Re: [Phys-L] Energy & Bonds

[Jeffrey Schnick <JSchnick@Anselm.Edu>]

The qs have opposite signs.  The potential is negative.  The smaller r, the lower (more negative) the value of the potential energy.

> -----Original Message-----
> From: Phys-l [mailto:phys-l-bounces@phys-l.org] On Behalf Of Paul Lulai
> Sent: Tuesday, November 12, 2013 1:21 PM
> To: Phys-L@Phys-L.org
> Subject: Re: [Phys-L] Energy & Bonds
>
> 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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