# Determine if the dissolving process is exo/endothermic and

1. Apr 15, 2015

### davidbenari

1. The problem statement, all variables and given/known data
The molar fraction of methane dissolved in water can be calculated from Henrys constant which are: 4.131 atm at 25ºC and 5.771 atm at 50ºC.

(a) Determine if the dissolution process of methane in water is exothermic or endothermic, and calculate the value of the enthalpy of dissolution for methane in water.

2. Relevant equations
$\Delta H_{ideal}=0$
$\Delta S_{ideal}=-nR\Bigg(\sum_i x_i lnx_i\Bigg)$
$H^{Excess}=n\xi RT x_A x_B$
$p=K_{Henry}X_A$

3. The attempt at a solution

I don't see any pattern in the constants provided. The fact that pressure increases with temperature doesn't reflect any information about the interactions between the different molecules, so I cant deduce anything about enthalpy.

I frankly have no idea what to do for this problem I find it impossible... :S

Last edited: Apr 15, 2015
2. Apr 15, 2015

### Staff: Mentor

Question refers to Henry's constant - where is it used?

3. Apr 15, 2015

### davidbenari

Could you tell me a hint? I know of no relation with henry's constant that will allow me to say if the process is endothermic or exothermic

4. Apr 15, 2015

### Staff: Mentor

What I wrote was already a hint. What is the relationship that uses Henry's constant? You have not listed it in relevant equations.

Another hint: how does the methane concentration change with the growing temperature? What does it tell you about the dissolution process?

There is also another important equation that you have failed to list, one that combines changes of the equilibrium constant with the temperature changes.

5. Apr 15, 2015

### davidbenari

My textbook only mentions the equation $p=K_H X_A$. I saw an equation on wikipedia that expresses the temperature dependence with an exponential , but I doubt that I'll have to use that one since It requires a special parameter for the dissolved thing.

The fact that the constant is higher for higher temperature I think tells me nothing new, because all substances have a tendency to become vapor when their temperature is increased.

6. Apr 15, 2015

### Staff: Mentor

If I remember correctly, the heat of mixing is related to the activity coefficient. That's about all that I remember, but if I had to do this problem, I would know where to start looking. How is the Henry's law constant related to the activity coefficient at infinite dilution?

Chet

7. Apr 15, 2015

### davidbenari

At infinite dilution the activity coefficient becomes zero. Doesn't it?

8. Apr 16, 2015

### Staff: Mentor

Good. That should be enough for you to find solubility for two different temperatures. Once you know them it is like being able to say "dissolution equilibrium constant for temperature T1 is s1, and for temperature T2 is s2. That gives you two points to plug into an equation. Question remains - what equation?

No idea what equation you mean. The one I am thinking about is not related specifically to dissolution, it is pretty universal and covers all equilibrium processes.

I think you are mistaking dissolution with evaporation here. Knowing solubilities at two different temperatures and applying Le Chatelier's principle should already give you an answer to the first part of the question.

9. Apr 16, 2015

### Staff: Mentor

No. What would make you think that?

Chet

10. Apr 16, 2015

### davidbenari

Well the limit as it approaches pure A makes the activity coefficient equal to 1. I thought that it would approach zero at infinite dilution.

Anyways I ended up using this equation $Ln(S_2/S_1)=\frac{H}{R}(1/T_1-1/T_2)$ without any real idea why. Anyone knows where this equation comes from? I'm not being lazy here, its just that I'm clueless about this topic. My "class textbooks" don't talk about similar problems to this one. Neither do they speak about Henrys Constant relationship to enthalpy and temperature.

11. Apr 17, 2015

### Staff: Mentor

http://en.wikipedia.org/wiki/Van_'t_Hoff_equation

Perhaps you are expected to connect the dots by yourself?

12. Apr 17, 2015

### Staff: Mentor

This is not correct either. The activity solution at infinite dilution refers to the solute, not the solvent. Infinite dilution means that the concentration of the solute is approaching zero. The activity coefficient of the solute does not approach unity as its concentration approaches zero. I challenge you to find a single reference that says that it does.

Chet

13. Apr 17, 2015

### davidbenari

Chet I was saying the activity coefficient of A approaches 1 when the solution approaches pure A. I'm not talking about solutes and solvents and binary systems. I'm talking about component A. As for the reference I will annex an image of Levine's book on physical chemistry for this:http://imgur.com/epSZvC7 http://imgur.com/epSZvC7 .. Although I'm sure you thought I was saying something else.

My argument then was that if $\gamma_A$ approaches unity as $x_A$ aproaches 1, then it $\gamma$ approached zero when $x$ became zero.

14. Apr 17, 2015

### Staff: Mentor

Of course you are considering a binary system with solute and solvent. What do you think you have when a gas dissolves in a liquid? That's what Henry's law is all about.

Chet

15. Apr 17, 2015

### davidbenari

No. I was making my statement about the general case for activity coefficients. Substitute whatever size component system you want. The activity coefficient of a certain substance reaches 1 as the solution becomes pure in that substance. According to the image I posted, at least.

It might be that for this problem the relevant case is a binary system. But that wasn't my point.

I think this thread is pointless; all I've been told is I need to "look", and I've done that.

Threads regarding the torque on a current loop of wire, or the work done to an apple that falls to the ground aren't filled with comments telling the OP's to "look".

This question is as mysterious for me as when I first saw it.

16. Apr 17, 2015

### Staff: Mentor

David, as you know, what we try to do in PF is to help members solve their problems on their own by giving hints and asking leading questions. In regard to Henry's law, there is a direct connection between the temperature dependence of the infinite dilution activity coefficient of the solute gas and the heat of mixing of the solute gas with the solvent liquid. That's why I was asking about the infinite dilution activity coefficient. It seemed obvious to me that you would know that I was talking about the solute. Apparently not. I also assumed that you knew the relationship between the infinite dilution activity coefficient and the Henry's law constant. Apparently not again. What are they teaching you in that thermo course of yours?

Chet

17. Apr 17, 2015

### davidbenari

Well its a chemical thermodynamics course so I definitely see why they should've taught me all that you've mentioned. I'm supposed to read from Peter Atkins. And I can safely say that in all the chapters on mixtures they don't mention these dependences.

Wikipedia is a good place to look at once you have a solid understanding of the subject, I think. Not a very good one when you're trying to learn the stuff. At least that has been my experience so far, when investigating scientific stuff there.

We already finished the mixtures topic in class, but activity coefficients were only mentioned superficially as some type of deviation from the ideal scenario.

Thanks.

18. Apr 17, 2015

### Staff: Mentor

It's regretful that the text being used in your course is not satisfying your needs. As a supplement, I would like to suggest that you consider using Introduction to Chemical Engineering Thermodynamics by Smith and van Ness.

Chet