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Enthalpy


GeneralOJB

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Am I right in saying that, in an exothermic reaction for example, the reactants have a higher enthalpy, and the products have a lower enthalpy, as seen from the energy profile diagram.
But the reactants have a lower bond enthalpy, and the products have a higher bond enthalpy, because a lower amount of energy is required to break the bonds in the reactants, but a higher amount of energy is released when the new bonds are formed. This just confused me as I thought enthalpy and bond enthalpy were the same thing!

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  • 2 months later...

You are right that reactants have a higher entalphy than products in an exothermic reaction. However,I think you have confused yourself a bit about bond enthalpy.

I will try to explain. Let me give an example:

be1a.gif

(sourcehttp://www.kentchemistry.com/links/Kinetics/BondEnergy.htm )

 

In this reaction you can calculate the bond enthalpy like this:

be1b.gif

 

Here, you can see that the product bond enthalpies are bigger than the reactant bond enthalpies (1852 is bigger than 1371). This means, as you said, that a lower amount of energy is required to break the reactant bonds than what is released when the product bonds are formed.

 

If you would have calculated without using bond enthalpy you would calculate (total reactant enthalpy - total product enthalpy). Here, you are thinking that 'the stored energy in these compounds/elements are different and so energy is either released or taken up'. But when using bond enthalpy you instead think that 'energy is released when the product bonds are created'.

 

So the one involves only using 'stored energy' while the other involves 'energy stored in the reactants minus energy released'.

 

That is to say, you are correct about what you have written. The confusion is just that one should think differently when calculating the enthalpy difference using bonds. This is what makes it seem that the product got 'higher enthalpy'; but in fact it is just showing that the product is 'releasing energy with a higher value than how much the reactant taking up energy'.

 

I hope that I have explained ok without writing too much.

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