# Phase transition problems

1. Sep 11, 2013

1. The problem statement, all variables and given/known data
http://icho2013.chem.msu.ru/materials/IChO-2013_Theoretical_Official_English_Version.pdf
Looking at problem 1.

2. Relevant equations

The integrated form of the Clausius-Clapeyron equation:

$$ln(\frac{P_{T2}}{P_{T1}}) = -\frac{\Delta H}{R} \cdot (\frac{1}{T_2} - \frac{1}{T_1})$$

3. The attempt at a solution

Part 1 is irrelevant, do not need to read. Part 2 is fine, Part 3 is fine. For Part 4, once you correctly guess the temperature, it's ok, because we rely on the same reaction as before, decomposition of methane hydrate into ice, and looking for minimum temperature at which it will not occur; we can use either of the 2 possible P,T sets (one given, one ascertained from Part 3) to calculate the P for Part 4.

However, Part 5 causes trouble because only using one of the 3 P,T sets of data we have now can we find the correct T from the P in Part 5; the other two give wrong readings. Why is this, and how should I tell which sets of P,T data can be used for future P, T calculations from the situation and which cannot?

2. Sep 17, 2013

### DrDu

Could you explicitly state the p, T sets you are referring too? Which values did you get?

3. Sep 26, 2013

I'll write out the problem again if that helps:

Real methane hydrate has a non-stoichiometric composition close to CH4•6H2O. At atmospheric pressure, methane hydrate decomposes at –81 °C. However, under high pressures (e.g. on the ocean floor) it is stable at much higher temperatures. Decomposition of methane hydrate produces gaseous methane and solid or liquid water depending on temperature.

2. Write down the equation of decomposition of 1 mole of CH4•6H2O producing solid water (ice) H2O(s).

The enthalpy of this process equals 17.47 kJ•mol-1. Assume that the enthalpies do not depend on temperature and pressure, the volume change upon decomposition of hydrate is equal to the volume of released methane, and methane is an ideal gas.

3. At what external pressure does decomposition of methane hydrate into methane and ice take place at –5 °C?

4. What is the minimum possible depth of pure liquid water at which methane hydrates can be stable? To answer this question, you should first deduce at which minimum temperature methane hydrate can coexist with liquid water, by choosing from 3 possibilities: 272.9 K, 273.15 K, 273.4 K.

Large methane hydrate stocks on the floor of Baikal lake, the largest freshwater lake in Russia and in the world, have been discovered in July 2009 by the crew of a deep-submergence vehicle «Mir-2». During the ascent from the depth of 1400 m methane hydrate samples started to decompose at the depth of 372 m.

5. Determine the temperature in Baikal lake at the depth of 372 m. The enthalpy of fusion of ice is 6.01 kJ•mol-1.

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3 is straightforward because it is asking for the conditions required to make a reaction happen, given that the reaction was previously spontaneous at a given p,T of (192.15 K, 101300 Pa). But the confusion starts at 4. I really don't understand the scenario being proposed, and how it relates with the CH4•6H2O -> CH4 + H2O (s) reaction for which the two p,T sets we have (the initial given p,T of 192.15 K, 101300 Pa, and the solution to part 3, at 268.15 K) apply. And the same confusion grows with 5.

I think the problem is a fundamental questioning of whether one understands multi-component phase diagrams, how the Clausius-Clapeyron equation can be used for spontaneity conditions in any reactions and with various different conditions (e.g. 4 proposes that liquid water must also be present), etc. Any help on these topics and understanding them better would be much appreciated. (I have seen the solutions, but they do not clear up my haze about how the Clausius-Clapeyron equation is to be applied in a case like this, and how I should understand and analyse each scenario in the problems, which is the sort of understanding I am trying to build.)

4. Sep 30, 2013

### DrDu

I think in 4 what is meant is that the minimum depth where the hydrate is stable is determined in a two step process: For a given depth (=pressure), the hydrate is most stable at the lowest possible temperature, i.e. the temperature where liquid water is stable and does not freeze. So you would have to find the freezing point curve T(P) of water and plug this temperature into the Gibbs Duhem relation in a second step. Apparently the authors of the question did consider this to be too difficult, and let you chose between 3 temperatures. As the volume of water decreases on melting, the melting point decreases with increasing pressure, so the correct T must be 272.9 K (as 273.15 is at normal pressure).
In question 5 you are given the depth (=pressure) where decomposition occurs. Apparently, lake Baikal is somewhat warmer than the freezing point at that depth. So the hydrate decomposes into methane and liquid water and you have to take into account melting enthalpy in calculating the change in equilibrium.

5. Oct 5, 2013

An understanding of the rest of the question is beginning to shape in my head but I will save those questions for later. First I'd like to know what the Gibbs Duhem relation is by which we could have calculated 272.9 K? How do we do this - and if all the needed information for predicting this value is given then why would the question-writer not have expected us to do this?

6. Oct 8, 2013

### DrDu

I think the problem is already very hard the way it is formulated, and giving the temperature was meant to make it more tractable.

Gibbs-Duhem:
ndμ=VdP-SdT
Along the solidus line $\mu_l=\mu_s$
hence
$V_l/n_l dP-S_l/n_l dT=V_s/n_s dP -S_s/n_s dT$.
V/n and S/n are the molar volume and entropy, respectively.
We can write this as
$\Delta V_m dP=\Delta S_m dT$
with $\Delta S_m =\Delta H/T$,
where $\Delta H$ is the enthalpy of melting
we get $dT/dP=T\Delta V_m/\Delta H$ which is the change of melting temperature with pressure.
As melting costs heat $\Delta H>0$, but volume decreases $\Delta V_m<0$, the melting temperature decreases with pressure.

7. Oct 12, 2013