## Half Life

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

Erik Buetow 1F
Posts: 96
Joined: Fri Aug 30, 2019 12:15 am

### Half Life

In my notes I have written down that half life depends on initial concentration for a second order reaction, but for a first order reaction, it makes no difference for initial concentration regarding half life. Why is this?

505316964
Posts: 95
Joined: Thu Jul 11, 2019 12:17 am

### Re: Half Life

If you look at how half life is derived from the original 1st order reaction equation the initial concentration is canceled out when deriving the half life, which is why it doesn't matter.

Ariana Iranmahboub1G
Posts: 114
Joined: Fri Aug 09, 2019 12:17 am

### Re: Half Life

Since we use the integrated rate law to compare the rate function over time, when we solve for the 1st order integrated rate law, the rate = k[A] so the the initial concentrations cancel out, leaving the half-life equal to 0.693/k. On the other hand, since the rate for 2nd order is k[A]^2, when we solve for the 2nd order integrated rate law, we are left with 1/[A]=k*t + 1/[A], which makes the half-life equal 1/k*[A].

Owen-Koetters-4I
Posts: 50
Joined: Fri Sep 28, 2018 12:16 am

### Re: Half Life

the initial concentration cancels in the derivation

Lauren Tanaka 1A
Posts: 109
Joined: Sat Aug 17, 2019 12:18 am

### Re: Half Life

This is because in a second-order reaction the initial concentrations don't cancel out like they do in a first-order reaction. Because of this it doesn't matter for a first-order reaction but does for a second-order reaction.

BeylemZ-1B
Posts: 95
Joined: Thu Jul 25, 2019 12:17 am

### Re: Half Life

505316964 wrote:If you look at how half life is derived from the original 1st order reaction equation the initial concentration is canceled out when deriving the half life, which is why it doesn't matter.

let's work this out:
we know ln([A]t/[A]0) = -kt
if we divide both sides by (-k), we get:
t = (1/k)*ln([A]0/[A]t) --- {{since ln(1/x) = -lnx}}

now if we set t = t(1/2) and [A]t = .5[A]0,
t(1/2) = (1/k)*ln([A]0/.5[A]0)
... initial concentrations cancel, and 1/0.5 =2...
t(1/2) = 1/k*ln2

Julia Holsinger_1A
Posts: 50
Joined: Tue Feb 26, 2019 12:16 am

### Re: Half Life

How do you determine time (t) when given half life and the percentage something has decreased?

EX: from self-test 7B.4A: In 1972, grain treated with methyl mercury was released for human consumption in Iraq, resulting in 459 deaths. The half-life of methyl mercury in body tissues is 70 days. How many days are required for the amount of methyl mercury to drop to 10% of the original value after investigation?

answer is 230 Days but I dont know how to get there.

Joanne Lee 1J
Posts: 100
Joined: Thu Jul 25, 2019 12:15 am

### Re: Half Life

When you derive the half life, the initial concentration will cancel out in that equation so therefore initial concentration makes no difference in first order reactions.

Michelle Xie 2B
Posts: 99
Joined: Sat Aug 17, 2019 12:18 am

### Re: Half Life

You can see the difference when you derive the half life formula from each rate law.