[Big-Daddy]

Problem:
http://icho2013.chem.msu.ru/materials/IChO-2013_Theoretical_Official_English_Version.pdf
Looking at problem 1.

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?

[curiouscat]

Type it out. It's a pain having to read a pdf and read a problem  in it.

[Big-Daddy]

Ok but it's long, because every part has to actually be read for my confusion to make sense, and since I have a general lack of clarity towards the end I'll write all the text so there's no risk of my mistakenly paraphrasing. Here goes:

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.

So the answer to 2 is trivial, 3 is fine; the confusion starts at 4, because now we've got liquid water, whereas we just have pressure/temperature readings for the decomposition to ice to become spontaneous, so at this point what's going on is unclear; and in 5, clearly we need to work out the pressure/temperature for the conversion of methane hydrate to liquid water (hence the enthalpy of fusion), but then how can we use the P,T values from before which correspond to the methane hydrate -> ice transition, and how come, when you try it, the P,T set from question (4) is ok to use, but from question (3) is not?

Sorry for all the questions. I feel an overall lack of mastery with this problem, I can do the textbook questions but unsure about this one.

[curiouscat]

Can you post how you solved 3?

[Big-Daddy]

Sure:

3) Question asks for what P the reaction CH4·6H2O (s) -> CH4 + 6 H2O (s) becomes spontaneous, at a temperature of 268.15 K, given that at 101300 Pa this reaction becomes spontaneous at temperature 192.15 K (-81 C). We also have the enthalpy change of -17,470 J/mol. So we just plug into the integrated Clausius-Clapeyron equation

[tex]p_{T_2} = p_{T_1} \cdot e^{- \frac{\Delta H}{R} \cdot (\frac{1}{T_2} - \frac{1}{T_1})}[/tex]

And get the answer. Can you help with the rest?

[curiouscat]

Sure:

3) Question asks for what P the reaction CH4·6H2O (s) -> CH4 + 6 H2O (s) becomes spontaneous, at a temperature of 268.15 K, given that at 101300 Pa this reaction becomes spontaneous at temperature 192.15 K (-81 C). We also have the enthalpy change of -17,470 J/mol. So we just plug into the integrated Clausius-Clapeyron equation

[tex]p_{T_2} = p_{T_1} \cdot e^{- \frac{\Delta H}{R} \cdot (\frac{1}{T_2} - \frac{1}{T_1})}[/tex]

And get the answer. Can you help with the rest?

Hmm. Interesting.

I am not sure but this doesn't sound right to me. I'm still thinking though; maybe you are indeed correct.

[Big-Daddy]

I am not sure but this doesn't sound right to me. I'm still thinking though; maybe you are indeed correct.

What's the issue with it? The logic seems airtight to me, and the mark scheme agrees with the answer so I'd assume it's correct. It's 4 & 5 I have trouble with. And they're also just simple manipulations of the same equation (along with pressure = P_atm + ρgh), except I don't feel I understand how to use the Clausius-Clapeyron equation for these reactions and situations properly...

[curiouscat]

I am not sure but this doesn't sound right to me. I'm still thinking though; maybe you are indeed correct.

What's the issue with it? The logic seems airtight to me, and the mark scheme agrees with the answer so I'd assume it's correct. It's 4 & 5 I have trouble with.

Ok, then you must be right.

[Big-Daddy]

Any ideas for 4?

I'm trying to think around it very carefully. As we go deeper into the liquid water, pressure increases so water's melting point decreases, thus water can remain liquid at lower temperatures than 273.15 K. I'm not sure how to go from this to conclude that the minimum temperature of coexistence between liquid water and methane hydrate is 272.9 K though? (It is 272.9 K according to solutions.)

So this minimum temperature and pressure condition is the same situation in which methane hydrate will be able to decompose into ice. (or, if we lower the T by an infinitesimally small amount, etc.) Thus finding the pressure at this temperature from the previously obtained P,T for the same reaction (decomposition of methane hydrate to form ice) makes sense, since this is the point at which the ice can begin to form as well. Ok then. So let's clear up the rest of 4 and then hopefully look at 5...

I get the feeling I am missing some sort of principle to do with manipulations of the Clausius-Clapeyron equation here

[curiouscat]

Any ideas for 4?

Sorry. None, until I first convince myself why your solution to #3 is right. #4 builds upon the easier #3
I think so I need to grok it first.

[curiouscat]

What's the issue with it? The logic seems airtight to me, and the mark scheme agrees with the answer so I'd assume it's correct.

What's the numerical answer you get for P for part #3? I was getting ~22 atm.

[curiouscat]

Here's one idea for #4. Not 100% sure though.

First find L' for the phase change CH4·6H2O (s)  CH4 + 6 H2O (l)

by applying Hess' Law to the known L for

 CH4·6H2O (s)  CH4 + 6 H2O (s)
and
H2O (s)  H2O (l)

Now using Clausius Clayperon for the H2O(s)  H2O(l) phase change get a P-T relation.

Roughly, I get:

P = -135 T (P in atm; T in °C).(Eq. 1)

Now use Clausius Clayperon in the form you previously had but with L':



Substitute P=fn(T) from Eq. 1

You ought to get dP / dT = fn(T, constants)

Now find the range for P on the feasable solution set.

[Big-Daddy]

For part 3 I got 2.247314·10⁶ Pa i.e. yes, 22.2 atm.

Here's one idea for #4. Not 100% sure though.

First find L' for the phase change CH4·6H2O (s)  CH4 + 6 H2O (l)

by applying Hess' Law to the known L for

 CH4·6H2O (s)  CH4 + 6 H2O (s)
and
H2O (s)  H2O (l)

I'm not sure about this solution ... the enthalpy for fusion is provided for question 5, not 4. In 4, we're still meant to be working with CH4·6H2O (s)  CH4 + 6 H2O (s). It's the circumstances the question is proposing that confuses me. I mean, what are we looking at here - minimum temperature at which water will be liquid, i.e. maximum temperature at which it would be solid at that P; and the pressure point at which methane hydrates are stable, whereas at lower pressures they would decompose - so the point at which liquid water, ice, and methane hydrate can all simultaneously coexist? Does it work to think like this?