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Duration of Equilibrium Reaction

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  1. Apr 25, 2017 #1
    I am wondering the answer to the following question, and what equation you would use to solve it:
    What is the duration of the following reversible reaction?

    A <--> B
    You have 10 moles of A and zero moles of B.
    After 1 minute, 25% of A turns into B, and 50% of B turns into A.

    My class calculated it to be 5 minutes by calculating how much A and B there will be after each minute until it reached equilibrium (6.667 moles A, 3.333 moles B), but I believe the correct answer will require calculus, but I don't know how.
     
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  3. Apr 26, 2017 #2

    Orodruin

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    This depends on what you mean by "duration". The reaction keeps occuring even in equilibrium and the approach to equilibrium is asymptotic (the difference from equilibrium decays exponentially).
     
  4. Apr 26, 2017 #3

    Borek

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    Actually there is not enough data to answer the problem. You are given two points - concentrations at the beginning and concentrations after one minute. That's not enough to determine the order of the reaction, and depending on what the order is, equation required to calculate reaction progress is different.

    In most cases the concentration changes - as Dale said - asymptotically. The only case when it is not true is for the 0th order reactions, where the reaction speed is not dependent on the concentration (although this can get complicated, as for low concentrations such reactions often stop following the 0th order kinetics).

    Compare https://en.wikipedia.org/wiki/Rate_equation
     
  5. Apr 27, 2017 #4
    I think what they meant for you to do was this:
    $$A^{n+1}=A^n+(-0.25A^n+0.5B^n)$$
    $$B^{n+1}=B^n+(0.25A^n-0.5B^n)$$
    where n is the number of minutes that have passed. These equations reduce to:
    $$A^{n+1}=0.75A^n+0.5B^n$$
    $$B^{n+1}=0.25A^n+0.5B^n$$subject to ##A^0=1## and ##B^0=0##, so

    After 1 minute:
    ##A=0.75##
    ##B=0.25##

    After 2 minutes:
    ##A=0.6875##
    ##B= 0.3125##

    ETC
     
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