## K limiting rate

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

Frenz Cabison 1B
Posts: 54
Joined: Fri Sep 29, 2017 7:07 am

### K limiting rate

How does K limit the rate of a reaction?

Rachel Brown 3A
Posts: 52
Joined: Fri Sep 29, 2017 7:06 am

### Re: K limiting rate

Because it's the rate constant so a large value of K will make the reaction faster where a lower K will cause a slower reaction

AnuPanneerselvam1H
Posts: 52
Joined: Fri Sep 29, 2017 7:07 am

### Re: K limiting rate

Each reaction has a unique rate. A higher k means a faster reaction and a lower k means a slower reaction. When you multiply k times the concentrations in the rate law, you get the unique reaction rate for the specific reaction.

Gurkriti Ahluwalia 1K
Posts: 37
Joined: Fri Sep 29, 2017 7:07 am

### Re: K limiting rate

how are the units for k the same as the rate for a zero order reaction?
if rate =k[A]^0 then wouldn't you divide both sides by mol/L to get s^-1=k? i know this is the answer to the units for k for a first order reaction, but i'm assuming for a zero order reaction they are the same because of the 0 exponent which makes the whole term [A]^0=1.

clarifications??