# Integrated Rate Law Expression & Half-Life

#### busupshot83

##### S.D.N. Vet
7+ Year Member
15+ Year Member
Integrated Raw Law Expression:
ln [A]t = -kt + ln [A]0

Are there any problems on the DAT which require the integrated rate law?

also,

Half Life:
0.693/k

I heard that the half-life problems on the DAT can be determined without the use of formulas. For instance: If the half-life of an isotopic nuclide is 24 hours, how long will it take for 75% of the sample to decay?

However, will there be half-life problems in which the formula above will be required to use?

L

#### laee3

if i were you, i would just memorize it. It's not like they will tell you not to use certain forumals on the test, which will be pretty rare case. Even if they did, I'm sure there is no way for them to check it, since people are using stupid whiteboards to solve their problems. #### Streetwolf

##### Ultra Senior Member
10+ Year Member
7+ Year Member
It's not hard to use it if you know the formula. But if they gave one to you like that, they'd probably give you easy numbers.

Most of the time it's something like, if the half life is 8 days then what % remains after 24 days?

#### ananda

##### Full Member
10+ Year Member
For the half-life problem, you just need to think of it this way. You are right! No formula is needed to solve this problem.

What is given: half life = 24 hours.
This means that half of the starting material will have decayed after the first 24 hours and for each 24 hours that go by... half of the remaining material will have decayed.

You start out with 100% of the material. In simple terms, we will pretend that 100% = 100 g.

So after one day (24 hours) half of the 100 grams are decayed. So you are left with 50 grams. At the end of the second day (24 additional hours have passed) half of the remaining material which was 50 grams is now decayed, so you are left with 25 grams.

If you have 25 grams left, that means that 75 grams are decayed. In other words 75% is lost and it took two days or 48 hours for that to happen.

I hope this helps and good luck!

I do not think they will get any fancier than this. I truly don't think the formula above is necessary.

#### busupshot83

##### S.D.N. Vet
7+ Year Member
15+ Year Member
thanks for the replies. as for the integrated rate law, does anyone know a specific problem in the DAT Destroyer, or elsewhere, that asks you to utilize this formula?

J

#### jackbauer!

thanks for the replies. as for the integrated rate law, does anyone know a specific problem in the DAT Destroyer, or elsewhere, that asks you to utilize this formula?

there is no destroyer problem using the integrated rate law...however, the straight line plots for 0, 1st, and 2nd order rxns you must know and they are in the destroyer. if you know those plots you can derive the integrated rate laws b/c the rate law can be translated in the eqn of a line! (y = mx +b)..

ex. 1st order rxns: ln[A]t = -kt + ln[A]0
y axis = ln[A]t
slope = -k
x axis = t
y intercept = ln[A]0

so if you know the plot you should be able to derive the integrated rate law or vice-versa.... therefore, you have less to memorize! jb! #### busupshot83

##### S.D.N. Vet
7+ Year Member
15+ Year Member
there is no destroyer problem using the integrated rate law...however, the straight line plots for 0, 1st, and 2nd order rxns you must know and they are in the destroyer. if you know those plots you can derive the integrated rate laws b/c the rate law can be translated in the eqn of a line! (y = mx +b)..

ex. 1st order rxns: ln[A]t = -kt + ln[A]0
y axis = ln[A]t
slope = -k
x axis = t
y intercept = ln[A]0

so if you know the plot you should be able to derive the integrated rate law or vice-versa.... therefore, you have less to memorize! jb! Good looking out JB! Thanks man.

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