## K in Rate Laws [ENDORSED]

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

Jared Smith 1E
Posts: 51
Joined: Fri Sep 29, 2017 7:04 am

### K in Rate Laws

I noticed that depending on the reaction order, the slope of the graph can be equal to positive or negative k. For first and zero-order reactions, the slope of the graph is equal to -k, but for second order reactions, the slope is equal to +k. Can someone please explain to me why one is different from the other two?

Nisarg Shah 1C
Posts: 54
Joined: Sat Jul 22, 2017 3:00 am

### Re: K in Rate Laws

It depends on the derivation of the integrated rate laws. When you integrate the zero and first-order laws you get a -kt term, and when you integrate the second order law you get a positive kt term. In this case, where we relate these integrated laws to y = mx+b, m (the slope) is equal to positive or negative k.

Justin Folk 3I
Posts: 43
Joined: Wed Sep 21, 2016 2:56 pm

### Re: K in Rate Laws  [ENDORSED]

Because for second order reactions the y-axis is 1/[A]. As time progresses the concentration of [A] decreases (assuming A is reactant) and thus 1/A increases, therefore slope of graph >0