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TiCl4 and O2 Thermodynamics

  1. Mar 13, 2014 #1
    1. The problem statement, all variables and given/known data

    If a feed of TiCl4 and O2 at 600 degrees is fed into a reactor and the product is 1500 degrees, how much heat is added or removed?


    2. Relevant equations

    Cp = A + BT + C/T^2 + DT^2


    3. The attempt at a solution

    I managed to find the heat of reaction for the whole process previously as well as the heat of formation for TiCL4 at 1500degrees. Im kind of stuck from here though.

    Do i integrate the Cp eqn again with T1 as 1500 and T2 as 600?
     
  2. jcsd
  3. Mar 13, 2014 #2

    BvU

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    Hello Sero,
    Apparently you do have a reaction taking place. That's a few equations too. Could you post them ?
    And if you want to do something with the Cps: which component has which Cp ?
    In ChE there are all kinds of reactors. Any idea what we are looking at here ?
     
  4. Mar 13, 2014 #3
    The reaction is TiCl4 + O2 --> TiO2 + 2 Cl2

    You need to get the enthalpies of the reactants at 600 C and the enthalpies of the products at 1500 C. Do you know how to do that?

    Chet
     
  5. Mar 13, 2014 #4
    Oh ok i get it! so It's just integration of the Cp formula for TiCl4 with T2 at 600 degrees and T1 at standard temp and i do the same for TiO2 except instead of 600 i use 1500 then i find the difference.
     
  6. Mar 13, 2014 #5
    Don't forget that you need to do O2 and 2Cl2 also, since, even though their heats of formation at standard temp are zero, you need to consider O2 at 600 and 2Cl2 at 1500.

    chet
     
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