Calculation of enthalpy and entropy of fusion of an unknown solid..?

The molar volume of a certain solid is 142.0 cm3/mol at 1.00 atm and 427.15 K, its melting temperature.
The molar volume of the liquid at this temperature and pressure is 152.6 cm3/mol.
At 1.2 MPa the melting temperature changes to 429.26 K. Calculate the enthalpy and entropy of fusion of the solid.

1 Answer
Nov 3, 2015

DeltaS_"fus" = "5.52 J/mol K"
DeltaH_"fus" = "2.36 kJ/mol"

Explanation:

Your tool of choice for this problem will be the Clapeyron equation used in the form

color(blue)((dp)/(dT) = (DeltaS_"fus")/(DeltaT_"fus"))" " " "color(purple)((1))

Now, you know that the following relationship exists between the enthalpy change of fusion, DeltaH_f, and the entropy change of fusion, DeltaS_f

color(blue)(DeltaS_"fus" = (DeltaH_"fus")/T)" " " " color(purple)((2))

This is derived from the Gibbds free energy change at equilibrium

DeltaG = DeltaH - T DeltaS

Since at equilibrium DeltaG = 0, it follows that you have

DeltaH = T * DeltaS implies DeltaS = (DeltaH)/T

Now, in your case, T would represent the meting temperature. A good rule of thumb to go by here is that you can use the average of the two given melting temperatures

T_"average" = ("427.15 K" + "429.26 K")/2 = "428.21 K"

Now, you should rearrange equation color(purple)((1)) to solve for dT, and then integrate, but you could skip that step if you go by the assumption that the temperature change, dT, is small enough.

You can get away with such an approximation because you're operating on the solid - liquid phase line, so you're bound to have small changes in temperature in such cases.

Now, if you take this route, you can say that

(dp)/(dT) ~~ (DeltaP)/(DeltaT) = (DeltaS_"fus")/(DeltaV_"fus")

At this point, you have everything you need to solve for DeltaS_"fus". More specifically, you know that

DeltaT = T_2 - T_1 = "2.11 K"

Here comes the tricky part - you need to convert DeltaV_"fus" and DeltaP to cubic meters per mole, "m"^3"/mol", and pascals, "Pa" - you'll see why in a minute!

DeltaV_"fus" = V_2 - V_1 = "152.6 cm"^3 - "142.0 cm"^3 = "10.6 cm"^3

This means that you have

10.6color(red)(cancel(color(black)("cm"^3)))/"mol" * "1 m"^3/(10^6color(red)(cancel(color(black)("cm"^3)))) = 10.6 * 10^(-6)"m"^3"/mol"

Finaly, you have

DeltaP = 1.2 * 10^6"Pa" - 1.01325 * 10^5"Pa" = 10.987 * 10^5"Pa"

So, plug in these values and solve for DeltaS_"fus"

DeltaS_"fus" = (DeltaP)/(DeltaT) * DeltaV_"fus"

DeltaS_"fus" = (10.987 * 10^5"Pa")/"2.11 K" * 10.6 * 10^(-6)"m"^3/"mol"

This will be equal to

DeltaS_"fus" = 55.195 * 10^(-1)"Pa m"^3/"K mol"

But "Pa" xx "m"^3 = "J", so the answer will be

DeltaS_"fus" = color(green)(5.52"J"/"mol K")

Now use equation color(purple)((2)) to get

DeltaH_"fus" = DeltaS_"fus" * T_"average"

DeltaH_"fus" = 5.52"J"/("mol" * color(red)(cancel(color(black)("K")))) * 428.21color(red)(cancel(color(black)("K"))) = 2363.72 "J"/"mol"

I'll leave this value rounded to three sig figs as well, but expressed in kilojoules per mole

DeltaH_"fus" = color(green)("2.36 kJ/mol")