Entropy of fusion; two methods, different result?

[standardflop]

"Calculate the entropy of fusion of A at 25deg C given that its enthalpy of fusion is 32kJ/mol at its melting point 146deg C. Also given is Cp,m(liquid)= 28J/mol/K and Cp,m(s)= 19J/mol/K."

I thought of two approaches. Both should be valid (to my knowledge), but only the first gives the correct result. The first one:
Using Kirchhoffs Law,
$$\Delta H (T2) = \Delta H(T1) + \int_{T_1}^{T_2}\Delta Cp,m dT\approx \Delta H(T1)+\Delta Cp,m (T2-T1)$$
and then applying the definition
$$\Delta S(T2)= \frac{\Delta H(T2)}{T2}$$
(gives the correct result)

But my second idea apparently failed:
Simply setting up the cyclic equality
$$\Delta S(s\to l,T2) = \Delta S(s,T2to T1)+\Delta S(s\to l,T1)+\Delta S(l,T1\to T2)$$

where the terms are of the form
$$\Delta S(i,T1\to T2) = Cp,m \ln (T2/T1)$$
Can someone tell me why this approach gives a different result, about 73J/mol/K compared with the first, and correct, result of approx. 100J/mol/K.?
Thank you in advance.

 

[anathema]

Hello, thank you for your contribution to the forum. I can help explain why your second approach did not give the correct result. The reason is that the cyclic equality you set up assumes that the process is reversible, meaning that the system is in thermal equilibrium at all times. However, in the case of melting, this is not true. When a solid melts, there is a temperature gradient between the solid and liquid phases, meaning that the temperature is not the same throughout the system. Therefore, the process is not reversible and the cyclic equality cannot be applied.

In the first approach, Kirchhoffs Law takes into account the change in enthalpy due to the temperature difference between the melting point and the desired temperature. This accounts for the non-reversibility of the process and gives the correct result.

I hope this helps clarify the issue. Let me know if you have any further questions. Thank you.