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chemistry

Discussion in 'DAT Discussions' started by arpitpatel86, May 25, 2008.

  1. arpitpatel86

    2+ Year Member

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    The following reaction is secnod order with respect to W
    and second order with respect to X. If the concentration
    of W and X are doubled what happens to the rate of
    the reaction?
    W + X
    → Y + Z
     
  2. userah

    7+ Year Member

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    damn it stupid thing messed up. gotta retype it now...

    ok i believe this is how it goes.

    This is how the reaction rate works.
    aA + bB -> cC + dD

    the reaction rate for this would be: rate = k [A]^a ^b

    so for the example you gave where concentrations are doubled for both: rate = k[W]^2 [X]^2

    if you increase ONE of the reactants by two, the rate increases by a factor of 4 (2^2). If you increase BOTH of the reactants by two like the problem states, the rate would increase by a factor of 16 (2^2) * (2^2)

    hope that helps =)
     
  3. doc3232

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    Well-explained.
     
  4. userah

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    i don't understand why the editing won't fix the issue but it should read
    rate = k[A]^a ^b

    in case it still doesn't work...the B in the bracket should be capitalized

    lol still not working...really odd =(
     
  5. Zerconia2921

    Zerconia2921 Bring your A-game!
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    Userah your awsome! :thumbup: great explanation.
     
  6. userah

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    hahaha thanks =) although i'll be honest, google helped a little =P
     

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