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Topic: Standard Gibbs energy change for a reaction  (Read 3915 times)

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Offline ussername

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Standard Gibbs energy change for a reaction
« on: June 20, 2014, 12:43:04 PM »
[tex]\Delta G_{eq}=G_{eq}-G_{0}=-RTlnK_{eq}=-RTln\frac{\prod [product]_{eq}}{\prod [reactant]_{eq}}[/tex]
where [itex]0[/itex] is the state before the reaction, [itex]eq[/itex] is the equilibrium state. Is the relation correct?

If so, I don't understand how the change of state variable can depend just upon the final state, rather than both initial and final state (R = constant, T = constant, K = describes the final state). ???
« Last Edit: June 20, 2014, 02:19:58 PM by Borek »

Offline mjc123

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Re: Standard Gibbs energy change for a reaction
« Reply #1 on: June 23, 2014, 05:23:28 AM »
-RTlnKeq = ΔG° = G°products - G°reactants where the superscript ° indicates standard states (e.g. [] = 1M)

The problem with your formulation is that Geq and G0 are arbitrarily (though not independently) variable. You can start with any values of initial concentrations, and the equilibrium concentrations will vary in a manner determined by the initial concentrations and Keq - think of how you use an ICE table. The above equation relates K to a fixed standard ΔG value. Note that this is the difference between the Gibbs energies of products and reactants in standard states. It is not the value of the Gibbs energy at equilibrium, i.e. the minimum of the G-α curve (where α is extent of reaction).

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I don't understand how the change of state variable can depend just upon the final state, rather than both initial and final state

I don't understand the question. State variable values are determined by the state; change in a state variable depends on the initial and final states.
Geq depends on the equilibrium concentrations, which as I have said depend both on the initial concentrations and on K - ICE table again.

Offline ussername

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Re: Standard Gibbs energy change for a reaction
« Reply #2 on: October 23, 2014, 03:24:43 PM »
Thank you.
Also ΔG = G°(P) - G°(R) is difference between Gibbs energies of products and reactants in the state of equilibrium of chemical reaction. What is the meaning of this difference? (Why is there some special meaning compared to e.g. difference between Gibbs energies of components in gaseous mixture)?

Offline ussername

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Re: Standard Gibbs energy change for a reaction
« Reply #3 on: October 23, 2014, 03:37:39 PM »
Well obviously, as Gibbs energy of products (resp. reactants) is a function of its concentration in reaction mixture, the difference of Gibbs energies is a function of reaction mixture composition, i.e. a function of an extent of the reaction.

Offline mjc123

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Re: Standard Gibbs energy change for a reaction
« Reply #4 on: October 24, 2014, 05:57:58 AM »
Did you get my point?
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ΔG = G°(P) - G°(R) is difference between Gibbs energies of products and reactants in the state of equilibrium of chemical reaction
No it is not! It is the difference between the Gibbs energies of products and reactants IN THEIR STANDARD STATES. Not in the equilibrium state.
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as Gibbs energy of products (resp. reactants) is a function of its concentration in reaction mixture, the difference of Gibbs energies is a function of reaction mixture composition, i.e. a function of an extent of the reaction
The Gibbs energy of the system is a function of the extent of reaction, but ΔG° (and K) is a constant at a given temperature. The position of equilibrium depends on K and the initial concentrations.
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What is the meaning of this difference? (Why is there some special meaning compared to e.g. difference between Gibbs energies of components in gaseous mixture)?
I don't understand what you're getting at here.

Offline ussername

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Re: Standard Gibbs energy change for a reaction
« Reply #5 on: October 24, 2014, 10:38:29 AM »
[tex]aA + bB \leftrightharpoons  cC + dD[/tex]
[tex]k_{1}[A]^{a}[ B]^{b} = k_{2}[C]^{c}[D]^{d}[/tex]
[tex]K_{eq}=\frac{k_{1}}{k_{2}}=\frac{[C]^{c}[D]^{d}}{[A]^{a}[ B]^{b}}[/tex]
[tex]\Delta G^{0} = -RT lnK_{eq}[/tex]
[tex]G_{P}^{0} = -RT ln([C]^{c}[D]^{d})[/tex]
[tex]G_{R}^{0} = -RT ln([A]^{a}[ B]^{b})[/tex]
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No it is not! It is the difference between the Gibbs energies of products and reactants IN THEIR STANDARD STATES. Not in the equilibrium state.
As I see, [itex][C]^{c}[D]^{d}[/itex] resp. [itex][A]^{a}[ B]^{b}[/itex] are concentrations in the equilibrium state. Then I don't understand why Gibbs energies ([itex]G_{P}^{0}[/itex] resp. [itex]G_{R}^{0}[/itex]) don't describe the equilibrium state.
By standard state you mean concentrations of pure compounds under ([itex]T=298.15K[/itex], [itex]p=101.325 kPa[/itex]) conditions?
« Last Edit: October 24, 2014, 12:32:24 PM by Borek »

Offline mjc123

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Re: Standard Gibbs energy change for a reaction
« Reply #6 on: October 24, 2014, 12:03:59 PM »
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Then I don't understand why Gibbs energies (G0P  resp. G0R) don't describe the equilibrium state
For the simple reason that at equilibrium GP = GR and ΔG = 0. It is ΔG° that tells us the equilibrium constant. (And GP° = -RTln([C]c[D]d) is just wrong.)
Let's work it out. (Standard states by the way, means standard conditions at the specified temperature - which can be whatever you want, not necessarily 298.15K. For pure substances, standard state at that temp (pure liquids and solids have activity 1); gases 1 atm, solutions 1M.) Let's consider a solution reaction
aA + bB  ::equil:: cC + dD
Molar Gibbs energy of  compound A at concentration [A] is given by
GA = GA° + RT ln([A]/[A]°) = GA° + RT ln[A] since [A]° = 1M
ΔG = cGC + dGD - aGA - bGB
= cGC° + dGD° - aGA° - bGB° + RT (cln[C] + dln[D] - aln[A] - bln[B ])     {Incidentally B in square brackets is code for bold text - put a space in if you want it to appear like conc of B}
= ΔG° + RT ln([C]c[D]d/[A]a[B ]b)
At equilibrium ΔG = 0, so
ΔG° = -RTln([C]eqc[D]eqd/[A]eqa[B ]eqb) = -RTlnK
So K depends on ΔG°, which refers to reactants and products in their standard states.

Offline Corribus

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Re: Standard Gibbs energy change for a reaction
« Reply #7 on: October 24, 2014, 02:03:42 PM »
@ussername

We've had some good discussions in the recent past about the difference between ΔG° and ΔG, if this is a point of confusion for you. For example, you might want to take a look at this post. This thread and this thread may also be relevant to you.
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

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