December 22, 2024, 11:02:59 PM
Forum Rules: Read This Before Posting


Topic: Gibb's Free Energy & Equilibrium confusion  (Read 6576 times)

0 Members and 1 Guest are viewing this topic.

Offline nelle178

  • New Member
  • **
  • Posts: 3
  • Mole Snacks: +0/-0
Gibb's Free Energy & Equilibrium confusion
« on: April 03, 2008, 02:25:12 AM »
Hello - I've been trying to figure this out for days. :)

The concept is Gibbs Free energy - I am attempting to understand the equation for it, DG=DG°+RTlnK.  D, the change in free energy, is defined as final minus initial. But what exactly is this final and initial for a reaction, what is this change being measured to and from? Since all reactions are driven towards equilibrium, is that “final”?

Holding that assumption, what is then DG° (delta G standard)? Is it the change in free energy from the point where all components of the system are at 1M to the point where all
components are at their equilibrium concentrations? OR, as I’ve read in a definition somewhere, is it the change in free energy as 1M concentrations of all the reactants are driven to 1 M concentrations of the products, with no reactants remaining at the end?

Let’s say you have a reaction of A converting to B. Would DG° be achieved when you take 1M A and convert it completely into B? Could DG° also be determined by measuring the DG of 1 M A (0M B) going to equilibrium, and subtracting the DG resulting from a separate reaction of 1 M B (0M A) going to equilibrium? These make sense, but I can’t see how putting 1 M concentrations of both A and B into solution and having them reach equilibrium will give the same value for DG°.

Now, K can only be put into the equation if you look at the last scenario – the first two give different K’s. Which brings me to – how can you ever have an equation relating a change in G with a static point in concentrations? Especially if you look at DG=DG°+RTlnK as a function, and try to see a mathematical relationship… For every ratio of concentrations you have… a change, even though the concentrations don't change? Is it a potential change of reaching equilibrium? A chemical potential? My first instinct was derivatives, but for those, don’t you need a change in both x and y, while here you only have a change in one?

Maybe what’s confusing me most is this graph I have of G vs. reaction progress (100% A at point 0 going to 100% B somewhere further along the axis). I have a lot of ideas on what might be happening, but I need some feedback on what might actually be true.
Thanks in advance!! Sorry for the huge post.

Offline Yggdrasil

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3215
  • Mole Snacks: +485/-21
  • Gender: Male
  • Physical Biochemist
Re: Gibb's Free Energy & Equilibrium confusion
« Reply #1 on: April 03, 2008, 06:26:07 PM »
OR, as I’ve read in a definition somewhere, is it the change in free energy as 1M concentrations of all the reactants are driven to 1 M concentrations of the products, with no reactants remaining at the end?

This is the correct definition of ΔGo

Quote
Let’s say you have a reaction of A converting to B. Would DG° be achieved when you take 1M A and convert it completely into B?

Yes.

Quote
Could DG° also be determined by measuring the DG of 1 M A (0M B) going to equilibrium, and subtracting the DG resulting from a separate reaction of 1 M B (0M A) going to equilibrium?

This would work as well.

Quote
These make sense, but I can’t see how putting 1 M concentrations of both A and B into solution and having them reach equilibrium will give the same value for DG°.

Consider your first two experiments.  In beaker A, you let 1M A go to equilibrium and in beaker B you let 1M B go to equilibrium, then mix the two solutions together.  Would this give the same result as if you had just mixed beaker A and beaker B together before the reaction occurred, rather than after?

Quote
DG=DG°+RTlnK

This equation is not correct.  The correct equation is ΔG = ΔGo + RT lnQ, where Q is the reaction quotient.  Q is not a static concentration, but it changes with the composition of your reaction mixture.  From thermodynamics, we know that systems will always try to minimize their free energy.  Therefore, at equilibrium, free energy can not go any lower, so ΔG = 0.  Also, K is defined as the reaction quotient at equilibrium, so we get the following formula at equilibrium:

0 = ΔGo + RT lnK

Which rearranges to K = e-ΔGo/RT, the correct relation between ΔGo and K.

Offline nelle178

  • New Member
  • **
  • Posts: 3
  • Mole Snacks: +0/-0
Re: Gibb's Free Energy & Equilibrium confusion
« Reply #2 on: April 04, 2008, 04:37:29 PM »
Thanks, that made a lot of sense. Finally.  :)

Sponsored Links