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[Noneq Thermodynamics] How to account for the energy released in a chemical reaction?

  1. Apr 22, 2012 #1
    Hello,

    So presume we have a system in which a chemical process A + B -> X + Y is happening. We allow it to be a non-equilibrium process (so there will be an entropy production inside the system) but for ease we presume the system is characterized by the usual variables E, V, N_A, ..., N_Y (and the homogenous T, P, \mu_i), i.e. no local densities.

    In a book I found that [itex]\mathrm d S = \frac{1}{T} \left( \mathrm d Q - \sum \mu_i \mathrm d N_i \right)[/itex] where they regard the first term as an entropy flux, i.e. an equilibrium process (I presume Q is simply the energy the system gets from an environment in equilibrium). Hence they explicitly draw the distinction [itex]\mathrm d_e S = \frac{\mathrm d Q}{T}[/itex] which is the entropy flux from the environment, and [itex]\mathrm d_i S = - \frac{1}{T} \sum \mu_i \mathrm d N_i[/itex], which is the entropy produced internally, by the chemical process (remember: non-equilibrium).

    But I was wondering: are they then neglecting energy production from the chemical reaction? Or am I overlooking something? For example, is it allowed, in a more general case, for there to be a ``[itex]\mathrm d_i Q[/itex]'' which would stand for the energy produced in the chemical reaction? Hence in that case [itex]d_e S[/itex] would go unchanged and we would have [itex]\mathrm d_i S = \frac{1}{T} \mathrm d_i Q - \frac{1}{T} \sum \mu_i \mathrm d N_i[/itex].

    Hence if we write [itex]\mathrm d_i S = \sum_j X_j J_j[/itex] (= entropy production in terms of thermodynamic forces X_j and currents J_j) we would have that the heat production would have the thermodynamic force [itex]\frac{1}{T}[/itex] (which is notably different from the thermodynamic force for heat conduction, being [itex]\nabla \frac{1}{T}[/itex] or sometimes written as [itex]\sim \nabla T[/itex] (Fourier's law!))

    The thing I'm also wondering about: I'm saying "the energy created by the chemical reaction" but of course there is no real energy created: the energy was there all along. So does it make sense to say that thermodynamically energy was created, but fundamentally there was not?
     
  2. jcsd
  3. Apr 23, 2012 #2

    DrDu

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    Re: [Noneq Thermodynamics] How to account for the energy released in a chemical react

    Obviously there is no energy "created" in course of the reaction. The internal energy change at fixed V and S (i.e. the energy change due to chemical reaction) is Sum mu_i dN_i, but it is energy being released which was stored in the chemical compounds.
     
  4. Apr 23, 2012 #3
    Re: [Noneq Thermodynamics] How to account for the energy released in a chemical react

    I didn't say though there was energy created, did I? But thermodynamically, we act as if new energy is entering the system, right? I'm a bit confused about this dichotomy. If E stands for internal energy, then I would expect it to be constant even in a thermodynamic sense, but apparently...

    So you say that the energy [itex]\mathrm d_i Q[/itex] (the heat released in a chemical reaction) that I describe in my OP is actually the [itex]\sum \mu_i \mathrm d N_i[/itex] term? For example, saying a reaction is exothermic, means [itex]\sum \mu_i \mathrm d N_i > 0[/itex]?
     
  5. Apr 24, 2012 #4

    DrDu

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    Re: [Noneq Thermodynamics] How to account for the energy released in a chemical react

    I fear I still don't understand exactly your question.
     
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