Exploring Entropy Production in Chemical Processes

In summary: Do you mean to ask if the energy released in the reaction is greater than the energy needed to initiate the reaction? If so, the answer is no. The energy released in the reaction is simply the sum of the energies needed to initiate the reaction.
  • #1
nonequilibrium
1,439
2
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?
 
Science news on Phys.org
  • #2


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.
 
  • #3


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...

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.
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]?
 
  • #4


I fear I still don't understand exactly your question.
 
  • #5



Hello,

Thank you for sharing your thoughts on exploring entropy production in chemical processes. It is clear that you have a good understanding of the thermodynamic principles involved in this topic.

To answer your question, yes, it is possible for there to be energy production from a chemical reaction. In fact, this is a key aspect of many chemical processes, as the conversion of one form of energy to another is often the driving force behind these reactions. However, in the context of thermodynamics, we do not consider this energy production as a separate term in the equation for entropy production. This is because, as you mentioned, the energy was already present in the system and is simply being converted from one form to another.

In terms of your notation, it is important to keep in mind that the thermodynamic force \frac{1}{T} (or \frac{1}{T} \mathrm d Q) is not the same as the thermodynamic force for heat conduction, which is \nabla \frac{1}{T}. The former represents a change in entropy due to heat transfer, while the latter represents a change in temperature due to heat transfer. They are related, but not the same.

In summary, while there may be energy production from a chemical reaction, it is not explicitly included in the equation for entropy production in thermodynamics. Instead, it is accounted for in the terms for heat transfer and chemical potential. I hope this helps clarify your doubts. Keep exploring and asking questions!
 

FAQ: Exploring Entropy Production in Chemical Processes

What is entropy production in chemical processes?

Entropy production refers to the measure of the amount of disorder or randomness in a chemical system. In chemical processes, entropy production occurs due to the irreversible loss of energy and the increase in disorder of the system.

Why is it important to study entropy production in chemical processes?

Entropy production is important in understanding the efficiency and sustainability of chemical processes. A high level of entropy production can result in energy wastage and environmental impacts. By studying it, scientists can identify ways to minimize entropy production and improve the efficiency of chemical processes.

How is entropy production quantified in chemical processes?

Entropy production is quantified using the second law of thermodynamics, which states that the total entropy of an isolated system always increases over time. In chemical processes, the rate of entropy production can be calculated by measuring the heat and work exchanged between the system and its surroundings.

What factors affect entropy production in chemical processes?

The degree of irreversibility, temperature, and the type of chemical reactions involved are some of the factors that affect entropy production in chemical processes. A higher temperature and irreversible reactions tend to result in higher entropy production.

How can we reduce entropy production in chemical processes?

There are several ways to reduce entropy production in chemical processes, such as using more efficient reactions and processes, optimizing reaction conditions, and recycling by-products. Another approach is to use renewable energy sources, which can reduce the overall entropy production and environmental impact of chemical processes.

Back
Top