken - Bonds formed

Since breaking bonds takes energy, and forming bonds releases energy, this basically means "Energy in - Energy out". This is easier to remember when you consider that a favorable enthalpy value is negative- in other words, you get more energy out than you put in. I think this is somewhat arbitrary: If a favorable delta H was positive, then it would be Bonds formed - Bonds broken, which probably makes a little more sense.

When you're summing enthalpy values it is Products - Reactants, just like everything else. You can do this because the enthalpy values you use already have the correct sign.
10. ### axp107UCLA 09': Italian Pryde

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Well the way I do it:

Delta H = bonds broken + bonds formed

I make the bonds broken positive, and bonds formed negative. That way I don't have to worry about which one comes first and I can use addition.

But I have never seen this type of question on an AAMC.. I've just seen simple H product - H reactant to give you enthalpy.