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Topic: chemical potential "u" and free Gibbs energy "G"  (Read 4050 times)

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Offline xshadow

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chemical potential "u" and free Gibbs energy "G"
« on: October 17, 2015, 05:36:45 AM »
Hi!

I've a question about this two concepts in the calculation of the ΔG_r°  (where ° means standard )
In my textbook (atkins) the calculation of  ΔG_r° is so defined for a generic reaction  aA+bB ::equil:: cC+dD :

ΔG_r°=[c*ΔG°_f(C)+ d*ΔG°_f(D)] - [a*ΔG°_f(A)+b*ΔG°_f(B)]

Then,when introduces the chemical potential "u", it's  also defined as:
ΔG_r°= [c*u°(C)+ d*u°(D)] - [a*u°(A)+b*u°(B)]
Where u(i)= (dG/dn_i).

So the two expressions above are equal and in particular is ΔG°_f(i)= u°(i)
But i don't understand how it is possible!!!
 
u°(i)  is a derivate,an infinitesimal variation of G, while  ΔG°_f(i) is the real variation  ΔG that there is in the production of 1 MOLE  of the specie "i"

Someone can clear up me this??
Thanks !!!!
: )

Offline Plontaj

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Re: chemical potential "u" and free Gibbs energy "G"
« Reply #1 on: October 17, 2015, 11:01:43 AM »
Everything depends on conditions:

ui = d(G)/d(ni) when T,p = const
ui = d(A)/d(ni) when T,V = const
ui = d(H)/d(ni) when S,p = const
ui = d(U)/d(ni) when S,V = const

So, your equation: ΔG_r°= [c*u°(C)+ d*u°(D)] - [a*u°(A)+b*u°(B)]

is right when:
- temerature is constant dT=0
- pressure is constant dp=0

Change of Gibbs energy:
dG = dH - TdS = (dU+pdV+Vdp+dpdV) - TdS = dU + pdV - TdS

dG/dni = dH/dni - TdS/dni = dU/dni +pdV/dni - TdS/dni
dA/dni = dU/dni - TdS/dni = dH/dni - pdV/dni - Tds/dni

The comparison of d(U)/dni with d(G)/dni for conditions: S,V=const (dS=0, dV=0), ideal gas.
First let's start with definition for considered conditions:
ui=d(U)/dni

d(G)/dni = ((dU+pdV+Vdp+dpdV) - TdS)/dni = d(U)/dni + pdV/dni = d(U)/dni + R*T = ui + R*T


Offline xshadow

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Re: chemical potential "u" and free Gibbs energy "G"
« Reply #2 on: October 17, 2015, 12:39:39 PM »
Hi!
i know this.

My question is why ΔG_f(A)= u(A) during the calculation of  ΔG_r.

Because the  chemical potential is a derivate:  u_(A)= (dG/dn_A)
While ΔGf (f=formation) for A is definied as  ΔG_r of the reaction   B+C  :rarrow: A  to form 1 mole of A; (where B,C are the constituent elements)


ΔG_f is the free Gibbs energy variation along the process of formation of 1 MOLE of A, while u(A) is the infinitesimal variation of G,dG,  as a result of an infinitesimal variation of the number of moles,dn_A
So, how can these two definitions be equivalent??

thanks.

Offline mjc123

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Re: chemical potential "u" and free Gibbs energy "G"
« Reply #3 on: October 17, 2015, 07:07:09 PM »
It is no more contradictory than saying (i) you drove for an hour at constant speed and travelled 50 miles (ii) your instantaneous speed dx/dt was 50 mph.

Offline xshadow

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Re: chemical potential "u" and free Gibbs energy "G"
« Reply #4 on: October 17, 2015, 07:53:08 PM »
Hi!!
your phrase is not so contradictory...
Because if you  draw a position vs time diagram you get a straight  line...so is obviosuly that the Δx/Δt=dx/dt


But in this case you don't get a linear graph because the chemical potential of a specie A ,u(A) is defined as:
dG/dn_A  where all others components are constants.

But chemical potential is also a function of composition that is NOT linear(i.e the variation of G is not a constant for every variation of n_a  but his numerical value is a function of the composition) ...so the only situation when  dG/d_na=ΔG_f =ΔG/Δn_a is when the composition is kept cnstant: but  in a reaction the composition changes continuosly...

so dG/d_na shouldn't be equal to ΔG_f =ΔG/Δn_a   , where Δn_a is 1 mole (for his definition) !!

So i don't understand :(
« Last Edit: October 17, 2015, 08:33:39 PM by xshadow »

Offline mjc123

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Re: chemical potential "u" and free Gibbs energy "G"
« Reply #5 on: October 20, 2015, 07:12:53 AM »
"Excuse me sir, did you know you were driving at 100 miles per hour?"
"That's impossible, officer, I haven't been driving for an hour."

ΔGf is the heat of formation per mole; it is equal to the heat of formation of 1 mole, but you don't have to be working with 1 mole for it to be meaningful. It may, for example, be derived from DSC measurements on < 1 mmol.

Consider the case when you have, say, a million litres of 1M solution of A, and you add an extra mole of A made from its elements (B and C):
1000000 A (aq) + B + C :rarrow: 1000001 A (aq); ΔG = 100 kJ
The change in concentration of A is negligible, we can regard it as effectively constant at 1M.
ΔG for this reaction is equal to ΔGf° for A (aq) = 100 kJ/mol
Now suppose you scale it down, you have 1 L and you add 1 μmol A. ΔG for this will be 100 mJ. μ° = (dG/dn)° = 100 mJ/1 μmol = 100 kJ/mol = ΔGf°

Quote
But chemical potential is also a function of composition that is NOT linear
Remember your original question was about ΔGr°. You asked whether ΔGf°(i) = μ°(i)
μ°(i) does not change with composition because it is a constant, defined by the definition of the standard state.

Offline xshadow

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Re: chemical potential "u" and free Gibbs energy "G"
« Reply #6 on: October 22, 2015, 01:16:26 PM »
"Excuse me sir, did you know you were driving at 100 miles per hour?"
"That's impossible, officer, I haven't been driving for an hour."

ΔGf is the heat of formation per mole; it is equal to the heat of formation of 1 mole, but you don't have to be working with 1 mole for it to be meaningful. It may, for example, be derived from DSC measurements on < 1 mmol.

Consider the case when you have, say, a million litres of 1M solution of A, and you add an extra mole of A made from its elements (B and C):
1000000 A (aq) + B + C :rarrow: 1000001 A (aq); ΔG = 100 kJ
The change in concentration of A is negligible, we can regard it as effectively constant at 1M.
ΔG for this reaction is equal to ΔGf° for A (aq) = 100 kJ/mol
Now suppose you scale it down, you have 1 L and you add 1 μmol A. ΔG for this will be 100 mJ. μ° = (dG/dn)° = 100 mJ/1 μmol = 100 kJ/mol = ΔGf°

Quote
But chemical potential is also a function of composition that is NOT linear
Remember your original question was about ΔGr°. You asked whether ΔGf°(i) = μ°(i)
μ°(i) does not change with composition because it is a constant, defined by the definition of the standard state.

I think to have understand!!!
Anyway is  an approximation supposing that  doesn't change ...but is a very good approximation because the variation is minimal so??

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