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Stoichiometry unsolvable problem

  1. Feb 19, 2013 #1
    1. The problem statement, all variables and given/known data

    Toluene (C6H5CH3) is hydrogenated according to the two simultaneous
    chemical reactions
    C6H5CH3 + H2 → C6H6 + CH4
    2C6H5CH3 + H2 → (C6H5)2 + 2CH4
    Initially, the reactor contains 40% toluene and 60% H2 (% mole). At the end of the operation, the reactor contains 10% toluene and 30% CH4 (% mole). Find:

    a. The mole fraction of the diphenyl, (C6H5)2, at the end of the operation
    b. The conversion of H2

    2. Relevant equations

    I came up with the following equations from the molar balances(a,b and c are the final mole fractions of H2, Benzene and Biphenyl, respectively, ε1 and ε2 are the extents of reactions):

    0.1=(0.4*n1-ε1-2*ε2)/n2
    a=(0.6*n1-ε1-ε2)/n2
    b=ε1/n2
    0.3=(ε1+2*ε2)/n2
    c=ε2/n2

    3. The attempt at a solution

    The final amount of moles n2 is related to the initial amount by the following equation:

    n2=n1=0.4*n1-ε1-2*ε2+0.6*n1-ε1-ε2+ε1+ε1+2*ε2+ε2

    Now I have five equations with six unknowns but I can suppose an initial amount of moles of 10 and then solve the equations with five unknowns. However this system of equations is unsolvable. I'm studying stoichiometry and found this problem inside a reactor design book. Any help would be welcome. Thanks.
     
    Last edited: Feb 19, 2013
  2. jcsd
  3. Feb 19, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I agree. Both reactions convert 1 toluene molecule to 1 CH4 molecule and the total number of molecules is conserved - the given concentration of CH4 is redundant.

    The two possible extreme cases are:
    Just reaction one, 30% H2, 30% C6H6.
    Just reaction two, 45% H2, 15% (C6H5)2
    And everything in between is possible, too.
     
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