Gibbs free energy -- mathematical expression

[rajendra123]

I am not able to understand the mathematical expression of "change in Gibbs free energy",

For a chemical reaction occurring at constant temperature and constant pressure,
(ΔS)_total = (ΔS)_system + (ΔS)_surrounding

Considering that reaction is exothermic, ΔH be the heat supplied by system to surrounding at constant pressure and temperature,

(ΔS)_total = (ΔS)_system + (-ΔH)/T

- T * (ΔS)_total = ΔH - T * (ΔS)_system

The term on left hand side is known as Change in Gibbs free energy ΔG. WHY?
also, the temperature in above expression is T_system and not T_surrounding , HOW?

(ΔS)_surrounding should be Q/T_surrounding, where Q is heat added to surrounding.

By intuition, ΔH amount of heat is available and T * ΔS_system is the unavailable energy. Thus (ΔH-T*ΔS_system) is the amount of energy that is available to be converted to work and should be Gibbs free energy right? I am not able to understand it from the mathematical equation.

Please explain what am I missing out? and where am I wrong? Please tolerate me for any mistakes, this is my first post and I do not have chemical engineering background. Thank you.

 

[DrDu]

The equation $\Delta S_\mathrm{surrounding}=-\Delta H/T$ only holds if the heat is transferred under equilibrium conditions, hence $T_\mathrm{system}=T_\mathrm{surrounding}$.

 

[Chestermiller]

Hi rajendra123,

Your question is very puzzling to me. The Gibbs Free Energy of a system is defined as G≡H-TS. I don't know why you have even introduced the surroundings into the discussion. For a chemical reaction at constant temperature and pressure, the change in Gibbs Free Energy is defined with respect to the following two thermodynamic equilibrium states of a specific system:

State 1: Pure reactants in stoichiometric molar proportions in separate containers, each at temperature T and pressure P

State 2: Pure products in corresponding stoichiometric molar proportions in separate containers, each at temperature T and pressure P

Now, how you determine the change in free energy between State 1 and State 2 might be a little complicated, and might involve devices like ideal semi-permeable membranes and a Van't Hoff equilibrium box, but I can be done. And it is done only using reversible steps.

Chet

 

[DrDu]

Hi rajendra123,

Your question is very puzzling to me. The Gibbs Free Energy of a system is defined as G≡H-TS. I don't know why you have even introduced the surroundings into the discussion.

How else would you prove that a reaction is spontaneous if you don't take the entropy of the surrounding into account?

 

[Chestermiller]

How else would you prove that a reaction is spontaneous if you don't take the entropy of the surrounding into account?

Huh?? All reactions are spontaneous if you start out with pure reactants. It's just a matter of where the system equilibrates. So, I have no idea how the entropy of the surroundings can matter.

The usual very rough rule of thumb for a reaction being considered spontaneous is if the change in Free Energy (as defined in my previous response) at 1 atm and 298 K is negative.

Chet

 

[DrDu]

Ok, so let me invert my question. How do you show that $\Delta G=0$ corresponds to chemical equilibrium? The argument of rajendra123 is that at $\Delta G=0$, also the total entropy of the system and it's surrounding doesn't change. Hence we are at equilibrium.

 

[Chestermiller]

Ok, so let me invert my question. How do you show that $\Delta G=0$ corresponds to chemical equilibrium? The argument of rajendra123 is that at $\Delta G=0$, also the total entropy of the system and it's surrounding doesn't change. Hence we are at equilibrium.

Wow. That's not how I interpreted his question. But certainly, that's a way of showing it. Now your original answer makes more sense to me.

 

[rajendra123]

First of all thank you for your quick reply Friends.

Hi rajendra123,

Your question is very puzzling to me. The Gibbs Free Energy of a system is defined as G≡H-TS. I don't know why you have even introduced the surroundings into the discussion.

Do you mean that Gibbs Free Energy is calculated for ISOLATED systems only? If the way I have mentioned in the question is not right then how else could we get mathematical relation.

For a chemical reaction at constant temperature and pressure, the change in Gibbs Free Energy is defined with respect to the following two thermodynamic equilibrium states of a specific system:

State 1: Pure reactants in stoichiometric molar proportions in separate containers, each at temperature T and pressure P

State 2: Pure products in corresponding stoichiometric molar proportions in separate containers, each at temperature T and pressure P

Yes, but to maintain temperature constant there must be heat interaction with something (e.g. surrounding) right?

 

[rajendra123]

Ok, so let me invert my question. How do you show that $\Delta G=0$ corresponds to chemical equilibrium? The argument of rajendra123 is that at $\Delta G=0$, also the total entropy of the system and it's surrounding doesn't change. Hence we are at equilibrium.

Wow. That's not how I interpreted his question. But certainly, that's a way of showing it. Now your original answer makes more sense to me.

YES, both of you are right, though I didn't mean that, it's also a way of showing it.

 

[rajendra123]

The equation $\Delta S_\mathrm{surrounding}=-\Delta H/T$ only holds if the heat is transferred under equilibrium conditions, hence $T_\mathrm{system}=T_\mathrm{surrounding}$.

If T_system = T_surrounding then how can heat transfer take place?

 

[DrDu]

If the two temperatures are equal, the heat exchange will be reversible but it will take infinite time. So you have to understand this as a limit of a very small temperature difference and a correspondingly slow heat exchange rate.

 

[Chestermiller]

Do you mean that Gibbs Free Energy is calculated for ISOLATED systems only?

No. I'm saying that the Gibbs Free Energy is a very precise quantity that we define as being equal to H-TS. It definitely doesn't only apply to isolated systems. It can be applied to any closed system.

If the way I have mentioned in the question is not right then how else could we get mathematical relation.

I don't know what relation you are referring to, but, as I said, the Gibbs Free Energy of a closed system is a very precise mathematical definition.

Yes, but to maintain temperature constant there must be heat interaction with something (e.g. surrounding) right?

Yes, but so what? I've always felt that it is a very bad idea to include the surroundings in the development because it confuses the issue and because one has much less control over what is happening in the surroundings in terms of reversibility. In my judgment, the focus should always be on the system.

 

[rajendra123]

If the two temperatures are equal, the heat exchange will be reversible but it will take infinite time. So you have to understand this as a limit of a very small temperature difference and a correspondingly slow heat exchange rate.

Okay, so to calculate ΔG we have to have such a device which will transfer heat at such a rate that T_system will be constant and heat transfer shall take place between infinitesimal temperature difference.

 

[rajendra123]

No. I'm saying that the Gibbs Free Energy is a very precise quantity that we define as being equal to H-TS. It definitely doesn't only apply to isolated systems. It can be applied to any closed system.

I don't know what relation you are referring to, but, as I said, the Gibbs Free Energy of a closed system is a very precise mathematical definition.

Yes, but so what? I've always felt that it is a very bad idea to include the surroundings in the development because it confuses the issue and because one has much less control over what is happening in the surroundings in terms of reversibility. In my judgment, the focus should always be on the system.

I have found this link which says "The whole Gibbs relationship or function is about entropy change."
http://2ndlaw.oxy.edu/gibbs.html

 

[Chestermiller]

Okay, so to calculate ΔG we have to have such a device which will transfer heat at such a rate that T_system will be constant and heat transfer shall take place between infinitesimal temperature difference.

Yes. And, if you're considering a chemical reaction (which can occur spontaneously, if permitted), identifying a physical system (at least conceptually) that only allows the reaction to process reversibly presents an interesting challenge.

 

[Chestermiller]

I have found this link which says "The whole Gibbs relationship or function is about entropy change."
http://2ndlaw.oxy.edu/gibbs.html

I have no idea what this article is talking about, but, whatever it is doesn't work for me.

 

[rajendra123]

I have no idea what this article is talking about, but, whatever it is doesn't work for me.

So, you mean to say, if a chemical reaction occurring in closed container releasing ΔH amount of heat somehow we convert (ΔH-T_system*ΔS_system) to work i.e. without interacting with surrounding. This work is ΔG?

 

[Chestermiller]

So, you mean to say, if a chemical reaction occurring in closed container releasing ΔH amount of heat somehow we convert (ΔH-T_system*ΔS_system) to work i.e. without interacting with surrounding. This work is ΔG?

No. I'm saying that we don't need to specify the details of how the interaction with the surroundings takes place. In this case, the heat transferred to the surroundings is ΔH, and the system may do work on the surroundings. But the details are not our concern. Our focus is on the system.

Incidentally, ΔG is not the reversible work done on the surroundings. It is the maximum non-PV work that can be done on the surroundings.

 

[rajendra123]

No. I'm saying that we don't need to specify the details of how the interaction with the surroundings takes place. In this case, the heat transferred to the surroundings is ΔH, and the system may do work on the surroundings. But the details are not our concern. Our focus is on the system.

Incidentally, ΔG is not the reversible work done on the surroundings. It is the maximum non-PV work that can be done on the surroundings.

okay. yes it's the maximum non-PV work for closed system.