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Topic: Gibbs free energy and Gibbs standard free energy  (Read 2292 times)

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Offline assaftolko

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Gibbs free energy and Gibbs standard free energy
« on: July 19, 2013, 05:47:09 PM »
I'm reading Lehninger's Principles of Biochemistrey and I'm having trouble to understand biochemistry's defenition for ΔG'. It stated in the book that this expression represents the change in gibbs free energy when we are at 298K, pH 7.0, 1 atm and when the initial molaric conc. of reactants and products are 1M each. But ther's something I don't quite understand:
for the general reaction: Aa+Bb ::equil::Cc+Dd
ΔG=ΔG'+RTln[C]c*[D]d/[A]a*(B)b
But in equillibrium: ΔG=0 so:

ΔG'=-RTlnKeq
But Keq is a constant value for a given temprature, and is independent of the initial conc. of the reactants and products. Thus - if the reaction is done at 298K, I will get the same value for ΔG', no matter what will be [A],(B),[C],[D] at the beginning. Sure - different initial conc. of reactents and/or products will give different equillibrium conc. for A,B,C,D - but the ratio of their conc. at equillibrium (which defines Keq) has to stay the same! If this is the case - why does it matter that ΔG' is calculated only if the initial conc. at standard conditions of both reactants and products are 1M???

Offline Corribus

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Re: Gibbs free energy and Gibbs standard free energy
« Reply #1 on: July 19, 2013, 07:24:51 PM »
The Standard Gibbs Energy Change is a reference point.  By comparing DG under a set of conditions to this reference point, the spontaneous direction of change can be determined.

Maybe this thread will help:

http://www.chemicalforums.com/index.php?topic=66737.msg241353#msg241353

(See my long post on the second page.)
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline assaftolko

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Re: Gibbs free energy and Gibbs standard free energy
« Reply #2 on: July 20, 2013, 01:10:36 AM »
But where do you reffer to the reason that the initial conc. for ΔG0 have to be 1M?

Offline magician4

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Re: Gibbs free energy and Gibbs standard free energy
« Reply #3 on: July 20, 2013, 05:40:17 AM »
Quote
(...) and when the initial molaric conc. of reactants and products are 1M each.

I don't know the exact paraphrasing Lehninger uses in the book, but I see the source for your misunderstanding of the "per mole" bit in it...

o, let's say we had a reaction 1 A + 2 B  :rarrow: 2,5 C  + energy

how much energy would that be?
let's say, we gain X kJ here if the total consumption of A was 1 mole A

so, at standard conditions, our ΔG0 would be X kJ/mole (A consumption)

likewise, it follows that there would be another ΔG0 related to the consumption of B:
ΔG0 would be Y kJ/mole (B consumption)
with Y being half of X , of course

...and a third one referring to the formation of C:
ΔG0 would be Z kJ/mole (C formation)
with Z being 0.4 of X

...and last not least a more general one referring to a general equation of conversion:

x*A + 2x*B  :rarrow: 2,5*x C

where one mole of conversion of the above type would result in ΔG0 = x*X kJ/mole "conversion as shown by equation"
now, you could choose x to be "1 mole" ...
... and that's what Lehninger must have meant, I take it

regards

Ingo
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