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Topic: Bond enthalpy  (Read 2938 times)

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Offline Rutherford

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Bond enthalpy
« on: June 29, 2012, 01:49:46 PM »
When bond dissociation energies are given, why is entalphy calculated so that the bonds that make the reactants are deducted from the bonds that the products are made of? When the heat of formation is given it is opposite. Why is this different?

Offline blaisem

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Re: Bond enthalpy
« Reply #1 on: June 29, 2012, 03:42:05 PM »
The bond dissociation energy is the energy needed to break a bond.  The enthalpy of formation is the energy needed in forming a bond.  A negative enthalpy of formation corresponds to a bond that is favorably formed; in other words, this is the energy that holds the molecule together and is the energy that must be overcome to break the bond.  This means that the enthalpy of formation is the negative value of the bond dissociation energy.  Essentially, both terms involve the energy of bonds, but the sign (positive or negative) is defined differently.

I believe this is a somewhat qualitative comparison of the bond dissociation energy and the heat of formation because the two values may be derived differently.  Someone else may be able to clarify this.

For the enthalpy of formation, the reactant values are subtracted from the product values.  This again is relevant to the sign used in the enthalpy of formation.  If a reactant has a negative enthalpy of formation, then it is formed favorably, and that energy must be overcome before it will be broken apart.  If the product also has a negative enthalpy of formation, then its formation is also favorable.  The question is, which one is more favorable?  The answer is the one with the larger negative enthalpy of formation because this will lead to a more negative enthalpy overall, and this is why you subtract the values.  It should be noted that by favorable, I mean the reaction is enthalpically favored.  Whether the overall reaction is favorable involves entropy as well.

If you use the same formula using bond dissociation energies, given that the sign of a bond dissociation energy is the negative of the enthalpy of formation (again the values may not be 100% exact, depending on how they are derived), you end up with the negative value of the enthalpy of the reaction.  That's why you multiply by negative one to get the enthalpy of the reaction.

I find this website helpful: http://www.science.uwaterloo.ca/~cchieh/cact/c120/heatreac.html

Offline Rutherford

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Re: Bond enthalpy
« Reply #2 on: June 30, 2012, 05:45:38 AM »
I understood when the enthalpy of formation is used: if the product has a more negative enthalpy of formation than the reactant it is more stable and ΔH will be negative indicating that the reaction is more favored (which can be seen from the formula:ΔG=ΔH-TΔS).

If I write a reaction:
1/2H2+1/2Cl2 :rarrow: HCl,how is the enthalpy of formation of HCl the same as the bond creation energy of H-Cl? When I calulate ΔH:
1. way using enthalpies: ΔH=ΔHf(HCl)
2. way using bond enthalpies: ΔH=ΔHd(H-H)+ΔHd(Cl-Cl)-ΔHd(H-Cl)
From these 2 equations ΔHf(HCl) doesn't seem equal to the ΔHd(H-Cl) (I used d as dissociation).

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