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Continuous process and equilibrium constant.

  1. Oct 24, 2015 #1
    Assuming that we have a chemical plant which produces methanol through the following equations:
    CH4 + H2O <--> CO + 3H2
    CO + 2H2 <--> CH3OH
    CO2 + 3H2 <--> CH3OH + H2O
    I know that with specific pressure, temperature and flow rates, I can produce reactions with specific equilibrium constants. Are these conditions maintained until water and methanol are separated, even in the pipelines? I'm guessing that I can calculate the theoretical yield by applying the equilibrium constant. Is that correct? Thank you for your time.
    Edit: Also, wouldn't I risk changing the position of equilibrium if I change the conditions to separate water and methanol?
     
    Last edited: Oct 24, 2015
  2. jcsd
  3. Oct 24, 2015 #2
    Are you referring to the conventional 2-step process using a catalyst for the final conversion of the syngas to methanol?
     
  4. Oct 25, 2015 #3
    Yes, but that does not matter. I'm interested in the equilibrium constant manipulation.
     
  5. Oct 25, 2015 #4
    I would think so, because the pipelines roughly maintain the same temperature and pressure.

    Yes, but actual yield could be somewhat less. There's usually a compromise between catalyst, residence time, and yield.

    Do you mean change the conditions of the reactions, or change the conditions of the water/methanol separation process?
     
  6. Oct 25, 2015 #5
    The conditions of the reactions; condensation then distillation comes to mind, although it may not be the one used by the industry.
     
  7. Oct 25, 2015 #6
    I'm confused by this reply. Condensation and distillation are physical operations where no reactions take place.
     
  8. Oct 25, 2015 #7
    Condensation is a cooling process. If we consider the third equation which is an exothermic reaction, the position of equilibrium would move to the right during condensation which would be ideal for this reaction since we are generating more methanol. But what if the reaction was actually endothermic? Wouldn't we be decreasing our potential output?
     
  9. Oct 25, 2015 #8
    Now I understand. No, the condensation and distillation will have no effect on the yield because the rates of the reactions are insignificant. (That is why a catalyst is necessary for the final reaction step.)
     
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