Question

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 constant-pressure molar heat capacity for the solid from `25^@ "C"` to `113.6^@ "C"` is given by

`barC_P = 13.07 + 3.21 xx 10^(-4)(T - 25) "cal/deg/mol"`

For the liquid, it is approximately constant at `barC_P = "19.5 cal/deg/mol"`.

For the fusion and vaporization, occurring at `113.6^@ "C"` and `184^@ "C"` respectively,

`DeltabarH_"fus" = "3.74 kcal/mol"`
`DeltabarH_"vap" = "6.1 kcal/mol"`

Original problem given here:

Answer 1

The total molar entropy change is:

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

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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")`