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Homework Help: 1st order, 2nd order Rate reactions HELP

  1. Jan 11, 2006 #1
    1st order, 2nd order Rate reactions HELP!!

    I have read the section of my book over and over and studied the practice problems, but I still do not understand 1st order, 2nd order, or 0 order rate reactions. What does it mean to be of any particular order?
     
  2. jcsd
  3. Jan 11, 2006 #2
    what are you talking about?

    what lesson are you reading????

    you information is incomplete, i'm sorry.... that's why i cannot understand the order that you're referring to....
     
  4. Jan 11, 2006 #3
    the order is just the relationship between the molarity of a reactant and the amount that it affects the rate of the reaction.
     
  5. Jan 11, 2006 #4
    oh!!! so that's what brandon meant... lol!!! i didn't get it first....
     
  6. Jan 12, 2006 #5

    Gokul43201

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    If the rate of a reaction is given by :

    [tex]rate= dP/dt = k*[A]^a \cdot ^b \cdot [C]^c \dots [/tex]

    then the order of the reaction is said to be a+b+c+...

    So, a first order reaction has a rate that is linearly proportional to the concentration of one of the reactants. A second order reaction can be one of the two following kinds :

    [tex]rate = k*[A]^2 [/tex]
    or
    [tex]rate = k*[A] [/tex]

    A zero order reaction has a rate that is independent of the concentration of any of the reactants.
     
  7. Jan 20, 2006 #6
    Let's say you have the following data:

    A + B ---> C

    [A]-----------reaction rate
    .010----.020-----.050
    .020----.020-----.100
    .020----.040-----.400

    Look at the first and second lines; [A] increases by a factor of 2 as does the reaction rate ( is held constant). 2^x = 2? x = 1, so [A] is a first order reactant. Now look at lines 2 and 3; [A] is held constant, increases by a factor of 2, and the reaction rate increases by a factor of 4. 2^x = 4? x = 2, so is a second order reactant.

    Now you right the differential rate law like this: rate = k[A]^2. The overall order of the reaction is 3.
     
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