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Difference between Reaction Rate and Rate Law?

  1. Aug 13, 2013 #1
    Lets say you have the reaction

    aA + bB => cC + dD

    The rate of the reaction is given by the change in initial concentration over the change in time.

    [itex]-\frac{\Delta \left[ A \right]}{a\Delta T}\; =\; -\frac{\Delta \left[ B \right]}{b\Delta T}\; =\; \frac{\Delta \left[ \mbox{C} \right]}{c\Delta T}\; =\; \frac{\Delta \left[ D \right]}{d\Delta T}[/itex]

    However, there is also the rate expression, which is of the form

    [itex]\mbox{Re}action\; Rate\; =\; k\; \left[ A \right]^{m}\left[ B \right]^{n}[/itex]

    What is the difference between these two, and what purpose does each one serve?
    Last edited: Aug 13, 2013
  2. jcsd
  3. Aug 13, 2013 #2


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    For a reaction with the indicated stoichiometry the first equation has always to be true, that is fairly obvious and you should be able to see that.

    For the second equation the m and n, or whether the equation is even exactly true, depends on the mechanism of the reaction. In simple examples m and n are often 1 or 0. You are evidently on the first pages of a chapter and it will soon show you examples.
  4. Aug 13, 2013 #3
    Yes, but do the two equations mean the same thing? Why are there two different equations to indicate rate?
  5. Aug 13, 2013 #4


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    The first equation just defines what the reaction rate is and how one would measure the reaction rate if performing an experiment.

    The second equation explains how different factors (namely the concentration of reactants) affects the reaction rate. If you were setting up a reaction, this is the equation you would use to calculate the rate at which the reaction will occur.
  6. Aug 13, 2013 #5


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    They do not mean the same thing. The first equations are almost trivially obvious - for that reaction for every a moles of A that reacts, how many of B react? How many of C or D are formed? (Assuming that is the only reaction going on).

    It does not say anything about how the reaction rate depends on [A], etc. - fundamentally there is here only one reaction and only one rate which can be expressed as any of d[A]/dt, d/dt, d[C]/dt or d[D]/dt since they are all simply related by that equation. Which does not say anything about how the rate depends on concentrations. The last equation is a possible or hypothetical law that might and sometimes does represent the rate and it can be investigated experimentally whether it does or not. The result says something about the mechanism of the reaction. Different reaction mechanisms will have different m, n or may not obey the equation at all, it is not a universal equation.
  7. Aug 13, 2013 #6
    I think I understand now:

    The reaction rate in the form [itex]\frac{d\left[ A \right]}{dt}[/itex] simply defines the reaction rate as the instantaneous change in concentration with respect to change in time. However, the rate law actually defines it in terms of the concentration of reactants, and thus allows you to actually determine the rate given the initial concentration of reactants, but reaction rate definition does not.

    In other words, it is like saying that for the function [itex]y=x^{2}[/itex] the instantaneous change in y with respect to change in x is [itex]\frac{dy}{dx}[/itex]. However, that alone does not help you much unless you actually define that to be [itex]2x[/itex].
  8. Aug 14, 2013 #7


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    I would not say the rate law defines anything, I would say what I said in my last post. Do not take the word 'law' over-seriously here - if that is a law it is often broken! - see my first post. I would not say your second paragraph puts it well. Frankly I think you need to get into that chapter before it is profitable to discuss further. :smile: You will surely find there enough about the integration of rate equations, possibly disproportionate to its real importance. Kinetics, though indispensable and necessary to grasp, is just a part of the study of chemical reaction mechanisms.
    Last edited: Aug 14, 2013
  9. Aug 14, 2013 #8
    Those two expressions are, in fact, saying the same thing (assuming of course that the Rate Law = K[A]mn is appropriate to the reaction being studied). If you google the derivation of the integrated rate laws, you will see that the overall rate law, if correct in any specific situation, is equal to the change in any of the species. Setting the change in [A] with respect to time equal to the rate law you must then solve some differential equations (you happened to have picked the one of the hardest rate laws to integrate with your example) and you can get an expression to figure out the concentration of a given species at any point in time.
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