Mean Bond Enthalpy

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Rachael_1H
Posts: 31
Joined: Fri Sep 25, 2015 3:00 am

Mean Bond Enthalpy

Postby Rachael_1H » Wed Jan 13, 2016 11:21 am

Using bond enthalpy to determine the enthalpy change of the rxn: H2(g) + (.5)O2(g) -----> H2O(l)

The equation used is sum of bonds broken minus sum of bonds formed. Using the book's given bond enthalpies of diatomic molecules and mean bond enthapies, the answer key has the equation used as (H2 bond enthalpy+(.5)(O2 bond enthalapy)-((2)(H-O mean bond enthalapy))

Why did they use the mean bond enthalpy of H-O bond, and then multiply it by 2 moles, when there is only 1 mole of H2O in the equation?

Srivinay_Irrinki
Posts: 20
Joined: Fri Sep 25, 2015 3:00 am

Re: Mean Bond Enthalpy

Postby Srivinay_Irrinki » Wed Jan 13, 2016 12:46 pm

The reason that the H-O bond enthalpy is multiplied by 2, is that there are two H-O bonds within the H2O molecule. That means that two H-O bonds are formed during this reaction, which is why you must subtract the H-O bond enthalpy value twice.

Aya Ghoneum 1H
Posts: 22
Joined: Fri Sep 25, 2015 3:00 am

Re: Mean Bond Enthalpy

Postby Aya Ghoneum 1H » Wed Jan 13, 2016 2:18 pm

You’re right, there is only one mole of water molecules in the reaction, however each water molecule is made up of TWO H-O bonds (think about the Lewis structure). That is why you need to multiply the bond enthalpy for the H-O bond by 2. I hope this helps!


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