college chem
posted by hymie .
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 
DrBob222
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 8016 = 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
how did you get 80  16?

college chem 
hymie
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 
DrBob222
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 8016  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
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 
DrBob222
You didn't do it right.
If you want to use 64, then 64/5 = 12.8 and 6412.8 = 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
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 
DrBob222
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 10020=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/(357) = 20k
ln(35/28) = 20k
k = 0.01116
nuff said?
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