Imagine to have a tube (to make a reactor simple) inside which a chemical reactions take place and reactants are constantly fed so overall can be considered in steady-state conditions.

Just to pick one, I assume the reaction to be exothermic.

Now in my mind there are the following two options, one of them is not clear.

  1. Assuming that the tube is cooled and kept to a constant temperature and that the reaction is not controlled by kinetics, I think that at the end of the tube (if long enough) it can be considered in thermodynamic equilibrium.

  2. If the tube is adiabatic then a distribution of temperature inside should arise. For each temperature there exists an equilibrium constant, and thus a different composition. In this case, would there be thermodynamic equilibrium at each single step (infinitesimal) along the tube? Again, this is considering an equilibrium-controlled reaction.

  • $\begingroup$ What is the actual question here? Is it are these descriptions correct? $\endgroup$ – bon Jan 12 '17 at 20:32
  • $\begingroup$ Sure. If the tube is long enough, the system will reach a final temperature and composition. You can even precisely calculate it by considering the same reaction mixture in an adiabatic batch reactor, and waiting until the system has equilibrated. $\endgroup$ – Chet Miller Jan 12 '17 at 21:59
  • $\begingroup$ @bon The actual question is related to the second point, if it is correct or not! $\endgroup$ – horowitz Jan 13 '17 at 9:06
  • $\begingroup$ @ChesterMiller yes, but how about the second point, each step of the reaction can be considered as in equilibrium? If so it is correct, there exist an optimal temperature to maximize the calories conversion $\endgroup$ – horowitz Jan 13 '17 at 9:07
  • $\begingroup$ What you describe is a 'flow reactor' commonly used to measure rate constants in gas phase reactions by placing the detector at different places along the tube. Question (1) yes; eventually equilibrium is reached at some point at a point distant from the source inlet. (2) If the temperature is not controlled the rate constant will change as the gases travel along the tube and equilibrium reached sooner if the temperature is raised. No reliable rate constant can now be obtained. $\endgroup$ – porphyrin Jan 13 '17 at 9:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.