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About Matthew99

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  1. 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.
  2. 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.
  3. @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.
  4. 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.
  5. 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.
  6. 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.
  7. 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
  8. 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.
  9. Very interesting! It's great to see when application of some theoretical maxwell relations actually gives us a practical result which is of great importance in many industries for sure. Given that walls with perfect heat insulation are rather a matter of theoretical considerations than of a real world system, would the result change significantly if we allow some heat transfer with the surroundings? Thanks a lot for your effort!
  10. Thank you very much for your answers! Considering the Van der Waals potential, it seems very reasonable that the JT coefficient changes with temperature and pressure. However, the formula that studiot posted suggests that the JT coefficient is a function of temperature only, as the other coefficients are constants and as far as I know the VDW coefficients do not change with pressure. How does that match?
  11. Dear community! Obviously, the Joule Thomson coefficient is a function of pressure and temperature, as inversion temperature plots show. Thus, I was wondering why this is the case. I found an explanation that says that depending on the pressure, either repelling forces or attractive forces predominate and according to conservation of energy, when the gas expands without a change in enthalpy the thermal energy and therefore the temperature has to decrease when the potential energy is increasing and vice versa - resulting in a positive JT-coefficient when attractive forces predominate and a negative JT-coefficient when repelling forces predominate. This seems reasonable. However, inversion temperature plots show that the JT-coefficient changes substantially when temperature is changed. So far, I have not figured out why the temperature of the gas has that much of an importance here. Anyone got an intuitive explanation? Thank you in advance for your help!
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