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

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SUMMARY

The discussion centers on the validity of the equation ##dH = dQ## as presented by Callen in the context of chemical reactions. Participants clarify that while Callen's derivation assumes constant mole numbers ##(dN_i = 0)##, this condition does not hold during chemical reactions where mole numbers change. The consensus is that for a process at constant pressure and temperature, the relationship ##\Delta H = Q## can still be applied, but it requires careful consideration of the system's state and the irreversibilities involved in real processes. The discussion emphasizes the importance of understanding the implications of the First and Second Laws of Thermodynamics in these scenarios.

PREREQUISITES
  • Understanding of the First Law of Thermodynamics
  • Familiarity with the concept of enthalpy (H) and its relation to heat transfer (Q)
  • Knowledge of chemical equilibrium and the role of mole numbers (N_i) in reactions
  • Basic principles of reversible and irreversible processes in thermodynamics
NEXT STEPS
  • Study the derivation of the First Law of Thermodynamics in detail
  • Learn about the implications of the Second Law of Thermodynamics on irreversible processes
  • Explore the concept of chemical equilibrium and its mathematical representation
  • Investigate the differences between reversible and irreversible processes in thermodynamic systems
USEFUL FOR

This discussion is beneficial for thermodynamics students, chemical engineers, and researchers in physical chemistry who are looking to deepen their understanding of heat transfer and enthalpy changes in chemical reactions.

  • #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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