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Topic: Enthalpy Reactions, +/-, Product/Reactant Favored Endo/Exo thermic  (Read 8292 times)

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Offline stev3n

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I'm just struggling with if a chemical reaction is product or reactant favored, endothermic and exothermic, and how to use +/- when calculating. I know this is a lot but I'm including an example that I think can get everything answered.

Calculating the enthalpy of combustion per gram for propane.

C3H8 + 5 O2 -> 3CO2 + 4 H2O

3(-393.5 kJ/mol) + 4(-241.8 kJ/mol) - (-104.7 kJ/mol) = 2043 kJ/mol.

(2043 kJ/mol) (1 mol C3H8 / 46.10 g C3H8) = -46.33 kJ/g

Okay so now my questions are:

I understand that one side of the equation has to be negative, but why is it the propane? Is it because the reactions is causing the decomposition of it?

Since the C3H8 is negative, does that mean the reaction is product favored?

Is this an endothermic reaction because the propane is gaining heat?

Offline macman104

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Re: Enthalpy Reactions, +/-, Product/Reactant Favored Endo/Exo thermic
« Reply #1 on: October 23, 2008, 12:28:57 AM »
 :delta: Hreaction =  :delta: Hproducts - :delta: Hreactants

That's why your propane "side" is the negative side.

Offline stev3n

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Re: Enthalpy Reactions, +/-, Product/Reactant Favored Endo/Exo thermic
« Reply #2 on: October 23, 2008, 12:39:05 AM »
:delta: Hreaction =  :delta: Hproducts - :delta: Hreactants

That's why your propane "side" is the negative side.

Ah thank you. I was over thinking it and forgot that principle.

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