Why is Entropy expressed in terms of Enthalpy?

Why does ${\Delta}S_{m(fusion)} = \frac{{\Delta}H_{m(fusion)}}{T}$ ?

I always thought ${\Delta}S = \frac{dQ}{T}$

In this case does it mean $dQ = {\Delta}H$ ??

Why is it so?

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As much as I know, this relation is for a reversible(slow) thermodynamic change. And dQ is the heat given, enthalpy is a different thing. The relation is: $\delta S=\frac{dQ_{rev}}{T}$. The heat given is at constant pressure equal to change in enthalpy, I think. – Ashish Gaurav Apr 1 '13 at 10:20
Can you give a source of $dQ = \Delta H$? – user12345 Apr 1 '13 at 11:35
$dQ=\Delta H$ only if $\Delta W$ is zero. – ABC May 1 '13 at 13:29

I guess you have a special case of constant pressure, as in general you have $$dH = TdS + Vdp,$$ if the number of particles is constant. Then if you assume constant pressure, i.e. $dp = 0$, as I believe was done in your problem, you get $$dH = TdS,$$ and rearrange this to $$dS = \frac{dH}{T}.$$ Hope this answered your question.

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${\rm d}Q$ is normally taken as heat into/out of a closed system. ${\rm d}H$ is the change of energy in an existing system when adding something to the system. Enthalpy takes into account the internal energy of that which you are adding, $T{\rm d}S$ and the work done in creating space in the system, $V{\rm d}P$.

It follows that where ${\rm d}P =0$ then the change in energy of the system is just that energy of the 'part' you have added $T{\rm d}S$.

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Phase transitions are commonly studied at constant pressure and temperature. The right thermodynamic potential for these constraints (i.e. fixed pressure and temperature) is the gibbs free energy sometimes called the free enthalpy $G=H-TS$.

Now, for a single component system, when two phases 1 and 2 are at coexistence in thermodynamic equilibrium the gibbs free energy per mol has to be the same in the two phases (I may recall here that this also makes sense because the Gibbs free energy happens to be proportional to the chemical potential of the component). This implies that

$\Delta_{12}G_m = \Delta_{12}H_m - T\Delta_{12}S_m = 0$

Overall, it turns out that one can then relate the latent heat $\Delta_{12}H$ to a change in entropy when the system changes from phase 1 to phase 2.

In the case of fusion (going from solid state to liquid state) for instance you get the formula

$\Delta S_{m(fusion)} = \frac{\Delta H_{m(fusion)}}{T}$

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