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Chemical Equilibrium of Reaction's constant?

  1. Feb 13, 2014 #1
    I'm currently studying for a test and I'm having problem with this concept:

    If I have 3 different Equilibrium constants for different reactions like Kc=103, Kc=1 and KC=10-4, which one's forward reaction is most favorable, and which one's reverse reaction is most favorable?

    Maybe it's the way the question is stated but I have a good understanding of most of the things in this chapter, this one doesn't make sense at all. Can anyone please help? Thank you very much.
     
  2. jcsd
  3. Feb 13, 2014 #2
    The equilibrium constant is the ratio of the kinetic rate constant for the forward reaction to the kinetic rate constant for the reverse reaction.
     
  4. Feb 13, 2014 #3
    Thank you for replying. I understand that part, but what I'm having a hard time with is how to determine from different constants which movement(forward or backward) will be favorable(or not so).
     
  5. Feb 13, 2014 #4
    If K is high, then the forward reaction is favored over the reverse reaction. If K is low, then the reverse reaction is favored over the forward reaction.
     
  6. Feb 13, 2014 #5
    But why is this so- perhaps algebraically? My teacher told me something like this but I couldn't make out the reasoning behind it. Thank you.
     
  7. Feb 14, 2014 #6

    Borek

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    Staff: Mentor

    Write the equilibrium constant for the reaction

    A -> B

    If it is higher than 1 - which substance is in the excess? Where does the equilibrium lie, on the left, or on the right?

    And if it is lower than 1?
     
  8. Feb 14, 2014 #7
    For the reaction that Borek wrote, the kinetic expression at equilibrium is:

    kf[A]-kr=0

    Now solve algebraically for the equilibrium constant /[A]. This should be in your textbook.

    Chet
     
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