# Matthew99

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1. ## Thought experiment about entropy

You are absolutely right, thank you very much for your answer. This is the solution to my thought experiment. From my point of view a very astonishing reminder of how theoretical physics apply to such a simple practical experiment.
2. ## Thought experiment about entropy

You just brought me to an idea, thank you for your input. Would it be correct to say that the increase in thermal energy inside the jar comes from the decrease of potential energy as the distance between magnet and solution decreases?
3. ## Thought experiment about entropy

Thank you very much for your answers! Indeed such an experiment would be very interesting but unfortunately I do not have any lab equipment at home & university labs are currently closed. You are right, my example probably isn't very appropriate. But as joigus mentioned, just think of some magnetic soluble substance / macro structure. Putting a magnet near the isolated jar, the entropy in the jar will decrease by the entropy of mixing (or maybe a bit less as even with the magnet on one side there will still be a few molecules in the other parts of the jar). Now, as total entropy cannot decrease in closed systems (as the mentioned perfectly isolated jar is), there should be an increase in temperature and my question is, where does the energy for this come from as the system itself is closed?
4. ## Thought experiment about entropy

Dear community! I've recently come across following intriguing thought experiment: Imagine a jar filled with water where infinitesimally small particles of iron dust are dissolved. The average particle diameter is small enough that they do not descend, therefore they are randomly distributed in the jar (high entropy). Now you take a magnet and hold it next to the jar - after some time has passed, the density of iron particles on the magnet side of the jar will be significantly higher than the density at the opposite side. Considering entropy, such a behavior should only be possible if the total amount of entropy increases or at least stays the same. However, this demixing/separation effect definitely has negative entropy, maybe not much but not zero. The only counteractive measure that would increase entropy I can think of is an increase in temperature. But if this is really the solution to this thought experiment, where would the energy necessary to heat up the system come from? Thanks a lot for every input
5. ## Confusion about activity coefficients

Dear studiot, thank you for your answer. I'll see what I can find. I've had look in my atkins as well, but they never mention that there are different conventions at all. I've found a copy of the book and this chapter tells very clearly what I was searching for. If anyone else who reads this is interested: Activity coefficients for molality are indeed no good tool to measure deviations from ideal behaviour. This is what Denbigh says:
6. ## Confusion about activity coefficients

Dear studiot, my calculation is meant to display the following exemplary scenario. I dissolve 0.5 mole of J in one liter of water. Therefore, I've got: 0.5 moles of J 1000g/18 g/mole = 55 moles of H2O Further, as you pointed out, I assume that the volume does not change. Thus, I can calculate the mole fraction x_J and the molarity of J. This in turn shows that the activity coefficients y_x and y_c are different, what leads to the confusions I pointed out above.
7. ## Confusion about activity coefficients

Dear community! I've just found a few contradictions in a topic I've never put much thought in. In order to calculate the thermodynamic activity, different formulas are being used. According to Wikipedia and common sense, these formulas should return the same value: According to IUPAC, the standard values for concentration and molality are 1 mole/l and 1 mole/kg respectively. Thus, this relation is only possible if the activity coefficients for x,b,c and so on are different. Consider a simple example, 0.5 mole of a substance J dissolved in 1 l of water (~55 mole): As you can see, the values of the activity coefficients depend on which formula I use. This leads to a few confusions: Often, the activity coefficient is interpreted in a quantitative way, for example, what a activity coefficient greater than 1 means. However, when I look at my calculations, there is no point in interpreting y as its value is quite random and depends on the equation I am using. When calculating the activity coefficient, for example with the Debye-Hückel equation, just looking at the equation, how do I know which y (y_x,y_c,y_b) it returns?
8. ## Derivative of Gibbs free Enthalpy

First of all, thank you very much for your detailed derivation of your equations @studiot Secondly, I've realized by now that I should start commenting with what I mean with the notations I use in my calculations as my assumption that they are standardized is obviously wrong. When I wrote , I meant the gibbs energy of reaction for Q = 1 (lnQ =0) and therefore the slope at this point. As this can be written as and the gibbs energy of formation is a function of temperature, is a function of temperature as well. What I meant, is that the reactions takes place in an open system with constant temperature (e.g. in a large room or outside). Usually only a small amount of moles react per reaction unit, therefore the temperature change of the system due to the reaction is negligible. (e.g. when ). On the contrary, what I want to quantify is how much will change when the reaction takes place at a different system temperature. Well, this is quite confusing. As I usually work with german lecture I've only seen A so far. Maybe we should agree upon a common symbol. Nevertheless, even though there are different names for G in german and english, at least the symbol is the same.
9. ## Derivative of Gibbs free Enthalpy

Dear studiot! I see what you wanted to point out, however, I do not think that this resolves the problem. As it says, standard reaction gibbs energy is the slope at xi = 0. Nevertheless, the slope is still a function of temperature: Therefore, I am not changing xi , I only want to quantify how much the reaction gibbs energy depends on T.
10. ## Derivative of Gibbs free Enthalpy

@joigus Sorry, I forgot to mention that v_J is negative for reactants and positive for products, that's some notation the Atkins has. @studiot I do not know where it is in the 5/6th edition, but in the 9th edition, there is a chapter called standard molar gibbs energies where the formula I used is introduced: Obviously, as joigus pointed out, this is only exactly the case when change in system temperature due to the reaction is constant, however, I've seen that this equation is commonly used for reactions in open systems as the change of system temperature per mole reacting is quite small compared to absolute system temperature. Lets take a closer look at this formula. As far as I understood, the superscript zero at G only indicates that products and reactants appear in same proportions, as the reaction Gibbs energy changes with the reaction quotient because the entropy of reaction is a function of the reaction quotient: When Q is 1, the reaction Gibbs energy is equal to the standard reaction gibbs energy, regardless of temperature. This is also stated in Atkins: As far as I know, the equation on the left does not make sense to use because finding the reaction entropy for any given reaction quotient seems difficulty. Thus, usually equation 3.42 is used to calculate the standard reaction Gibbs energy because we know the standard reaction entropy. Furthermore, the enthalpy of reaction should not change with reaction quotient (at least as far as I know). Therefore, this should be valid: So, lets consider a simple reaction with reactant A and product B: Now, a different approach: This is the equation I posted in the first place and the results I get from these two approaches do not mach up. I can't see any mathematical / logical error in my calculations.
11. ## Derivative of Gibbs free Enthalpy

Well, I suppose this is valid for reactions that happen under normal conditions and in large systems, I haven't seen any calculation where S*deltaT isn't neglected yet. The deltaT here stands for the temperature change that the system undergoes when the reaction happens, and this is usually low compared to the absolute temperature of the system. On the other hand, the integral you just posted is something completely different. This is the difference in reaction enthalpy with respect to standard reaction enthalpy at 298. When the reaction happens at temperature T, the enthalpy of the reaction can be higher/lower depending on the heat capacity of reactants/products than it would be when it would happen at 298K (where H0 is defined). When you take a close look at the picture I posted in my 3rd post, this is the graphical representation of what I'm trying to explain.
12. ## Derivative of Gibbs free Enthalpy

Hello studiot, I've got another version so there are some differences but the involved chapter is called chemical equilibrium. The formula I posted (the first one) comes into action when they derive the gibbs-helmholz equation.
13. ## Derivative of Gibbs free Enthalpy

Thanks a lot for the detailed answer, however, from my point of view, this does not cancel the discrepancy. I forgot to mention that S*deltaT is usually ignored here (at least in Atkins physical chemistry, they say that as the change in system temperature is very low in open systems, we can ignore the last term). On the other hand, the integration over the heat capacity of the reaction from 298K to T takes into account that the reaction enthalpy at temperature T can be different that the standard reaction ethalpy, which is defined at 298K.
14. ## Derivative of Gibbs free Enthalpy

Dear Joigus, thank you for your reply! First of all, I meant gibbs free energy, not gibbs free enthalpy, that was a mistake that occurred due to translation as I am learning physical chemistry in german and it is called free enthalpy there. Secondly, sub R stands for reaction - for example, G sub R would mean the molar free energy of products minus reactants. When the formula is valid for one side of the equation it must be for the whole equation too. 3rd, the total differential of G should indeed be the formula that I used in the beginning, I'll post the calculation via legendre transform of H down below. Concerning Edit 2/3: The superscript zero should suggest that the reaction happens at standard conditions in terms of standard pressure and equal concentrations of reactants/products. H sub r therefore is the enthalpy of the reaction as a function of temperature (as reactants and products can have different heat capacities, the enthalpy of the reaction can vary with different temperatures). Edit: The formula should be valid for fixed temperatures, however, as the temperature chosen can be anything and the total enthalpy is defined by standard conditions at 298K, we have to integrate from 298 to T
15. ## Derivative of Gibbs free Enthalpy

Dear community! I just stumbled upon a strange behavior of the Gibbs free enthalpy when fiddling about with Gibbs Helmholtz equation which I cannot explain. The first formula in the attached file is the one that I've been using all the time, the derivative of G with respect to temperature at constant pressure. The second one just came to my mind when using an alternative derivative. As cp of the reaction is, generally speaking, not equal to zero for reactions with no change in pressure, it seems that these two formulas contradict each other. Surely that cannot be. Anyone got a solution that resolves that contradiction? Thanks in advance.
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