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How does selectivity work in chemical reactions again?

  1. Feb 16, 2014 #1
    I got a bit of a question that needs answering.
    See, I'm still not understanding how basic selectivity in organic chemistry works. I'm not looking to know the mechanism or anything. Just one simple thing. I'll give the description given from some textbook first:

    1. The problem statement, all variables and given/known data

    The feed to the reactor contains 88wt% isopropyl alcohol and the rest water (at 1 atm, 25degC). Three reactions take place; conversion to acetone, some other reaction forming di-isopropyl ether, and dehydration of IPA into propene. Outlets are at the temperature of the reactor. The relation when taking place at 300degC is given as:

    Acetone conversion: 46.3%
    Selectivity (mol di-iso-ether/ mol Ac): 0.01
    Selectivity (mol propene/ mol Ac): 0.08


    3. The attempt at a solution

    Okay, so let's establish a basis. Let's put in, for simplicity's sake, one mol of pure IPA in the reactor, after heating it up to 300C of course. So, 46.3% is converted, leaving 53.7% of the IPA as unreacted.

    But how does it get affected by the side reactions? Does that mean 0.01 x 46.3% becomes di-iso-ether and 0.08 x 46.3% becomes propene, while the rest is the actual acetone (that is to say, 0.91 x 46.3% becomes acetone)?
     
  2. jcsd
  3. Feb 16, 2014 #2

    Borek

    User Avatar

    Staff: Mentor

    It depends on how the selectivity is defined - please check your notes or the textbook.

    Could be there is a commonly accepted definition that I am not aware off, and the answer is obvious - but even if so, this commonly accepted definition must be in your book. Check it, and it should answer your question.
     
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