[AzeoPoint20]

Hi everyone!

I am an Italian Pharmaceutical Chemistry student and I am preparing for the Physical Chemistry exam. While solving some exercises I got in trouble with one in particular, I'll translate it from Italian:

"A compound melts at 146 °C (419.15 K) and its enthalpy of fusion is 32 kJ/mol. Thermal capacities (Cp) of the solid and liquid are respectively 19 J/(K mol) and 28 J/(K mol). Calculate the entropy of fusion at 25 °C (298.15 K)."

The first thing I thought to solve the exercise is to use Kirchhoff's Equation (relation between enthalpy and temperature) to find the enthalpy of fusion at 298.15 K and then divide it by the temperature (in this case 298.15 K) to get the ΔS of fusion I need. See attached image at point (1) – the result appears to be incorrect.

Fortunately my book has solved exercises and they propose the method of point (2) of the image.

My question is - really something that doesn't let me sleep at night - why can't I use my method, or better, why the two methods do not give the same result? Am I missing something about theory? (I must admit that we have not been given a lot of instruments to understand what we are doing, in many cases they just told us: "It is so").

The only explanation I could give myself is that Kirchhoff's equations doesn't work in this cases because the entropy of heating and cooling aren't calculated just by dividing by T the enthalpies of cooling and heating.

Thank you in advance!

Alberto

[Corribus]

Where did you find this formulation of Kirchhoff's Law? Typically it relates the change in reaction enthalpy over temperature to the change in heat capacity.

https://en.wikipedia.org/wiki/Gustav_Kirchhoff#Kirchhoff's_law_of_thermochemistry

[AzeoPoint20]

Hi Corribus, thank you for the reply.

My answer might sound a little idiot...our teacher gave us that form of the Kirchhoff's equation and we demonstrated it by means of a cycle of processes relying on the Hess's Law.

We have been told that we can use Kirchhoff, for example, to find a ΔH at a different temperature than the one given; for example: you need to calculate the ΔH of sublimation of water at 273.15 K so you have to add the ΔH of fusion @ 273.15 K to the ΔH of vaporisation @ 273.15 K, but on the handbooks you'll only find the ΔH of vaporisation at 373.15 K.

So we use Kirchhoff to convert ΔH_vap @ 373.15 K to ΔH_vap @ 273.15 K and then:

ΔH_sub(273.15 K) = ΔH_fus(273.15 K) + ΔH_vap(273.15 K)

And....sorry for my broken English

[Corribus]

ok, but I'm not sure how you're getting entropy in there, since your examples are only to calculate enthalpy at one temperature from enthalpy from another.

There are relationships between entropy and heat capacity, so you maybe could make some substitutions, but they generally make some assumptions about the way the system is changed (constant volume, constant temperature, reversibility, etc.) that may not apply here. I'd have to work it out on paper to see how that version of Kirchhoff's equation is derived, then maybe you could understand why it isn't giving you the correct answer.

[AzeoPoint20]

We have been told that there are some cases (constant pressure, volume and temperature) where we can get ΔS by substituting dq in the dS=dq/T with dH (const. P), dU (const. V) etc. but they definitely didn't give us analytic relations between thermal capacities and entropy.

By the way, I attach to this post the demonstration of our Kirchhoff's Law, even though it is in Italian many words are very similar to English. The demonstration of the same law for a generic reaction has been explained in the same exact way.

Thank you very much once again

[mjc123]

You can't use ΔS = ΔH/T at any arbitrary temperature, only at the equilibrium melting point. When, and only when, the phases are in equilibrium, ΔG = 0 = ΔH - TΔS. So your method 1 is wrong. You need to find the entropy change at the melting point, and adjust to T₂ using dS/dT = C_p/T.

[AzeoPoint20]

Thank you very much for your answer!
I knew my mistake had to do with something alike....and now everything is clearer.

Bye for now, Alberto