## Rate 1st Order Rxn

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

Christina_F_3F
Posts: 21
Joined: Fri Jul 22, 2016 3:00 am

### Rate 1st Order Rxn

The equation for the rate of a first order reaction is k [A] = k [A](naught)e^-kt

I was just wondering if there is a difference between the two k values on the right side of the equation?
Does the k being multiplied by [A] have a different meaning/value than the one in e^-kt?

Julia Nakamura 2D
Posts: 33
Joined: Wed Sep 21, 2016 2:57 pm

### Re: Rate 1st Order Rxn

The rate law for a first order reaction is k[A]. However, the integrated rate law is ln[A]= -kt + ln[A]naught, which means that there is only one value for k in the integrated rate law. It looks like you may have just mixed up the equations.

Christina_F_3F
Posts: 21
Joined: Fri Jul 22, 2016 3:00 am

### Re: Rate 1st Order Rxn

Whoops, I did mean the integrated rate law.
On page 64 of the course reader, ln [A] = -kt + ln [A]naught is then exponentiated and reads "RATE 1ST ORDER RXN = k[A] = k [A]naught e^-kt. So I was just curious if the k value that is next to [A] and the one that is being expoentiated are the same/represent the same thing.

Tiffany_Hoang_3C
Posts: 20
Joined: Wed Sep 21, 2016 2:59 pm

### Re: Rate 1st Order Rxn

Yes, k is the rate constant so they are both the same. k[A] = k[A]oe-kt is not meant to be seen as an equation, but rather as two different ways to express the rate of a 1st order reaction.