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Non Life Sciences => Chemistry => Topic started by: lancenti on 25/02/2009 08:32:05

Title: Why is the Gibbs Free Energy Curve of an Equilibrium Reaction U-Shaped?
Post by: lancenti on 25/02/2009 08:32:05

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 (http://www.oup.com/uk/orc/bin/9780199249732/freelecturer/figures/ch08/)

Title: Re: Why is the Gibbs Free Energy Curve of an Equilibrium Reaction U-Shaped?
Post by: Chemistry4me on 25/02/2009 08:33:26

What picture are you trying to show?

Title: Why is the Gibbs Free Energy Curve of an Equilibrium Reaction U-Shaped?
Post by: lancenti on 25/02/2009 08:36:02

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

Title: Why is the Gibbs Free Energy Curve of an Equilibrium Reaction U-Shaped?
Post by: Chemistry4me on 25/02/2009 08:39:06

This should be better.

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Title: Why is the Gibbs Free Energy Curve of an Equilibrium Reaction U-Shaped?
Post by: lancenti on 25/02/2009 09:19:26

Much better, but I still need an answer.

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

A [revarrow] B

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

ΔG_state = xΔG_A + (1-x)ΔG_B

x: Percentage of A at a certain state.

But since ΔG = ΔH - TΔS then,

ΔG_state = x(ΔH_A-TΔS_A) + (1-x)(ΔH_B-TΔS_B)

Expanding and rearranging, we get...

ΔG_state = xΔH_A - xTΔS_A -TΔS_B - xΔH_B + xTΔS_B + ΔH_B

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

So, using implicit differentiation we get...

δΔG_state/δx = ΔH_A - Tδ(xΔS_A)/δx -TδΔS_B/δx - ΔH_B + Tδ(xΔS_B)/δx

δΔG_state/δx = ΔH_A - TΔS_A - TxδΔS_A/δx -TδΔS_B/δx - ΔH_B + TΔS_B + TxδΔS_B/δx

Taking ΔH = ΔH_A - ΔH_B

δΔG_state/δx = ΔH + T(ΔS_B - ΔS_A - δΔS_B/δx) + Tx(δΔS_B/δx - δΔS_A/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 ΔS_A and ΔS_B. I've found the equation ΔS_state = ΔS^o - Rln[C], but have no idea how to use it and clearly this is for a one-element system.

Title: Why is the Gibbs Free Energy Curve of an Equilibrium Reaction U-Shaped?
Post by: lancenti on 25/02/2009 11:23:54

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...

Title: Why is the Gibbs Free Energy Curve of an Equilibrium Reaction U-Shaped?
Post by: lightarrow on 27/02/2009 13:38:43

Quote from: lancenti on 25/02/2009 11:23:54

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 ΔG_reaction 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).

Title: Why is the Gibbs Free Energy Curve of an Equilibrium Reaction U-Shaped?
Post by: lancenti on 27/02/2009 15:29:17

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?

Title: Why is the Gibbs Free Energy Curve of an Equilibrium Reaction U-Shaped?
Post by: lightarrow on 27/02/2009 17:44:36

Quote from: lancenti on 27/02/2009 15:29:17

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.