## After Integration [ENDORSED]

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

nelms6678
Posts: 53
Joined: Fri Sep 29, 2017 7:07 am

### After Integration

For the formula after we integrate it yields ln[A] = -k x t
Why is the k negative here, and should it be negative for a second order derivation as well?

Justin Lau 1D
Posts: 51
Joined: Sat Jul 22, 2017 3:00 am

### Re: After Integration

The k is negative here because it is left over from the unique average rate law equation, which says Rate=-(1/a) x d[A]/dt. Therefore, when you rearrange the equation to integrate, you get -kt on the right. Because the negative denotes a -1 basically, it is a constant and remains. For the second order rate law, the same principle applies. However, because you are doing -(1/a) x d[A]/dt= k[A]^2 for the second order, the integral of d[A]/[A]^2 is -[A]^-1. Therefore, the negative is moved over to the -kt side which nullifies the negative.

Justin Chang 2K
Posts: 53
Joined: Fri Sep 29, 2017 7:04 am

### Re: After Integration

Yeah, zero and first order reactions both have a -kt, but second order reactions have +kt.

Emily Mei 1B
Posts: 50
Joined: Fri Sep 29, 2017 7:04 am

### Re: After Integration  [ENDORSED]

Also you can think about k as the slope of the graph for (in this case) ln[A] vs t. The slope of the line will be -k.