Gibbs free energy change at constant pressure is zero?

In summary: And I am also sure that you are aware of the fact that I am writing to the audience, so that the response is broader than the question. In summary, the value of Gibbs free energy change at constant pressure is zero when the system is at equilibrium and there is no chemical reaction taking place. However, when a chemical reaction is involved, the value of ΔG can change as a result of the reaction taking place. This is because the formula ΔG = ΔH - TΔS only applies at equilibrium and when the temperature is constant. The confusion may arise from the use of Δ for both simple change and a derivative quantity.
  • #1
jd12345
256
2
gibbs free energy change at constant pressure is zero??

IS gibbs free energy change at constant pressure zero?
ΔS = q / T. At constant pressure q = ΔH so ΔS = ΔH / T
So ΔG = ΔH - ΔH/T . T
= 0

I think i am wrong but where? My friend tells me that q involved in enetropy is different from the one we consider in ΔH. I don't understand this. IS heat due to entropy different from the one involved in enthalpy change
I'm all messed up
 
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  • #2


In my opinion you unwittingly made two assumptions:

1) heat is exchanged reversibly (in the definition of ΔS there is reversible heat)
2) temperature is constant (otherwise ΔG = ΔH - TΔS does not hold, generally ΔG = ΔH - Δ(TS) )

So you have an equilibrium situation with constant T and P in which case G is indeed constant.
 
  • #3


When a chemical reaction is involved
then
[itex] dG=Vdp-SdT+\Delta G d\xi [/itex]
where [itex]\xi[/itex] is the reaction coordinate. So even if p and T are hold constant, G will change as a result of the chemical reaction taking place.
The problem with using [itex]\Delta S=q/T[/itex] is that it presuposes equilibrium. However, when there is no chemical equilibrium, the formula is not applicable.
 
  • #4


asym said:
I
2) temperature is constant (otherwise ΔG = ΔH - TΔS does not hold, generally ΔG = ΔH - Δ(TS) )

Not quite,
[itex]dU=TdS-pdV+\sum_i \mu_i dn_i[/itex]
[itex]dH=TdS+Vdp+\sum_i \mu_i dn_i[/itex]
[itex]dG=-SdT+Vdp+\sum_i \mu_i dn_i[/itex]
so [itex]\mu_i=\partial G/\partial{n_i}|_{T,p}=\partial H/\partial {n_i}|_{S,p} [/itex] etc.

For a reaction [itex] \sum_i \nu_i X_i=0[/itex] the reaction coordinate is defined as
[itex] dn_i/\nu_i=d\xi [/itex] and [itex]\Delta G=\sum_i \mu_i \nu_i [/itex].
Hence [itex]\Delta G=\partial G/\partial \xi |_{p,T}=\partial H/\partial \xi|_{p,S}[/itex]
Now by definition
[itex]\Delta H\equiv \partial H /\partial \xi|_{p,T}=\partial H /\partial \xi|_{p,S}+\partial H/\partial S|_{p,\xi}\partial S/\partial \xi|_{p,T}=\Delta G+T\Delta S[/itex] as [itex]\partial S/\partial \xi|_{p,T}\equiv \Delta S[/itex]. So ΔG = ΔH - TΔS holds always.
 
  • #5


DrDu said:
Not quite,
[itex]dU=TdS-pdV+\sum_i \mu_i dn_i[/itex]
[itex]dH=TdS+Vdp+\sum_i \mu_i dn_i[/itex]
[itex]dG=-SdT+Vdp+\sum_i \mu_i dn_i[/itex]
so [itex]\mu_i=\partial G/\partial{n_i}|_{T,p}=\partial H/\partial {n_i}|_{S,p} [/itex] etc.

For a reaction [itex] \sum_i \nu_i X_i=0[/itex] the reaction coordinate is defined as
[itex] dn_i/\nu_i=d\xi [/itex] and [itex]\Delta G=\sum_i \mu_i \nu_i [/itex].
Hence [itex]\Delta G=\partial G/\partial \xi |_{p,T}=\partial H/\partial \xi|_{p,S}[/itex]
Now by definition
[itex]\Delta H\equiv \partial H /\partial \xi|_{p,T}=\partial H /\partial \xi|_{p,S}+\partial H/\partial S|_{p,\xi}\partial S/\partial \xi|_{p,T}=\Delta G+T\Delta S[/itex] as [itex]\partial S/\partial \xi|_{p,T}\equiv \Delta S[/itex]. So ΔG = ΔH - TΔS holds always.

You confuse Δ for the total change of a state variable (what I use in my posts) with the derivative (slope) of the variable with respect to the reaction extent, which is unfortunately denoted by Δ as well, nonetheless, there should be subscript r (like reaction) in that case - to distinguish from mere change.

My arguments are independent on the process happening in the system, be it chemical reaction or anything else. If Δ is just change (what is it's original meaning), then ΔG = ΔH - TΔS cannot hold always which is seen immediately from the mere fact that with varying temperature we wouldn't know what value from the range should be used for T:-)

To sum up: we are not in disagreement, we just interpret the question differently. I believe the asker was referring to simple change by Δ, not to the above-mentioned differential quantity, am I right?
 
  • #6


Asym,

of course you are right. My impression is that this kind of questions mostly arise because people don't know the difference of the simple [itex] \Delta G[/itex] and the partial free energy change [itex]\Delta_r G[/itex], with chemists having mostly the second one in mind.
 

What is Gibbs free energy change at constant pressure?

Gibbs free energy change at constant pressure is a thermodynamic quantity that measures the amount of energy available to do useful work in a system at a constant pressure.

Why is Gibbs free energy change at constant pressure important?

Gibbs free energy change at constant pressure is important because it helps determine whether a chemical reaction will occur spontaneously or not. If the change is negative, the reaction is thermodynamically favorable and will occur spontaneously.

How is Gibbs free energy change at constant pressure calculated?

Gibbs free energy change at constant pressure is calculated using the equation ΔG = ΔH - TΔS, where ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy.

Can Gibbs free energy change at constant pressure ever be zero?

Yes, Gibbs free energy change at constant pressure can be zero under certain conditions. This can occur when the change in enthalpy and the change in entropy are equal and opposite, resulting in a net change of zero.

How does Gibbs free energy change at constant pressure relate to equilibrium?

Gibbs free energy change at constant pressure is closely related to equilibrium. At equilibrium, the Gibbs free energy change is zero, indicating that the system is in a state of minimum energy and maximum stability.

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