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Rate of a Chemical Reaction

  1. Jan 18, 2012 #1
    Hi,

    I am reading about rates of reactions, and I am confused about why we multiply the rate equation of a particular product or reactant by its respective coefficient, only taking the reciprocal of it. Could someone explain to me why this is done? Is it done merely out of convenience?
     
  2. jcsd
  3. Jan 19, 2012 #2

    Borek

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    Please show an example. In what context was it done?
     
  4. Jan 19, 2012 #3
  5. Jan 19, 2012 #4

    Borek

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    This way it doesn't matter which substance you observe - for a given reaction reaction rate is identical for every reactant and product.
     
  6. Jan 19, 2012 #5
    So then it is done out of convenience? We divide it by its stoichiometric coefficient so that we have a one-to-one rate?
     
  7. Jan 19, 2012 #6
    Like Borek said, it's done so you can look at the rate of consumption or production of any particular component and the answer will remain the same.

    If AB decomposes to form A2 and B2, you have the following reaction:

    2AB → A2 + B2

    The rate at which AB decomposes will be twice as fast as the formation of the products. If you just looked at the raw rate of decomposition of AB or of production of either product, it would be inconsistent.
     
  8. Jan 19, 2012 #7
    I'm sorry, could you possibly explain why it would be inconsistent?
     
  9. Jan 19, 2012 #8

    Borek

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    How many moles of A2 will appear while 1 mole of AB is consumed? It happens in the same time - is the ratio of number of moles to time constant?
     
  10. Jan 19, 2012 #9
    Oh, one mole of A2 will appear.
     
  11. Jan 19, 2012 #10

    Borek

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    No, not 1 mole.
     
  12. Jan 19, 2012 #11
    Is it 1/2? And is that where the inconsistency is?
     
  13. Jan 19, 2012 #12

    Borek

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    Yes, 1/2 mole of A2.

    Now, let's assume it happened in - say - 1 minute. Rate of the AB consumption was 1 mole per minute, and rate of A2 production was 1/2 mole per minute. Same reaction - and potentially two different rates. Can you calculate both rates using the definition with stoichiometric coefficients?
     
  14. Jan 21, 2012 #13
    Well, Borek, I just attempted to calculate what you told me to do, and, unfortunately, I could not. To calculate it I need a change in concentration over a change in time, correct; so would that change in time 0 min - 1 min?
     
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