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Temperature101

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  1. Medical Student
    A first order reaction has a rate constant of k=0.7*10^-3. At what time would you have 1/8 of your original concentration of reactant?

    A. 8*0.7x10^-3 s

    B. 3x10^3 s

    C. 3x10^-3 s

    D. 8*0.7x10^3 s


    I know the answer is B, but I dont know how to get it.
     

    BoxVersionAce

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    Jun 26, 2011
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    1. Pre-Medical
      You probably already knew this, but its just a typical half-life question. First, there is the general equation for finding the concentration of a species that undergoes first-order decay in any time period:

      Ct=Coe^(-kt) --> (where Co is the original concentration and Ct is the concentration at time t)

      So, you want the time it takes to get to half the concentration you just argue that Ct/Co=1/2 in the equation and solve for time. This leads to the simpler equation that people memorize for first-order half-life (remember, it is constant):

      t=.693/k

      Now math. Plug in the reaction constant they give you. Round up the .693 to .7. The time you get is 1/10^-3 or 10^3 seconds. Remember, this is the time it takes to go to half the concentration. Realize that 1/8 of it's concentration means three half-lives (1/ (2 x 2 x 2)). Then, multiply the half-life you got by three.

      You could have gotten the same answer with the general decay equation from above, but this requires that you know your natural logs. It's arguably easier to just memorize the .693 equation and calculate from there.
       

      Temperature101

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      May 27, 2011
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      1. Medical Student
        You probably already knew this, but its just a typical half-life question. First, there is the general equation for finding the concentration of a species that undergoes first-order decay in any time period:

        Ct=Coe^(-kt) --> (where Co is the original concentration and Ct is the concentration at time t)

        So, you want the time it takes to get to half the concentration you just argue that Ct/Co=1/2 in the equation and solve for time. This leads to the simpler equation that people memorize for first-order half-life (remember, it is constant):

        t=.693/k

        Now math. Plug in the reaction constant they give you. Round up the .693 to .7. The time you get is 1/10^-3 or 10^3 seconds. Remember, this is the time it takes to go to half the concentration. Realize that 1/8 of it's concentration means three half-lives (1/ (2 x 2 x 2)). Then, multiply the half-life you got by three.

        You could have gotten the same answer with the general decay equation from above, but this requires that you know your natural logs. It's arguably easier to just memorize the .693 equation and calculate from there.
        Got it...Thanks.
         

        sciencebooks

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        Jul 27, 2009
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        Michigan
        1. Pre-Medical
          You probably already knew this, but its just a typical half-life question. First, there is the general equation for finding the concentration of a species that undergoes first-order decay in any time period:

          Ct=Coe^(-kt) --> (where Co is the original concentration and Ct is the concentration at time t)

          So, you want the time it takes to get to half the concentration you just argue that Ct/Co=1/2 in the equation and solve for time. This leads to the simpler equation that people memorize for first-order half-life (remember, it is constant):

          t=.693/k

          Now math. Plug in the reaction constant they give you. Round up the .693 to .7. The time you get is 1/10^-3 or 10^3 seconds. Remember, this is the time it takes to go to half the concentration. Realize that 1/8 of it's concentration means three half-lives (1/ (2 x 2 x 2)). Then, multiply the half-life you got by three.

          You could have gotten the same answer with the general decay equation from above, but this requires that you know your natural logs. It's arguably easier to just memorize the .693 equation and calculate from there.

          Hm, really didn't know this! Thanks!
           
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