I Gibbs Energy and chemical equilibrium

[dRic2]

Hi, i'll apologize for my english in advance, so here's the question.

I was wondering about the equilibrium condition for a chemical reaction. We know that a closed system is in equilibrium if the Gibbs free energy's function has a minimun in that point. So, taking Temperature and Pressure as constants (to ease the calculations) the first differential of Gibbs function would be:

dG = ∑u_idn_i = 0 (where $u_i$ is the chemical potential and $dn_i$ is the infinitesimal variation of moles )

Defying the Extent of Reaction as

d\lambda = dn_i/v_i (where $v_i$ is the stoichiometric coefficient)

the formula can be re-arranged as

dG = ∑u_i(d\lambda v_i) = d\lambda(∑u_iv_i) = 0

then I don't get why my book says that $d\lambda$ is an "arbitrary variation" thus it simplifies in $∑u_iv_i = 0$
I mean if it is an equilibrium why would $dn$ be "arbitrary"? It should be zero because we can not have a variation in the composition of the system because, at the equilibrium, the properties of the system should remain unchanged.

 

[Lord Jestocost]

Hi, i'll apologize for my english in advance, so here's the question.

I was wondering about the equilibrium condition for a chemical reaction. We know that a closed system is in equilibrium if the Gibbs free energy's function has a minimun in that point. So, taking Temperature and Pressure as constants (to ease the calculations) the first differential of Gibbs function would be:

dG = ∑u_idn_i = 0 (where $u_i$ is the chemical potential and $dn_i$ is the infinitesimal variation of moles )

Defying the Extent of Reaction as

d\lambda = dn_i/v_i (where $v_i$ is the stoichiometric coefficient)

the formula can be re-arranged as

dG = ∑u_i(d\lambda v_i) = d\lambda(∑u_iv_i) = 0

then I don't get why my book says that $d\lambda$ is an "arbitrary variation" thus it simplifies in $∑u_iv_i = 0$
I mean if it is an equilibrium why would $dn$ be "arbitrary"? It should be zero because we can not have a variation in the composition of the system because, at the equilibrium, the properties of the system should remain unchanged.

Read it in a different way: You are searching for the minimum of the Gibbs free energy function.

(dG/dλ)_p,T = 0 = Σv_i μ_i

 

[dRic2]

Thank you, I didn't see that... :)

Anyways, can you explain the mathematical meaning of:

"$d\lambda$ is an arbitrary variation thus it simplifies in $\sum u_i v_i = 0$" (Hope i translated it correctly from my book)

I still don't get it...

 

[DrClaude]

If $d\lambda$ is arbitrary, then $d\lambda \times x = 0$ means $x = 0$.

 

[dRic2]

Ok, but why $d\lambda$ is arbitrary? that's my question. Sorry if I wasn't clear

 

[Lord Jestocost]

Ok, but why $d\lambda$ is arbitrary? that's my question. Sorry if I wasn't clear

The wording is somehow ambiguous. The quantity dλ is not arbitrary, it is by definition of infinitesimal size, viz. dλ is thought to approach zero as a limit, but doesn’t become zero.

 

[dRic2]

Yes, but i don't get why dn (or $d\lambda$) is infinitesimal (or whatever) and is not zero. Just I said, temperature and pressure are considered constants and so dT=dP=0, but at the equilibrium also dn should be zero because i can't have material transport from A to B. I think I miss something of the definition of differential...

 

[Lord Jestocost]

Yes, but i don't get why dn (or $d\lambda$) is infinitesimal (or whatever) and is not zero. Just I said, temperature and pressure are considered constants and so dT=dP=0, but at the equilibrium also dn should be zero because i can't have material transport from A to B. I think I miss something of the definition of differential...

What it all means: You simply determine the minimum of the function G(T, p, n₁, n₂, …., n_n) with respect to variations in the n_i’s; by this, you find the necessary condition for the Gibbs free energy to be a minimum: Σv_i μ_i = 0