
#37
Apr1510, 07:24 AM

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Entropy is defined by the second law. That may not sound helpful, but recall the second law, properly written, is a statement on the rate of entropy *production* during a process. SM is a theory of equilibrium states, nothing more. Thus, the "definition" of entropy in SM (k ln (W)) can only correspond to thermo*statics*. In thermodynamics, the entropy is related to the maximum amount of heat it is possible to generate during a process. Specifically, [tex]T \dot{S} \geq Q[/tex] It's important to recall that Q is not the 'heat' but the *rate* of heating, as seen from conservation of energy: [tex] \dot{E} = W + Q[/tex] Similarly, W is the *rate* of work (not the partition function). In thermostatics and SM, the time derivatives are gotten rid of, and differentials added (often times very confusingly). The problem of thermodynamics, from this point forward, is first how to determine the constitutive functionals W, S, and F (the free energy) and second, to derive thermostatics from the thermodynamic functionals previously assigned. Let me emphasize again that just because one has a "rigid" definition of thermostatic functionals, one *cannot* extrapolate to thermodynamics. k ln W is not a 'more fundamental' definition of entropy, since the domain of validity of SM is limited to only equilibrium states. Does this help? 



#38
Apr1510, 07:28 AM

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#39
Apr1510, 07:29 AM

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#40
Apr1510, 08:04 AM

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P: 1,395

The most fundamental equation in chemical kinetics, the Arrhenius equation, is an empirically derived formula, and the field remains firmly grounded in using empirical models for reaction rates. Chemical engineers certainly rely primarily on empirical data and models when determining things like how scaling up a chemical reactor will affect dynamic quantities like heat and material transport, and how those will then affect reaction rates. In almost all cases, the microscopic details of what is going on at the molecular scale is far too lowlevel and far too "fuzzy" to be worth considering. In any case, these microscopic theories are not wellenough understood or developed to be useful to a chemical engineer in the first place. Try telling them, "I have a model that describes in detail the molecular scale processes that must be contributing to the largescale process you are trying to model." They might say, "Great! How accurately can it predict changes in reaction rates over suchandsuch a range of temperatures and pressures." If one then responded, "It does really well, and is never off by more than an order of magnitude or so.", then one would be lucky if all they do is laugh in one's face. For their purposes, such a large uncertainty is almost certainly completely untenable. 



#41
Apr1510, 08:07 AM

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#42
Apr1510, 08:32 AM

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At first I thought to myself
Perhaps they teach a different brand of reaction kinetics in American universities from that put forward in Oxford and Cambridge in the Uk. But no, wait, I find on Page 274 of what is probably the foremost American text on the subject of physical chemistry by Moore, a chapter entitled 'Collision theory of gas reactions', followed by statistical mechanics maths. So I see all is OK and it is just that some at PhysicsForums have not read this book or its brothers. For the record The link between chemical kinetics and statistical mechanics is to do with the probabilities of two molecules meeting. This must obviously happen for those two molecules to react chemically. This application of SM is obviously different from the application Andy et al are describing, so can be expected to have a different appearance. In particular they are not necessarily about equilibrium. SM can be and indeed is employed in non equilibrium situations. Mix 1 mole of sodium hydroxide with 1 mole of hydrochloric acid. You have a definite non equilibrium situation. I wasn't thinking of the Arrhenius equation when I mentioned SM, but it is true you can get to it from there. However this is not the fundamental equation of chemical kinetics; it merely describes the temperature dependance of what is known as the 'rate constant'. This, of course, is only useful with simple order reactions that have a single rate constant. 



#43
Apr1510, 09:08 AM

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#44
Apr1510, 09:20 AM

P: 2,159

The second law does not define entropy at all. Rather one has a general definition of entropy applicable for general systems that does not assume thermal equilibrium. Then the second law can be derived under some assumptions (like the equal prior probability assumption and time reversibility). When we do thermodynamics what we actually do is approach a complicated situation as closely as we can from within thermostatics. The typical undergraduate textbook method is to consider only initial and final states which are accuratley describred by thermostatics. What engineers do in practice when considering the actual dynamics of a system in terms of fluid velocity field, temperature and pressure distribution, is also strictly speaking approaching the real problem from within thermostatics. What happens here is that you add a lot of external variables in the description of the system. So, you just use a different coarse graining level and make statistical assumptions about those desgrees of freedom that you do not keep in your description. Those assumptions do not have to be that it is in equilibrium, you can e.g. describe the situation using some Boltzmann distribution function whose evolution is given by a collision integral. But it should be clear that whenever you describe a system consisting of 10^23 degrees of freedom formally in terms of formulas that can be described in, say, 100 bits, you are necessarily making statistical assumptions about all those degrees of freedom. 



#45
Apr1510, 09:25 AM

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I provided 3 major chemical engineering references, and SM is nowhere to be found in them. Now you provide a text on physical chemistry, and make some more claims (which I cannot substantiate; amazon does not allow me to see the table of contents). However, other texts on physical chemistry that I can see the TOC (Atkins is one, Silbey is another) again have *nothing* on SM. Now you claim SM can be applied to nonequilibrium situations, and again, do not provide a reference. You do not see how kinetics is a linearized situation. Just as I can't make you eat your vegetables, I can't make you learn. I can, however, try and prevent you from spreading misinformation to others. 



#46
Apr1510, 09:27 AM

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#47
Apr1510, 09:37 AM

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#48
Apr1510, 10:43 AM

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Or appears to deny that if you mix some of the (nearly) stongest acid with some of the (nearly) strongest alkali you have a non equilibrium situation. 



#49
Apr1510, 10:49 AM

P: 5,462

However cast first the beam from thine own eye. 2) Atkins is an excellent book: My copy has two chapters about Statistical Thermodynamics, Ch19 entitled the concepts and Ch20 entitled the machinery. 3) I do not know Silbey so cannot comment. 4) I did provide a gentle comment on your bibliography in reply to your question, which is more than you did for mine when I provided an example of a non equilibrium situation in the chemical reaction. Do you deny that standard chemical reaction kinetics can be applied to this reaction? 



#50
Apr1510, 12:26 PM

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My point is that, while chemical kinetics can be rationalized qualitatively in terms of stat mech, you can't work the other way and derive quantitatively correct expressions from first principles...empirical adjustments are required. 



#51
Apr1510, 01:11 PM

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#52
Apr1510, 01:52 PM

P: 5,462

SpectraCat
Why do you keep harping on about the 'rate constant' when I presume you know that it applies strictly to equilibrium situations? Are you actually asking me to set up and solve the relevant differential equations (which involve this constant) that do lead to the actual rate of reaction? I wonder if, as you yourself observed so well in your post #29 that we are working on different interpretations of Statistical Mechanics? Taking Andy’s definition form his post #31 (I agree with this) This is not about partition functions, energy surfaces or whatever – though we could discuss those. 



#53
Apr1510, 02:13 PM

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#54
Apr1510, 02:16 PM

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The reason steadystate can be modeled in SM is because it can be transformed to appear like *equilibrium*. Kinetics is a linearized theory steadystate conditions are a linear condition. You never responded to my example: calculate Q(t) for the OP's problem. 


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