# college chem

posted by on .

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.
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?