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Author Topic: Why is the Gibbs Free Energy Curve of an Equilibrium Reaction U-Shaped?  (Read 7382 times)

Offline lancenti

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I know it's totally intuitive, since the reaction proceeds towards the equilibrium state therefore it must have the most negative Gibbs Free Energy but how do we actually get the curve?

http://www.oup.com/uk/orc/bin/9780199249732/freelecturer/figures/fig8_5.tif

Source: Oxford University Press
« Last Edit: 25/02/2009 08:34:42 by lancenti »


 

Offline Chemistry4me

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What picture are you trying to show?
 

Offline lancenti

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The picture of the curves. Unfortunately I have no idea how to use the image thing for the forum.
 

Offline Chemistry4me

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This should be better.

 

Offline lancenti

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Much better, but I still need an answer.

Considering the equilibrium reaction starting with either 1.00 mol of either A or B,

A reversible arrow B

I have thus far I've worked out that...

ΔGstate = xΔGA + (1-x)ΔGB

x: Percentage of A at a certain state.

But since ΔG = ΔH - TΔS then,

ΔGstate = x(ΔHA-TΔSA) + (1-x)(ΔHB-TΔSB)

Expanding and rearranging, we get...

ΔGstate = xΔHA - xTΔSA -TΔSB - xΔHB + xTΔSB + ΔHB

I know to that...
1. ΔGequilibrium is when ΔGstate is a minimum.
2. δΔGstate/δx should still have an x-term in it, and would be equal to zero at equilibrium

So, using implicit differentiation we get...

δΔGstate/δx = ΔHA - Tδ(xΔSA)/δx -TδΔSB/δx - ΔHB + Tδ(xΔSB)/δx

δΔGstate/δx = ΔHA - TΔSA - TxδΔSA/δx -TδΔSB/δx - ΔHB + TΔSB + TxδΔSB/δx


Taking ΔH = ΔHA - ΔHB

δΔGstate/δx = ΔH + T(ΔSB - ΔSA - δΔSB/δx) + Tx(δΔSB/δx - δΔSA/δx) 

From here I'm stuck. I'm not aware of how ΔS varies w.r.t. x. I know I probably have to re-express it before I can differentiate to show that the curve has a minimum, or am I missing something that is totally obvious? I'm also not sure if there's a relation between ΔSA and ΔSB. I've found the equation ΔSstate = ΔSo - Rln[C], but have no idea how to use it and clearly this is for a one-element system.
« Last Edit: 25/02/2009 09:48:43 by lancenti »
 

Offline lancenti

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Alternatively, for the same reaction, I thought that maybe this equation would help:

ΔG = -RT ln K

ΔG = -RT ln ([B']/[A])

Using B' since the forum turns it into the bold command.

Since we're using 1.00 mol in the beginning, then

ΔG = -RT ln ((1-[A])/[A])

Since we're using x = [A] to make it look neater,

ΔG = -RT ln ((1-x)/x)

Differentiating w.r.t. x,

δΔG/δx = - RT x/(1-x) * δ((1-x)/x)/δx

δΔG/δx = - RT x/(1-x) * - x-2

δΔG/δx =  RT /[x(1-x)]

But this expression does not have a δΔG/δx = 0 so obviously I'm missing something...


« Last Edit: 25/02/2009 11:25:32 by lancenti »
 

Offline lightarrow

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Alternatively, for the same reaction, I thought that maybe this equation would help:

ΔG = -RT ln K

ΔG = -RT ln ([B']/[A])

Using B' since the forum turns it into the bold command.

Since we're using 1.00 mol in the beginning, then

ΔG = -RT ln ((1-[A])/[A])

Since we're using x = [A] to make it look neater,

ΔG = -RT ln ((1-x)/x)

Differentiating w.r.t. x,

δΔG/δx = - RT x/(1-x) * δ((1-x)/x)/δx

δΔG/δx = - RT x/(1-x) * - x-2

δΔG/δx =  RT /[x(1-x)]

But this expression does not have a δΔG/δx = 0 so obviously I'm missing something...
You can't evaluate the variation of G in that way because ΔGreaction is constant at constant temperature (ΔG = -RT ln K and K depends *only* on the kind of reaction, it can't vary), it's a number. At constant temperature you have to use dG = dH - TdS. To compute dH is easy, you sum the dH of the two separated components according to their amounts, but to compute dS you can't do it, because the entropy of a mix is not the sum of entropies of the two chemicals (it depends on how they are mixed, for example).
« Last Edit: 27/02/2009 13:40:33 by lightarrow »
 

Offline lancenti

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Hmm, that's right but once re-expressed isn't it that the Gibbs Free Energy at the state before equilibrium can be found this way?

As in, when I read the equation, I see it as Delta G depends on the logarithm of the ratio of B to A. So in a sense, it does vary but I may have made that doubtful by using K since K is the equilibrium constant.

Any suggestions how we could then evaluate this?
 

Offline lightarrow

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Hmm, that's right but once re-expressed isn't it that the Gibbs Free Energy at the state before equilibrium can be found this way?

As in, when I read the equation, I see it as Delta G depends on the logarithm of the ratio of B to A. So in a sense, it does vary but I may have made that doubtful by using K since K is the equilibrium constant.

Any suggestions how we could then evaluate this?
Not at the moment. In case I'll post something.
 

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