Freezing/Melting of water and gibbs free energy

In summary: Delta H-\frac{\Delta S}{T}$$We can see that, in this instance, the free energy is negative, indicating that the system is spontaneous in moving from State 1 to State 2.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.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 atmState 2: 1 kg water ice at 0 C and
  • #1
tonyjk
227
3
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
 
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  • #2
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.
 
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  • #3
DrClaude said:
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.
 
  • #4
tonyjk said:
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).
 
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  • #5
Borek said:
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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  • #6
tonyjk said:
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.
 
  • #7
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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  • #8
Chestermiller said:
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?
 
  • #9
tonyjk said:
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.
 
  • #10
Chestermiller said:
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.
 
  • #11
DrDu said:
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
 
  • #12
Chestermiller said:
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?
 
  • #13
tonyjk said:
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?
 
  • #14
Chestermiller said:
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.
 
  • #15
tonyjk said:
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?
 
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  • #16
Chestermiller said:
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.
 

1. What is the freezing/melting point of water?

The freezing/melting point of pure water is 0 degrees Celsius or 32 degrees Fahrenheit. This means that at this temperature, water will either freeze into ice or melt into liquid form.

2. How does freezing/melting of water affect its density?

When water freezes, it expands and becomes less dense. This is because the molecules arrange themselves into a crystalline structure, which takes up more space. On the other hand, when water melts, it becomes more dense as the molecules are able to move more freely.

3. What is the significance of Gibbs free energy in the freezing/melting of water?

Gibbs free energy is a measure of the energy available to do work in a system. In the case of water freezing/melting, changes in Gibbs free energy can indicate whether the process will occur spontaneously. When the Gibbs free energy is negative, the process will occur spontaneously, while a positive value indicates a non-spontaneous process.

4. How does pressure affect the freezing/melting point of water?

Pressure has a significant effect on the freezing/melting point of water. As pressure increases, the freezing point decreases, meaning it takes a lower temperature for water to freeze. On the other hand, increasing pressure raises the melting point, requiring a higher temperature for water to melt.

5. Can the freezing/melting of water be reversed?

Yes, the freezing/melting of water is a reversible process. When water freezes, it absorbs heat energy from the surroundings. This energy can be returned to the system, causing the ice to melt back into liquid water. This process can be repeated indefinitely as long as the temperature and pressure conditions are appropriate.

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