Freezing/Melting of water and gibbs free energy

Hello,
I get that ΔG measures the spontaneity/capacity of a system to do non-mechanical work, and that if:
ΔG>0, the reaction is not spontaneous
ΔG<0, the reaction is spontaneous
ΔG=0, the reaction is at equilibrium
So why is Gibbs free energy zero for phase changes at constant temperature and pressure?
If it is negative or positive why we say that freezing water releases heat to the surrounding and it is equal to ΔH ? If at equilibrium ΔH is used to form liquid how we say it is released to the surrounding. I know I miss something. Can anyone tell me?

Thanks

DrClaude
Mentor
If you are exactly at the transition temperature, nothing happens. A mixture of liquid water and ice at 0°C will, at all times, have the same proportion of liquid and solid. To actually complete the phase transition, you need to put the system in contact with an environment either above or below the transition temperature, in which case ΔG≠0.

tonyjk
If you are exactly at the transition temperature, nothing happens. A mixture of liquid water and ice at 0°C will, at all times, have the same proportion of liquid and solid. To actually complete the phase transition, you need to put the system in contact with an environment either above or below the transition temperature, in which case ΔG≠0.
Thanks. Why then it is said that freezing releases ΔH of heat to the surrounding. This ΔH is used at equilibrium by the liquid/solid and not the surrounding.

Borek
Mentor
Why then it is said that freezing releases ΔH of heat to the surrounding.
Because it does. You can't increase amount of ice without removing heat from the mixture, just like you can't increase amount of liquid during melting without adding heat.

Please note when the mixture is isolated and doesn't exchange heat with the surroundings its composition doesn't change - amounts of solid and liquid stay the same. That's more or less equivalent to a statement "there is no freezing nor melting taking place" (although it is not entirely correct, as both processes do take place, just the same amount melts and freezes at the same time).

tonyjk
Because it does. You can't increase amount of ice without removing heat from the mixture, just like you can't increase amount of liquid during melting without adding heat.

Please note when the mixture is isolated and doesn't exchange heat with the surroundings its composition doesn't change - amounts of solid and liquid stay the same. That's more or less equivalent to a statement "there is no freezing nor melting taking place" (although it is not entirely correct, as both processes do take place, just the same amount melts and freezes at the same time).
Ah I thought the surrounding is meant here not the liquid/solid mixture. If macroscopically nothing is happening, than why there's change of entropy in the system?

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DrDu
Thanks. Why then it is said that freezing releases ΔH of heat to the surrounding. This ΔH is used at equilibrium by the liquid/solid and not the surrounding.
Consider melting. If you melt frozen soup in a pot and stirr it, you will ad lots of heat from outside the system but the temperature will stay the same until most (under ideal conditions all) of the soup is molten.

Chestermiller
Mentor
For the ##\Delta G## you are talking about, the initial and final thermodynamic equilibrium states of your system are as follows:

State 1: 1kg liquid water at 0 C and 1 atm

State 2: 1 kg water ice at 0 C and 1 atm

You want to find the change in enthalpy, entropy, and Gibbs free energy between these two states.

Now, we know that, to get the change in entropy between two thermodynamic equilibrium states of a closed system, we need to identify a reversible path between these states and then calculate the integral of dq/T for that path. For the present situation, this can be accomplished by putting the system into contact with a constant temperature reservoir at a temperature only *slightly* lower than 0 C. As the liquid water freezes, its molecules lock into place (losing potential energy) and this results in a release of heat to the reservoir. So the temperature of the system never deviates significantly from 0 C. Since the process is at constant pressure, the change in enthalpy is equal to the heat transferred from the surroundings to the system:$$\Delta H = q$$where both q and ##\Delta H## are negative for this process. Since the process is also at constant temperature, the change in entropy, which is given by the integral of dq/T, is:$$\Delta S=\frac{q}{T}=\frac{\Delta H}{T}$$Thus, for this change from State 1 to State 2, ##\Delta S## is also negative. If we now use the change in enthalpy and the change in entropy to calculate the change in free energy between the two thermodynamic equilibrium states, we obtain:
$$\Delta G = \Delta H-T\Delta S=\Delta H-T\frac{\Delta H}{T}=0$$
We have shown that ΔG=0 whenever we have phase equilibrium. The liquid is not reforming. We are removing heat until all the liquid has changed to ice. Look back at the initial and final states that we identified. Whenever you are trying to determine changes in thermodynamic functions, it is important to precisely identify the initial and final states that you are considering. The system is not unchanged. Only the free energy is unchanged. The entropy and enthalpy both change between the initial and final equilibrium states.

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tonyjk
For the ##\Delta G## you are talking about, the initial and final thermodynamic equilibrium states of your system are as follows:

State 1: 1kg liquid water at 0 C and 1 atm

State 2: 1 kg water ice at 0 C and 1 atm

You want to find the change in enthalpy, entropy, and Gibbs free energy between these two states.

Now, we know that, to get the change in entropy between two thermodynamic equilibrium states of a closed system, we need to identify a reversible path between these states and then calculate the integral of dq/T for that path. For the present situation, this can be accomplished by putting the system into contact with a constant temperature reservoir at a temperature only *slightly* lower than 0 C. As the liquid water freezes, its molecules lock into place (losing potential energy) and this results in a release of heat to the reservoir. So the temperature of the system never deviates significantly from 0 C. Since the process is at constant pressure, the change in enthalpy is equal to the heat transferred from the surroundings to the system:$$\Delta H = q$$where both q and ##\Delta H## are negative for this process. Since the process is also at constant temperature, the change in entropy, which is given by the integral of dq/T, is:$$\Delta S=\frac{q}{T}=\frac{\Delta H}{T}$$Thus, for this change from State 1 to State 2, ##\Delta S## is also negative. If we now use the change in enthalpy and the change in entropy to calculate the change in free energy between the two thermodynamic equilibrium states, we obtain:
$$\Delta G = \Delta H-T\Delta S=\Delta H-T\frac{\Delta H}{T}=0$$
We have shown that ΔG=0 whenever we have phase equilibrium. The liquid is not reforming. We are removing heat until all the liquid has changed to ice. Look back at the initial and final states that we identified. Whenever you are trying to determine changes in thermodynamic functions, it is important to precisely identify the initial and final states that you are considering. The system is not unchanged. Only the free energy is unchanged. The entropy and enthalpy both change between the initial and final equilibrium states.
If the system (mixture) is isolated from its surrounding and we have equilibrium between the 2 phases, do we have liquid forming and solid reforming at the same rate?

Chestermiller
Mentor
If the system (mixture) is isolated from its surrounding and we have equilibrium between the 2 phases, do we have liquid forming and solid reforming at the same rate?
At the molecular scale, yes. Macroscopically, we see no change.

DrDu
Macroscopically, we see no change.
That's not quite true. As smaller crystals have a higher surface energy than smaller ones, large crystals tend to grow while smaller ones disappear. While this takes some time, it is clearly visible.

Chestermiller
Mentor
That's not quite true. As smaller crystals have a higher surface energy than larger ones, large crystals tend to grow while smaller ones disappear. While this takes some time, it is clearly visible.
Good point. But the mass of ice and the mass of liquid water don't change significantly, correct? And, in the final thermodynamic equilibrium state, the ice will be just one large crystal. My post #7 and the ##\Delta G =0## in that post apply to this final thermodynamic equilibrium state. The ice with the small crystals in it is not a thermodynamic equilibrium state, because it is still changing with time to larger crystals.

Chet

Good point. But the mass of ice and the mass of liquid water don't change significantly, correct? And, in the final thermodynamic equilibrium state, the ice will be just one large crystal. My post #7 and the ##\Delta G =0## in that post apply to this final thermodynamic equilibrium state. The ice with the small crystals in it is not a thermodynamic equilibrium state, because it is still changing with time to larger crystals.

Chet
Even if the system is isolated, we will have at the end of the process only ice? or the isolated system will remain at equilibrium between liquid/ice?

Chestermiller
Mentor
Even if the system is isolated, we will have at the end of the process only ice? or the isolated system will remain at equilibrium between liquid/ice?
What do you think?

What do you think?
If the system is isolated then the system will remain unchanged. I asked this question because I am confused about ΔG=0 for both isolated and non isolated system and how in both case the system behaves. If my answer is right (hopefully), I think I've understood it.

Chestermiller
Mentor
If the system is isolated then the system will remain unchanged. I asked this question because I am confused about ΔG=0 for both isolated and non isolated system and how in both case the system behaves. If my answer is right (hopefully), I think I've understood it.
Tonyjk,

Please go back to my post #7 and re-read what I wrote in bold. I put it in bold because it is important. I'm going to define two changes of state, each of which gives ##\Delta G=0##. Please tell me whether you think these constitute the same change of state.

Change 1 in non-isolated system:
State 1: 5 kg of ice floating in 10 kg of liquid water
State 2: 6 kg of ice floating in 9 kg of liquid water

Change 2 in an isolated system:
State 1: 5 kg of ice floating in 10 kg of liquid water
State 2: 5 kg of ice floating in 10 kg of liquid water

Well, what do you think. Do they constitute the same change of state? Even though the change in free energy is the same for both these changes, is the change in enthalpy and the change in entropy the same?

tonyjk
Tonyjk,

Please go back to my post #7 and re-read what I wrote in bold. I put it in bold because it is important. I'm going to define two changes of state, each of which gives ##\Delta G=0##. Please tell me whether you think these constitute the same change of state.

Change 1 in non-isolated system:
State 1: 5 kg of ice floating in 10 kg of liquid water
State 2: 6 kg of ice floating in 9 kg of liquid water

Change 2 in an isolated system:
State 1: 5 kg of ice floating in 10 kg of liquid water
State 2: 5 kg of ice floating in 10 kg of liquid water

Well, what do you think. Do they constitute the same change of state? Even though the change in free energy is the same for both these changes, is the change in enthalpy and the change in entropy the same?
Thank you.