Wednesday

September 2, 2015
Posted by **hymie** on Sunday, February 21, 2010 at 10:31pm.

a) Prove that the reaction has a constant 1/5 life.

b) find the half life.

- college chem -
**DrBob222**, Sunday, February 21, 2010 at 10:39pmCan 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/t_{1/2}and solve for t_{1/2}

- college chem -
**hymie**, Sunday, February 21, 2010 at 10:44pmhow did you get 80 - 16?

- college chem -
**hymie**, Sunday, February 21, 2010 at 10:47pmoh 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 -
**DrBob222**, Sunday, February 21, 2010 at 10:52pmNo, 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/t

_{1/2}and solving for t_{1/2}.

I get something like 60 seconds for the half life (but that isn't an exact number).

- college chem -
**hymie**, Sunday, February 21, 2010 at 10:58pmwhen 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 -
**DrBob222**, Sunday, February 21, 2010 at 11:05pmYou 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 -
**hymie**, Sunday, February 21, 2010 at 11:09pmi 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 -
**DrBob222**, Sunday, February 21, 2010 at 11:23pmIf 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?