## Half-lives

Lizzie Zhang 2C
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am

### Half-lives

When can we use logarithm to calculate the time required to reach a certain concentration when we know the half-life of the reaction, and when are we supposed to use the original rate laws?

For example, hw question 15.83: The first-order decomposition of compound X, a gas,
is carried out and the data are represented in the following pictures. The green spheres represent the compound; the decomposition products are not shown. The times at which the images were taken are shown below each ask. (a) Determine the half-life of the reaction. (b) Draw the appearance of the molecular image at 8 s.

There are 12 molecules in the box at t = 0, and then there are 6 molecules when t = 5. I understand that the half-live is 5 seconds, when it asks about the number of molecules at 8s, why can't we simply do log 8/log 5 and then do (1/2) ^ Ans?
Why do we still need to use the rate law when it's a first-order reaction?

Remi Lathrop 1G
Posts: 51
Joined: Fri Sep 29, 2017 7:07 am

### Re: Half-lives

You still need to use the first order integrated rate law because in this case you know t (1/2) and you are given a time, therefore you can use the equation: A(t) = A(0)e^-kt to find the number of molecules at 8 seconds, given by A(t). Because you found t (1/2) to be 5 seconds, you can plug 5 into our first-order expression for half-life, which is t(1/2) = ln(2)/k or 0.693/k. Once you've solved for k you can use that, and the given 8 seconds, to find A(t) after 8 seconds.

A first-order reaction always follows the equations proposed in class, and I don't think you can use any logarithmic short-cuts on this one :(

Akash_Kapoor_1L
Posts: 52
Joined: Thu Jul 13, 2017 3:00 am
Been upvoted: 1 time

### Re: Half-lives

Apart from what Remi said, the main reason you need to use the integrate rate laws instead of the logarithmic shortcut is because the rate of change in the concentrations of the reactants follows a curve which isn't depicted in the half-life formula. We find the half-life as ln2/k, but the only reason we get the ln2 is because the concentration of A @t is exactly half (or a multiple thereof) of A @0. You would use your half-life formula to help you find k and then manipulate the formula
t = (1/k) * ln([Ai]/[At]) to get the formula that Remi showed above: [At] = [Ai] e^-kt