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Re: [Phys-L] Energy & Bonds



Hi Paul

Well, the way I understand it (and I'd be interested to see if I'm in a minority) the convention is that the nominal zero is infinite separation (zero force acting), and the formula you give relates to how far below the nominal zero - so the formula as you have it is the energy needed to separate to infinity (which of course does get larger as r gets smaller).

Remember that in your formula you have q and q - these need to have opposite sign (giving a negative product) for the force to be attractive - so the closer (smaller r) the larger NEGATIVE value you have - the lower the energy level.

Does that help at all?

Best wishes

Keith



On 12/11/2013 18:21, Paul Lulai wrote:
How do I reconcile this with U=kqq/r ?
Isn't r getting smaller as they approach and bond? Then U is getting larger.

What is wrong with my model here?

-----Original Message-----
From: Phys-l [mailto:phys-l-bounces@phys-l.org] On Behalf Of Dr Keith S Taber
Sent: Tuesday, November 12, 2013 11:56 AM
To: Phys-L@Phys-L.org
Subject: Re: [Phys-L] Energy & Bonds

Hi Paul

In your terms

The products are more bound because they must be more bound and have LESS potential energy

i.e. they require more energy to disrupt the bonds as there is stronger binding

Best wishes

Keith


On 12/11/2013 15:01, Paul Lulai wrote:
Hi.
I am more aware of my conceptual shortcomings after the conversation about energy, reactions, and misconceptions.
I would appreciate clarification from the group on:
How do we reconcile the traditional Exothermic & Endothermic graphs of Energy vs reaction process with the fact that the products must be more bound and have greater potential energy? Typically the endo / exo graphs show (for an exothermic reaction) that the reactants have more energy than the products.

I may have more follow up questions.
Thanks.
Paul.
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