Calculate the change in molar entropy when 1 mol of solid iodine goes from 25º C to 184º C at a constant pressure of 1 atm?

The total molar entropy change is:

#DeltabarS = "124.50 J/mol"cdot"K"#

I'm assuming you mean to go all the way through the vapor phase. In that case, the overall process is:

#"I"(s) (25^@ "C") stackrel("heat"" ")(->) "I"(s) (113.6^@ "C") stackrel("melt"" ")(->) "I"(l) (113.6^@ "C") stackrel("heat"" ")(->) "I"(l) (184^@ "C") stackrel("vaporize"" ")(->) "I"(g) (184^@ "C")#

We know that at constant pressure:

#dS(T,P) = ((delS)/(delT))_PdT + cancel(((delS)/(delP))_TdP)^(0)#

And we know that #((delbarH)/(delT))_P = barC_P#, where #barY# indicates a molar quantity. Therefore, and you should already know this definition,

#((delbarS)/(delT))_P = (barC_P)/T = 1/T((delbarH)/(delT))_P#

We break this into the four steps shown above.

#DeltabarS = DeltabarS_(25^@ "C" -> 113.6^@ "C") + DeltabarS_"fus" + DeltabarS_(113.6^@ "C" -> 184^@ "C") + DeltabarS_"vap"#

HEATING THE SOLID

#DeltabarS_(25^@ "C" -> 113.6^@ "C") = int_(25^@ "C")^(113.6^@ "C") barC_P/TdT#

I assume the equation for #barC_P# is in terms of #""^@ "C"# and not #"K"#. You'll have to verify whether that is true or not...

[By the way, I compared with a proper heat capacity curve... this #barC_P# expression is only good up to #32^@ "C"#, and then it underestimates #barC_P# by up to #"9 J/mol"cdot"K"# at #113.6^@ "C"#...]

#DeltabarS_(25^@ "C" -> 113.6^@ "C") = "4.184 J"/"cal"int_("298.15 K")^("386.75 K")13.07/T + 3.21 xx 10^(-4)(1 - 25/T + 273.15/T)dT#

I'll leave this integral for you to do. This would give #"14.434 J/mol"cdot"K"#.

MELTING THE SOLID

At any phase equilibrium, #DeltaG = 0# so that #DeltabarH_"fus" = TDeltabarS_"fus"#. Thus,

#DeltaS_"fus" = "3.74 kcal/mol"/(113.6+"273.15 K") xx "4.184 J"/"cal" xx "1000 J"/"1 kJ"#

#= "40.461 J/mol"cdot"K"#

HEATING THE LIQUID

Here we assumed that the heat capacity was a constant in the temperature range, so...

#DeltabarS_(113.6^@ "C" -> 184^@ "C") ~~ barC_P int_("386.75 K")^("457.15 K") 1/TdT#

#= barC_Pln("457.15 K"/"386.75 K")#

#= "19.5 cal/mol"cdot"K"cdotln("457.15 K"/"386.75 K") xx "4.184 J"/"cal"#

#= "13.771 J/mol"cdot"K"#

VAPORIZING THE LIQUID

And of course, this is the same idea as before.

#DeltaS_"vap" = "6.1 kcal/mol"/(184+"273.15 K") xx "4.184 J"/"cal" xx "1000 J"/"1 kJ"#

#= "55.829 J/mol"cdot"K"#

TOTAL CHANGE IN ENTROPY

And so, what I get is:

#color(blue)(DeltabarS = "124.50 J/mol"cdot"K")#