Gibbs free energy change at constant pressure is zero?

[jd12345]

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

 

[asym]

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.

 

[DrDu]

When a chemical reaction is involved
then
$dG=Vdp-SdT+\Delta G d\xi$
where $\xi$ 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 $\Delta S=q/T$ is that it presuposes equilibrium. However, when there is no chemical equilibrium, the formula is not applicable.

 

[DrDu]

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

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

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

 

[asym]

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

For a reaction $\sum_i \nu_i X_i=0$ the reaction coordinate is defined as
$dn_i/\nu_i=d\xi$ and $\Delta G=\sum_i \mu_i \nu_i$.
Hence $\Delta G=\partial G/\partial \xi |_{p,T}=\partial H/\partial \xi|_{p,S}$
Now by definition
$\Delta H\equiv \partial H /\partial \xi|_{p,T}=\partial H /\partial \xi|_{p,S}+\partial H/\partial S|_{p,\xi}\partial S/\partial \xi|_{p,T}=\Delta G+T\Delta S$ as $\partial S/\partial \xi|_{p,T}\equiv \Delta S$. 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?

 

[DrDu]

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 $\Delta G$ and the partial free energy change $\Delta_r G$, with chemists having mostly the second one in mind.