Question

Calculation of enthalpy and entropy of fusion of an unknown solid?

Answer 1

Entropy is the amount of intrinsic disturbance within a compound, while enthalpy is the amount of internal energy present in the compound. For elemental compounds like hydrogen and oxygen, enthalpy is zero; however, enthalpy is nonzero for water (regardless of phase).

Complete answer:
$\Delta {S_{fus}} = 5.52J/mol\,K$
$\Delta {H_{fus}} = 2.36kJ/mol$
Explanation:
The Clapeyron equation in the form will be your tool of choice for this problem.
$\dfrac{{dP}}{{dT}} = \dfrac{{\Delta {S_{fus}}}}{{\Delta {T_{fus}}}}$ (i)
You now realise that the enthalpy change of fusion, $\Delta {H_f}$, and the entropy change of fusion, $\Delta {S_f}$have the following relationship.
$\Delta {S_{fus}} = \dfrac{{\Delta {H_{fus}}}}{T}$ (II)
The Gibb’s free energy transition at equilibrium is used to derive this.
$\Delta G = \Delta H - T\Delta S$
Since $\Delta G = 0$ in equilibrium, it follows that you have
$\Delta H = T \cdot \Delta S \\\ \Rightarrow \Delta S = \dfrac{{\Delta H}}{T} \\\$
$T$ will now reflect the melting temperature in your case. The average of the two melting temperatures is a reasonable rule of thumb to follow in this situation.
${T_{average}} = \dfrac{{427.15K + 429.26K}}{2} = 428.21K$
You should now rearrange equation (i) to solve for $dT$ and then integrate, but if you assume that the temperature shift, $dT$ , is small enough, you can skip this stage.
You can get away with such an estimate because you're working on the solid-liquid phase line, which means minor temperature shifts are inevitable. If you go this way, you'll be able to tell that
$\dfrac{{dP}}{{dT}} \approx \dfrac{{\Delta P}}{{\Delta T}} = \dfrac{{\Delta {S_{fus}}}}{{\Delta {V_{fus}}}}$
You have everything you need to solve $\Delta {S_{fus}}$ at this stage. You are aware, in particular, that
$\Delta T = {T_2} - {T_1} = 2.11K$
Here comes the tricky part - you need to convert $\Delta {V_{fus}}$ and $\Delta P$ to cubic meters per mole, ${m^3}/mol,$and $pascals,{\text{ }}Pa$
$\Delta {V_{fus}} = {V_2} - {V_1} \\\ \,\,\,\,\,\,\,\,\,\,\,\, = 156.6c{m^3} - 142.0c{m^3} \\\ \,\,\,\,\,\,\,\,\,\,\,\, = 10.6c{m^3} \\\$
This implies that you've
$10.6\dfrac{{c{m^3}}}{{mol}} \cdot \dfrac{{1{m^3}}}{{{{10}^6}c{m^3}}} = 10.6 \times {10^{ - 6}}{m^3}/mol$
Finally, you have
$\Delta P = 1.2 \times {10^6}Pa - 1.01325 \times {10^5}Pa \\\ \,\,\,\,\,\,\,\, = 10.987 \times {10^5}Pa \\\$
As a result, plug in these numbers and solve for $\Delta {S_{fus}}$
$\Delta {S_{fus}} = \dfrac{{\Delta P}}{{\Delta T}} \cdot \Delta {V_{fus}}$
$\Delta {S_{fus}} = \dfrac{{10.987 \times {{10}^5}Pa}}{{2.11K}} \times 10.6 \times {10^{ - 6}}\dfrac{{{m^3}}}{{mol}}$
This will equal to
$\Delta {S_{fus}} = 55.195 \times {10^{ - 1}}{\dfrac{{Pam}}{{K\,mol}}^3}$
But, $Pa \times {m^3} = J$ , so the answer will be
$\Delta {S_{fus}} = 5.52\dfrac{J}{{mol\,K}}$
Now, use the equation (ii) to get
$\Delta {H_{fus}} = \Delta {S_{fus}} \cdot {T_{average}} \\\ \Delta {H_{fus}} = 5.52\dfrac{J}{{mol \cdot K}} \cdot 428.21K \\\ \,\,\,\,\,\,\,\,\,\,\,\, = \,\,\,2363.72\dfrac{J}{{mol}} \\\$
This value will also be rounded to three sig figs, but expressed in kilojoules per mole.
$\Delta {H_{fus}} = 2.36kJ/mol$

Note:
Now, the question can arise: Does enthalpy increase as entropy increases? As a result, an enthalpy shift will affect entropy. The external entropy (entropy of the surroundings) increases in an exothermic reaction. The external entropy (entropy of the environment) decreases in an endothermic reaction.