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Meaning of the rate in the Rate Law

  1. Nov 4, 2011 #1
    Hi all, I don't fully grasp definition of the rate as stated in the rate law. I do hope you guys could enlighten me! Here is what I don't understand:

    What does the rate in the rate law refer to? So lets say 3A + 2B -> 4C + D. Does the instantaneous rate derived from the rate law, given that I already have the rate equation and the instantaneous [A] and , refer to the rate of disappearance of A or B, or the rate of appearance of C or D.

    What I'm confused about is how the word rate as used in the rate law is simply referred to as the "instaneous rate of reaction". So if i get a figure from inputting all the data into the rate equation, what does this figure actually represent?
  2. jcsd
  3. Nov 4, 2011 #2
    In addition to my above question, I was wondering if the rate equation was as such:


    This means its a general second order reaction overall. How since its first order with respect to A and B, does it mean both A and B have constant half lifes each?

  4. Nov 5, 2011 #3

    It is the rate of the reaction what your wrote. Therefore it is the rate of disappearance of A and B and of appearance of C and D.
  5. Nov 5, 2011 #4


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    By convention the reaction rate is defined as the rate of disappearance of reactants or the rate of appearance of products divided by their stoichiometric coefficients. So, in your example, the rate would be given by:
    [tex]\text{rate} = \frac{-1}{3} \frac{d[A]}{dt} = \frac{-1}{2} \frac{d}{dt} = \frac{1}{4} \frac{d[C]}{dt} = \frac{d[D]}{dt} [/tex]
    Note that all four expressions for the rate are equivalent. In other words, the rate of appearance of C is four times greater than the rate of appearance of D, the rate of disappearance of B is half the rate of appearance of C, etc. (Please let me know if you are not familiar with calculus and differential equations, otherwise this explanation might not make sense to you yet).

    This represents the initial rate at which your reaction will proceed. The reaction, however, will not always go at the rate you calculate. As the reaction goes on, the concentration of reactants decreases, decreasing the rate of the reaction over time. This is why we refer to the rate as the "instantaneous rate."

    No. Although the reaction is first-order with respect to A and first order with respect to B, A and B will not have constant half-lives. The half-life of A will depend on the initial concentrations of A and B, and the same goes for the half-life of B.

    You can, however, have certain reaction conditions where this reaction acts like a first-order reaction. For example, if you have much more B present than A, even though B gets used up in the reaction, it's concentration hardly changes (for example, if you have 1 mole of A and 100 moles of B, the concentration of B decreases by only 1% when the reaction goes to completion). Under these conditions, we can treat the concentration of B as a constant, and the reaction acts like its a first-order reaction with the rate law: rate = k'[A]. Still, the half-life of A will depend on the initial amount of B that was present (although it will not depend on the concentration of A).
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