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Question that I was looking up on Google

  1. Jan 31, 2009 #1
    The following reaction was monitored as a function of time:
    AB---> A + B
    A plot of 1/[AB] versus time yields a straight line with slope 5.2×10−2 M \s.

    If the initial concentration of AB} is 0.210 M, and the reaction mixture initially contains no products, what are the concentrations of {A} and {B} after 80 s?


    Here's what I did:
    1/A = kt + 1 / A initial

    *When I did this equation it was wrong. The correct answer was that the concentration of both A and B=9.8×10−2,9.8×10−2. Please tell me what I am doing wrong. Thanks!
     
  2. jcsd
  3. Jan 31, 2009 #2
    How about 1/AB=kt+1/AB0? As your given graph.
    so AB=1/(kt+(1/AB0))
    in terms of moles I think
    A=B=1/(kt+(1/AB0))
     
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