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Equilibrium conversion of batch vs. flow reactor

  1. Oct 18, 2014 #1

    Maylis

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    Gold Member

    Hello,

    I went through a problem in my reaction engineering textbook, and solved them correctly by basically mindlessly plugging into the equation, but now I wanted to get some help interpreting the results.

    So we had a reversible, elementary gas phase reaction that is isothermal and isobaric in a flow system

    ##A \leftrightarrows 3C##

    The equilibrium constant, ##K_{C}##, as well as initial pressure and initial temperature (same as final T and P) are given in the problem. I went ahead and calculated ##X_{eq,flow} = 0.58## in this situation.

    Next, I calculated for a batch reactor that is carried out at constant volume and pressure with the same information given and found ##X_{eq,batch}^{V} = 0.39##. So of course, the conclusion is that ##X_{eq,flow} > X_{eq,batch}^{V}##. Now I want to think about why this is the case. Well, a flow reactor can change it's volume. Since for every mole of A, we are producing 3 moles of C, I figure the volume of the gas is expanding, which by Le Châtelier's principle will favor the side with more moles, hence C. Therefore, the conversion at equilibrium for a flow system will be larger than the batch system at constant volume, without any math to justify it.

    Next, they ask about a constant pressure batch reactor, and it turns out ##X_{eq,batch}^{P} = 0.58##, hence ##X_{eq,batch}^{P} = X_{eq,flow}##. I was surprised by this result for a moment. Then, I thought since the pressure is constant, but the volume is not, so using the same argument then equilibrium will lie to the right and favor production of C.

    I think one of the issues for the constant volume batch reactor is that one must assume that it is also constant pressure (final pressure is not given in the problem statement). Suppose I had a constant volume batch reactor that was not constant pressure, how might this change the solution? I suppose since for every mole of A, 3 molecules of C are being produced, then the pressure should increase since there will be more molecules slamming into each other, so equilibrium will favor the side with less moles, which is A. Hence, the equilibrium conversion for a isochoric, but not isobaric batch reactor will be less than that of a flow system. It would be even less than an isochoric and isobaric batch reactor.

    So now I want to compare ##X_{eq,flow}##, ##X_{eq,batch}^{V}##,##X_{eq,batch}^{P}##, and ##X_{eq,batch}^{V,P}##

    I think it would go ##X_{eq,flow} = X_{eq,batch}^{P} > X_{eq,batch}^{V,P} > X_{eq,batch}^{V}##

    Now I want to flip this problem the other way, since I noticed none of the textbook problems have the stoichiometric coefficients of the reactants larger than the products, so I figure that would be a good test problem. If we have ##3A \leftrightarrows C##, then how will the ordering change? I use the same arguments (don't know how to compare isobaric and isochoric)

    ##X_{eq,batch}^{P} = X_{eq,flow} > X_{eq,batch}^{V}##

    I want to do the calculation for this problem by making it up myself and convince myself, but I am hesistant because in the original problem ##K_{C} = 0.25 L/mol##, if I were to make it ##3A \leftrightarrows C##, would ##K_{C}## just be the inverse, 4? I figure that I might not get a good result if I just make up an equilibrium constant, and I am unsure how the constant would be affected by the stoichiometry being reversed, so I might be lead astray from the correct conclusion because I don't know how to handle the equilibrium constant..

    Sorry for such a long post, but I tend to pick apart every possible scenario and figure out what is going on!
     
    Last edited: Oct 18, 2014
  2. jcsd
  3. Oct 20, 2014 #2
    I only read part of what you wrote, but got very confused. My main issue is for a batch reactor. How can it be both isobaric and isochoric simultaneously, if the number of moles increases as a result of the reaction?

    Chet
     
  4. Oct 20, 2014 #3

    Maylis

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    Gold Member

    Yes, I was troubled by that too. I just went ahead and assumed it was isobaric, even though it was not specifically mentioned, and when I made that assumption I got the ''correct'' answer. I addressed it in my post, which is part of my confusion.

    I uploaded the problem statement and solution.

    Since this is a gas in a constant volume batch reactor, the equation for concentration of a species is
    [tex] C_{i} = C_{A0}(\theta_{i} + \nu_{i} X) [/tex]
    For species A, this simplifies to
    [tex] C_{A} = C_{A0}(1 - X) [/tex]
    [tex] C_{C} = C_{A0}(0 + 3X)[/tex]

    And to find the equilibrium conversion, we know the reaction rate should be zero,

    [tex] -r_{A} = 0 = k_{A}C_{A} - k_{C}C_{C}^{3} [/tex]
    [tex] k_{C}C_{C}^{3} = k_{A}C_{A} [/tex]
    Knowing that the equilibrium constant, ##K_{C} = \frac {k_{A}}{k_{C}}##,
    [tex] C_{C}^3 = K_{C}C_{A}[/tex]

    Notice how nowhere in here does the pressure change come into play. Maybe I am mistaking the lack of pressure in the equations as meaning it is isobaric, rather than the pressure simply not being a factor in finding the equilibrium concentration in a constant volume batch reactor.
     

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    Last edited: Oct 20, 2014
  5. Oct 20, 2014 #4

    Maylis

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    Gold Member

    Here is my work

    ImageUploadedByPhysics Forums1413809923.544011.jpg
    ImageUploadedByPhysics Forums1413809941.477332.jpg
    ImageUploadedByPhysics Forums1413809959.046947.jpg
     
    Last edited: Oct 20, 2014
  6. Oct 20, 2014 #5
    I looked over your discussion and your interpretations, and they looked correct to begin with. The only thing that you missed was that the pressure in the constant volume batch reactor is not constant. It increases by a factor of (1+2X); work it out for yourself. This produces higher concentrations of both A and C for a given conversion in the constant volume batch reactor than in the case of either a constant pressure batch reactor or a constant pressure flow reactor. But the effect of C wins out because it is cubed in the equilibrium expression, rather than to the first power as with A. This does not allow the reaction to go as far.

    Chet
     
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