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

## The molar volume of a certain solid is 142.0 cm3/mol at 1.00 atm and 427.15 K, its melting temperature. The molar volume of the liquid at this temperature and pressure is 152.6 cm3/mol. At 1.2 MPa the melting temperature changes to 429.26 K. Calculate the enthalpy and entropy of fusion of the solid.

Nov 3, 2015

$\Delta {S}_{\text{fus" = "5.52 J/mol K}}$
$\Delta {H}_{\text{fus" = "2.36 kJ/mol}}$

#### Explanation:

Your tool of choice for this problem will be the Clapeyron equation used in the form

color(blue)((dp)/(dT) = (DeltaS_"fus")/(DeltaT_"fus"))" " " "color(purple)((1))

Now, you know that the following relationship exists between the enthalpy change of fusion, $\Delta {H}_{f}$, and the entropy change of fusion, $\Delta {S}_{f}$

color(blue)(DeltaS_"fus" = (DeltaH_"fus")/T)" " " " color(purple)((2))

This is derived from the Gibbds free energy change at equilibrium

$\Delta G = \Delta H - T \Delta S$

Since at equilibrium $\Delta G = 0$, it follows that you have

$\Delta H = T \cdot \Delta S \implies \Delta S = \frac{\Delta H}{T}$

Now, in your case, $T$ would represent the meting temperature. A good rule of thumb to go by here is that you can use the average of the two given melting temperatures

${T}_{\text{average" = ("427.15 K" + "429.26 K")/2 = "428.21 K}}$

Now, you should rearrange equation $\textcolor{p u r p \le}{\left(1\right)}$ to solve for $\mathrm{dT}$, and then integrate, but you could skip that step if you go by the assumption that the temperature change, $\mathrm{dT}$, is small enough.

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, if you take this route, you can say that

$\frac{\mathrm{dp}}{\mathrm{dT}} \approx \frac{\Delta P}{\Delta T} = \left(\Delta {S}_{\text{fus")/(DeltaV_"fus}}\right)$

At this point, you have everything you need to solve for $\Delta {S}_{\text{fus}}$. More specifically, you know that

$\Delta T = {T}_{2} - {T}_{1} = \text{2.11 K}$

Here comes the tricky part - you need to convert $\Delta {V}_{\text{fus}}$ and $\Delta P$ to cubic meters per mole, $\text{m"^3"/mol} ,$ and pascals, $\text{Pa}$ - you'll see why in a minute!

$\Delta {V}_{\text{fus" = V_2 - V_1 = "152.6 cm"^3 - "142.0 cm"^3 = "10.6 cm}}^{3}$

This means that you have

10.6color(red)(cancel(color(black)("cm"^3)))/"mol" * "1 m"^3/(10^6color(red)(cancel(color(black)("cm"^3)))) = 10.6 * 10^(-6)"m"^3"/mol"

Finaly, you have

$\Delta P = 1.2 \cdot {10}^{6} \text{Pa" - 1.01325 * 10^5"Pa" = 10.987 * 10^5"Pa}$

So, plug in these values and solve for $\Delta {S}_{\text{fus}}$

$\Delta {S}_{\text{fus" = (DeltaP)/(DeltaT) * DeltaV_"fus}}$

$\Delta {S}_{\text{fus" = (10.987 * 10^5"Pa")/"2.11 K" * 10.6 * 10^(-6)"m"^3/"mol}}$

This will be equal to

$\Delta {S}_{\text{fus" = 55.195 * 10^(-1)"Pa m"^3/"K mol}}$

But $\text{Pa" xx "m"^3 = "J}$, so the answer will be

DeltaS_"fus" = color(green)(5.52"J"/"mol K")

Now use equation $\textcolor{p u r p \le}{\left(2\right)}$ to get

$\Delta {H}_{\text{fus" = DeltaS_"fus" * T_"average}}$

$\Delta {H}_{\text{fus" = 5.52"J"/("mol" * color(red)(cancel(color(black)("K")))) * 428.21color(red)(cancel(color(black)("K"))) = 2363.72 "J"/"mol}}$

I'll leave this value rounded to three sig figs as well, but expressed in kilojoules per mole

DeltaH_"fus" = color(green)("2.36 kJ/mol")