1. Questions

Chem

suppose that you place exactly 100 bacteria into a flask containing nutrients for the bacteria and that you find the following data at 37 °C:

t (min):0 15.0 30.0 45.0 60.0
Number of bacteria: 100 200 400 800 1600

What is the order of the rate of production of the bacteria?

How many bacteria will be present after 1.20 × 102 min?

What is the rate constant for the process?

please help! I'm not sure where to start!

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Bri

  1. The rate is doubling every 15.0 minutes so the half life is 15.0 min.
    k = 0.693/t1/2. Solve for k and substitute into the below equation.
    ln(No/N) = kt
    No = 100
    N = ?
    k = from above
    t = 1.2E2 min
    Solve for N.

    I pulled the k = 0.693/t1/2 our of a hat but you can get it this way.
    ln(No/N) = kt
    Pick out No = 200 and
    N = 100
    then t = 15 min and solve for k. You will get the same k as in the above formula BECAUSE
    ln 200/100 = k(15)
    0.693/15 = ?.
    I don't pay any attention the the signs; if your prof does the k is a negative sign meaning the bacteria are expanding instead of decomposing.

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    DrBob222

  2. This is a first order process since rate = k(A)

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    DrBob222

  4. i got N=0.83 does the seem right?

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    Bri

  5. how do i find the rate constant?

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    Bri

  6. Does it make sense to you that the bacteria started with 100, is doubling every 15 minutes and you HAVE LESS THAN YOU STARTED WITH after 120 minutes? Doesn't make sense to me.
    k is 0.693/t1/2

    Look at this table from your post, then extend it to 120 minutes.
    min....count
    0........100
    15.......200
    30.......400
    45.......800
    60......1600
    75........?
    90........?
    105.......?
    120.......?

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    DrBob222

  8. THANKS. I figured it out.

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    Bri

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