Why is the Gibbs free energy for phase changes zero at constant temperature and pressure?

Question

I get that $\Delta G$ measures the spontaneity/capacity of a system to do non-mechanical work, and that if:

$\Delta G > 0$, the reaction is not spontaneous

$\Delta G < 0$, the reaction is spontaneous

$\Delta G = 0$, 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 $\Delta H$? If at equilibrium $\Delta H$ is used to reform liquid how we say it is released to the surrounding. I know I miss something, anyone can tell me?

Answer 1

The initial and final thermodynamic equilibrium states of your system are as follows:

State 1: $\pu{1kg}$ liquid water at $\pu{0 ^\circ C}$ and $\pu{1 atm}$

State 2: $\pu{1 kg}$ water ice at $\pu{0 ^\circ C}$ and $\pu{1 atm}$

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 $\mathrm dq/T$ 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 $\pu{0 ^\circ C}$. 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 $\pu{0 ^\circ C}$. 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 $\mathrm dq/T$, 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: $$\Delta G = \Delta H-T\Delta S=\Delta H-T\frac{\Delta H}{T}=0$$

Answer 2

So why is Gibbs free energy zero for phase changes at constant temperature and pressure?

It isn't necessarily.

For example if you supercool liquid water to -10 degrees C, let the water freeze and do this in a way that the final state is solid water at -10 degrees C, the change in Gibbs free energy will be negative. This is a spontaneous process.

See W. R. Salzman's Gibbs Free Energy and Temperature: The Gibbs-Helmholtz Equation for further explanation.

Answer 3

Here is a simpler answer. $\Delta G$ is zero for a phase change at the normal phase change temperature. For instance, for the melting of ice,

• $\Delta G<0$ if at $T>\pu{0 ^\circ C}$ (spontaneous)
• $\Delta G=0$ if at $T=\pu{0^\circ C}$
• $\Delta G>0$ if at $T<\pu{0^\circ C}$ (not spontaneous)

At $\pu{0^\circ C}$ there is no preference for $\ce{H2O}$ being ice vs. water, so $\Delta G=0$.

For melting ice, $\Delta H=\pu{+6 kJ/mol}$ and $\Delta S=\pu{+0.022 kJ/(mol K)}$. Using $\Delta G=\Delta H-T\Delta S$, you can plug in temperatures in Kelvin (eg. $263, 273, \pu{283 K}$) and see the $\Delta G$ variations (eg. $+0.2, 0.0, \pu{-0.2 kJ/mol}$).

Freezing always releases heat ($\Delta H = \pu{-6kJ/mol}$ for freezing water); it just sometimes isn't spontaneous. If spontaneous, this lost $\pu{6 kJ/mol}$ of heat is being sucked out of the system by the sub-zero surroundings.