Verified answer

When a substance freezes at its equilibrium melting point, the solid and liquid phases are in equilibrium. That means that the free energy change for the reaction:

liquid -> solid

is equal to zero.

Noting that the enthalpy change for the reaction is minus the enthalpy of fusion (the enthalpy of crystallization), we can write:

ΔG = 0 = -ΔH_fusion - T_melting*ΔS

ΔS = -ΔH_fusion/T-melting

If you look up the enthalpy of fusion for water at 1 bar pressure, you will find that ΔH_fusion = 334,000 J/kg.

The density of water at 0°C is 999.8 kg/m^3, so 1 m^3 of water has a mass of 999.8 kg

The entropy change of 1 m^3 of water that freezes at 0°C is then:

ΔS = -(999.8 kg/m^3)*(1m^3)*(334 kJ/kg)/(273.15 K)

ΔS = -1233 kJ/K