Why is ##dQ = dH## still valid for chemical reaction (Callen)?

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Discussion Overview

The discussion revolves around the validity of the equation ##dH = dQ## in the context of chemical reactions, particularly as presented in Callen's work. Participants explore the implications of this equation when the number of moles of reactants and products changes during a reaction, questioning the assumptions underlying its application in such scenarios.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants argue that the derivation of ##dH = dQ## relies on the assumption that the number of moles ##dN_i## is constant, which is not the case during chemical reactions.
  • Others propose that if the system is treated as a black box in contact with a constant pressure environment, the first law implies that ##\Delta H = Q##, but this relationship may depend on changes in reactants and products.
  • A participant questions how to reconcile the fundamental relation for ##dH## when ##dN_i## are not constant, suggesting that the derivation does not hold in such cases.
  • Some responses emphasize the irreversibility of spontaneous chemical reactions and the impact of finite temperature gradients on heat transfer, complicating the relationship between ##Q## and ##\Delta S##.
  • There is a discussion about the conditions under which chemical reactions can be considered quasistatic, with some participants suggesting that idealized situations are often assumed in introductory texts.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of ##dH = dQ## in the context of chemical reactions, with no consensus reached on whether the assumptions made by Callen are valid in these scenarios. The discussion remains unresolved regarding the conditions necessary for the equation to hold true.

Contextual Notes

Participants note that the derivation of ##dH = dQ## may not account for changes in mole numbers during reactions, and the implications of irreversibility and temperature gradients are highlighted as significant factors that complicate the analysis.

  • #31
DrDu said:
Just to give some other examples: You could run a reaction far from equilibrium quasistatically if it requires some homogeneous catalyst and you are using very small concentrations of it. In case of enzymatic reactions, reactions will run very slowly at lower temperatures.
You could also run a reaction reversibly using a van't Hoff equilibrium box.
 
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  • #32
Chestermiller said:
You could also run a reaction reversibly using a van't Hoff equilibrium box.
Yes, but I understood that the question was whether you can run a non-equilibrium reaction quasistatically, not reversibly.
 
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