Where can I find a table showing the entropies of fusion of metallic elements? Or at least how can I calculate them?

You can calculate them. In some texts, or on NIST, you may be able to find the enthalpy of formation of the liquid metal, which for anything but #"Hg"# corresponds to:

#"M"(s) -> "M"(l)#

Hence, that gives the enthalpy of fusion at #"298.15 K"# (which is hypothetical, I know).

For a normal phase transition, #DeltaG_(trs) = 0#, so:

#DeltaH_(fus) = TDeltaS_(fus)#

#=> color(blue)(DeltaS_(fus) = (DeltaH_(fus))/T)#

Here's the full idea:

#DeltaH_(fus)^("High T") = int_(T_"hot")^("298.15 K") C_(P, "solid")(T)dT + DeltaH_(fus)^"298.15 K" + int_("298.15 K")^(T_"hot") C_(P, "liquid")(T)dT#

and then you would use the solid molar heat capacity function of temperature (for that the Shomate equation is common). These functions, when available, can be found on NIST (#"Mg"# is used as an example below).

This is a thermodynamic cycle .

1. Step 1 is cooling the solid metal from its melting point.
2. You want the true #DeltaH_(fus)# at the proper melting point, so by using #DeltaH_f^@#, we get the hypothetical enthalpy of fusion at #"298.15 K"#. That is step 2.
3. Step 3 is heating the liquid metal from #"298.15 K"# back to its melting point.

The net result is that you have indirectly calculated #DeltaH_(fus)# at the true melting point. That then allows you to calculate #DeltaS_(fus)# at the true melting point.

For example, #"Mg"(l)# has #DeltaH_(f)^@("298.15 K") = "4.79 kJ/mol"#. If we assume that #DeltaH_f^@("298.15 K") ~~ DeltaH_(fus)^(650^@ "C")#, like some people might do, then:

#color(red)(DeltaS_(fus)^("298.15 K")) = (4.79 cancel"kJ""/mol")/(650+"273.15 K") xx "1000 J"/cancel"1 kJ"#

#= color(red)("5.19 J/mol"cdot"K")#

This is not a good approximation. Also, #"Mg"# obviously doesn't melt at #"298.15 K"#. Instead, #"Mg"(s)# actually has for the Shomate equation:

#C_P(T) = 26.54083 - 1.533048(T/1000) + 8.062443(T/1000)^2 + 0.572170(T/1000)^3 - 0.174221/(T/1000)^2#

Integrating that from #"923.15 K"# to #"298.15 K"# gives #-"17753.1 J/mol"#, or #-"17.753 kJ/mol"#.

Then, #"Mg"(l)# has #C_P(T) = "34.309 J/mol"cdot"K"# (apparently constant with temperature). So,

#int_("298.15 K")^("923.15 K") C_PdT ~~ C_PDeltaT#

#= "34.309 J/mol"cdot"K" cdot ("923.15 K" - "298.15 K")#

#=# #"21443.1 J/mol"#

#=# #"21.443 kJ/mol"#

Including these contributions,

#DeltaH_(fus)^(650^@ "C") ~~ -"17.753 kJ/mol" + "4.79 kJ/mol" + "21.443 kJ/mol" = "8.48 kJ/mol"#

and so:

#color(blue)(DeltaS_(fus)^(650^@ "C")) = (8.48 cancel"kJ""/mol")/"923.15 K" xx "1000 J"/cancel"1 kJ"#

#= color(blue)("9.19 J/mol"cdot"K")#