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March 25, 2017

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a first order reaction has a fifth life of 20 seconds ( the amount of time needed to use up 1/5)
a) Prove that the reaction has a constant 1/5 life.
b) find the half life.

  • college chem - ,

    Can you do this?
    ln(No/N) = kt.
    Just start with a convenient number like 100 for No; therefore, at the end of 20 s we should have 80 left. Calculate k.
    ln(100/80) = k(20).
    solve for k.
    Then try the second 20 s span. That should leave you with 80-16 = 64. Plug
    ln(80/64) = k(20) and see if k is the same thing.

    For the half life, use
    k = 0.693/t1/2 and solve for t1/2

  • college chem - ,

    how did you get 80 - 16?

  • college chem - ,

    oh i see 1/5 times 80 is 16 so that is the fifth life of the rxn. for the half life usage should i take 50 seconds using the convenient number of 100 seconds?

  • college chem - ,

    No, I think you misunderstood. If the fifth life (I've never heard it used that way) is 20 s and that's the tim for 1/5 of it to be used, then start with No = 100 atoms/whatever it is so at the end of 20 s we will have 80 remaining. Then 1/5 x 80 = 16 so we have 80-16 - 64 remaining. The 100 atoms/whatever is the number we are choosing for convenience. I chose 100 because that's easy to divide evenly. You could choose any number and it will work. The half live is calculated by using the k you get from those original calculations and pluging that into k = 0.693/t1/2 and solving for t1/2.
    I get something like 60 seconds for the half life (but that isn't an exact number).

  • college chem - ,

    when i chose the next interval of a fifth i did 64 x 1/5 = 12.8 and then i took ln(64/12.8) and that did not give me the same constant as the first two so does this mean that the fifth life is not constant

  • college chem - ,

    You didn't do it right.
    If you want to use 64, then 64/5 = 12.8 and 64-12.8 = 51.2, then
    ln(64/51.2) = k(20)
    and k DOES remain the same.
    No is what you start with.
    N is what remains.
    12.8 is what was used, not what remains.

  • college chem - ,

    i think you made a mistake in the first answer you take ln(100/20) not 100/80 then it will prove it has a constant fifth life

  • college chem - ,

    If your definition of a fifth life is correct (the amount of time it takes to use up 1/5 of the material), I didn't make a mistake.
    If we take 100 as No, then 1/5 of that is 20 and 100-20=80 so
    ln(100/80) = k(20)
    and it IS a constant. You get different numbers if you use 100 and 20.

    ln(100/80) = 20k
    k = 0.01116.

    ln(80/64) = 20k
    k = 0.01116

    ln(64/51.2) = 20k
    k = 0.01116

    ln(35/(35-7) = 20k
    ln(35/28) = 20k
    k = 0.01116

    nuff said?

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