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

Answer:

#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")#