OK. 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: $$\Delta G = \Delta H-T\Delta S=\Delta H-T\frac{\Delta H}{T}=0$$

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: $$\Delta G = \Delta H-T\Delta S=\Delta H-T\frac{\Delta H}{T}=0$$