## half-life

$\frac{d[R]}{dt}=-k[R]^{2}; \frac{1}{[R]}=kt + \frac{1}{[R]_{0}}; t_{\frac{1}{2}}=\frac{1}{k[R]_{0}}$

Jane Burgan 1C
Posts: 73
Joined: Fri Sep 28, 2018 12:15 am

### half-life

For problems where the textbook gives us the half-life of a second order reaction of a substance and it asks us to calculate the time needed for the concentration of the substance to decrease by 1/16, why can't we just multiply the half-life by 4 to get the time needed?

Sierra Cheslick 2B
Posts: 61
Joined: Fri Sep 28, 2018 12:27 am

### Re: half-life

The speed of the reaction depends on how much reactant is left, so it is not directly proportional to a quarter of the half-life.

Sarah Zhao 4C
Posts: 64
Joined: Fri Sep 28, 2018 12:28 am

### Re: half-life

You can't multiply it by four but you can square it by four!
Break down the math:
1/2 x 4 = 2
vs
(1/2)^4 = 1/16

Shivangi_2J
Posts: 65
Joined: Fri Sep 28, 2018 12:15 am

### Re: half-life

Sarah Zhao 4C wrote:You can't multiply it by four but you can square it by four!
Break down the math:
1/2 x 4 = 2
vs
(1/2)^4 = 1/16

This doesn't give you the correct answer :( I think it's because, as someone said above, the half-life of a second-order reaction depends on how much reactant is left

Ethan Breaux 2F
Posts: 63
Joined: Sat Sep 29, 2018 12:16 am

### Re: half-life

you have to get the formula for t from the second rate equation which causes you to get t = (1/[A]t - 1/[A]o)/k and in this case you can replace [A]t with (1/16) x [A]o causing the equation to be t = (16/[A]o - 1/[A]o)/k and you just plug stuff in :)