Impact of Pressure on equilibrium of Tin allotropes

Question

The pressure at the bottom of the Mariana Trench in the Pacific Ocean is $1090$ bar. What temperature will the two allotropes of tin be at equilibrium? Assume that the molar volume, energy, and entropy change does not vary with temperature.

Relevant data: Density of white tin: $7.287$ g/mL; Density of grey tin: $5.766$ g/mL

At $1$ bar: The enthalpy change from white tin to grey tin is $-2.016$ kJ/mol and the entropy change is $-7.04$ J/mol K.

My attempt: Since the two species are in equilibrium, I thought first to use the Clausius Clapeyron equation. Assuming that the molar volume is independent with temperature, I calculated the molar volume change to be $4.26\cdot 10^{-6} \frac{\pu{m^3}}{\pu{mol}}$ from white tin to grey tin. Now, I substituted these values into the equation to find that $$\frac{dP}{dT}=\frac{\Delta S}{T\Delta V} =-\frac{1.6\cdot 10^7}{T}$$ upon integration, I find that $$P-P^\circ = 1.6\cdot 10^7 \ln(\frac {T^\circ}{T}).$$ Substituting the temperature of equilibrium at one bar from the data given I find that the temperature is $286.4 \pu{ K},$ thus $T^\circ = 286.4\pu{K}$ and $P^\circ = 10^5\pu{Pa}.$ Thus at $P = 1090\cdot 10^5\pu{Pa},$ $T = 0.317\pu{K}.$ However, the temperature given in the answer is much higher ($220.35$ K). The given answer was derived from a completely different method not involving the Clausius Clapeyron equation.

Hence I have the following question:

Question: Why is it not valid to use the Clausius Clapeyron equation in this scenario? (Or did I make a mistake in my derivation that caused my values to be off?)

Answer 1

First (as you already did) you need to determine the equilibrium temperature at $1\ \mathrm{bar}$. This happens to be when $\Delta G^\circ = 0$, so that $T^\circ_{trans}= \frac{\Delta H^\circ}{\Delta S^\circ}=286.36\ \mathrm K$.

Next you can apply the Clausius equation in integrated form

$$ T = T^\circ_{trans} + (P-P^o) \frac{\Delta V_m^\circ}{\Delta S^\circ} $$

I obtain the following value for $\Delta V_m^\circ$:

$\Delta V_m^\circ = (1/5.766-1/7.287)\times \pu{ 118.71\times10^-6 m3/mol} = \pu{4.2973\times10^-6 m3/mol}$

Plugging this and the values you provide for $\Delta P$ ($1089\ \mathrm{bar}$) and the standard entropy changes $\Delta S_m^\circ$ gives $T=\pu{219.89 K}$. Not exactly what you say is the right answer, but, closer.