## Relation between rate law and zero order [ENDORSED]

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

Mary Becerra 2D
Posts: 53
Joined: Fri Sep 29, 2017 7:06 am

### Relation between rate law and zero order

I noticed that in lecture we were given an experimentally determined rate law equal to k[NO2]^2 and made the connection that since the other reactant, CO, when not present in this given rate law it was of zero order. Can we always make this connection? Also, how would this look in a reaction with more than one reactant?

Posts: 58
Joined: Fri Sep 29, 2017 7:06 am
Been upvoted: 1 time

### Re: Relation between rate law and zero order

Yup you can always make that connection because if you were to look at it using the formula, you see that like it becomes [CO]^0 which would be 1. So in this example it would look somewhat like Rate = k [NO2]^2 (1) and the 1 would come from the CO which is 0 order. Hopefully that helps!

Andy Nguyen 1A
Posts: 56
Joined: Thu Jul 27, 2017 3:00 am

### Re: Relation between rate law and zero order

Since the CO wasn't present we were able to assume it was a 0 order because it wasn't in the rate law. I think Dr. Lavelle mentioned that not all reactants actually affect the rate so in those cases they should have a 0 order as well.

Yashaswi Dis 1K
Posts: 56
Joined: Fri Sep 29, 2017 7:04 am

### Re: Relation between rate law and zero order  [ENDORSED]

Yeah for similar problems like that, I suggest looking at the given experimentally determined rate law and then look at all the reactants in the equation and if any reactant is not shown in the rate law shown, then it's most likely zero order because [Reactant]^0 = 1. Hope that helps!