## 15.25

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

Lorie Seuylemezian-2K
Posts: 62
Joined: Fri Sep 29, 2017 7:04 am

### 15.25

Can someone explain where we get the reaction A= e^-kt * ao

Vasiliki G Dis1C
Posts: 53
Joined: Fri Sep 29, 2017 7:04 am

### Re: 15.25

This comes from the integrated rate law for a first order reaction. The equation is ln[A]=-kt+ln[A]o. If we take everything and raise e to that power, we get e^ln[A]=e^-kt+e^ln[A]o. the terms on the right side can be added (since they have the same base) to become e^(-kt + ln[A]o). e^ln[A] becomes just [A], and e^ln[A]o becomes just [A]o, so the equation is changed into [A]=[A]o(e^-kt). Hope this helps!

Nickolas Manipud 1C
Posts: 60
Joined: Fri Sep 29, 2017 7:07 am

### Re: 15.25

The above poster is right. It's a shortcut but you can still use the 1st order integrated rate law to solve the problem.

Lorie Seuylemezian-2K
Posts: 62
Joined: Fri Sep 29, 2017 7:04 am

### Re: 15.25

Got it this makes sense now!

Jeremiah Samaniego 2C
Posts: 33
Joined: Fri Sep 29, 2017 7:05 am

### Re: 15.25

Adding on to this, we want to use the exponential (non-linear form) when we are discussing a reactant that is decomposing.