# Impact of Pressure on equilibrium of Tin allotropes

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?)

• One problem with your approach is that the C-C equation is $\frac{dP}{dt}=\frac{\Delta S}{\Delta V}=\frac{\Delta H}{T\Delta V}$. You've mixed up $\Delta S$ and $\Delta H$. – Andrew Jan 25 '19 at 17:47

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 $$T^\circ_{trans}= \frac{\Delta H^\circ}{\Delta S^\circ}=286.36\ \mathrm K$$.
$$T = T^\circ_{trans} + (P-P^o) \frac{\Delta V_m^\circ}{\Delta S^\circ}$$
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.