The enthalpy is represented through the following equation. There are some molecules which take part in this change are called “internal enthalpy” and the molecules that do not are referred to as “external enthalpy”. For example, it increases when heat is added and decreases when heat is withdrawn from that system. Thereby it changes when heat enters or leaves a system. It deals with the heat contained in any system. Relying on these two factors, a new product is formed through a standard reaction of several compounds.Įnthalpy is defined as a change in internal energy and volume at constant pressure. Both of them are partly related to each other in a reaction because the fundamental rule of any reaction is releasing or absorbing heat or energy. #DeltaH_"fus" = color(green)("2.Enthalpy and Entropy are two significant terms related to thermodynamics. I'll leave this value rounded to three sig figs as well, but expressed in kilojoules per mole #DeltaH_"fus" = DeltaS_"fus" * T_"average"# Now use equation #color(purple)((2))# to get #DeltaS_"fus" = (DeltaP)/(DeltaT) * DeltaV_"fus"# So, plug in these values and solve for #DeltaS_"fus"# 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! #(dp)/(dT) ~~ (DeltaP)/(DeltaT) = (DeltaS_"fus")/(DeltaV_"fus")#Īt this point, you have everything you need to solve for #DeltaS_"fus"#. Now, if you take this route, you can say that 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, 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. A good rule of thumb to go by here is that you can use the average of the two given melting temperatures
Now, in your case, #T# would represent the meting temperature.
#DeltaH = T * DeltaS implies DeltaS = (DeltaH)/T# Since at equilibrium #DeltaG = 0#, it follows that you have
MOLAR ENTROPY FREE
This is derived from the Gibbds free energy change at equilibrium
Now, you know that the following relationship exists between the enthalpy change of fusion, #DeltaH_f#, and the entropy change of fusion, #DeltaS_f# Your tool of choice for this problem will be the Clapeyron equation used in the form
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