## Pseudo 1st order rate laws

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

taywebb
Posts: 60
Joined: Fri Sep 28, 2018 12:15 am

### Pseudo 1st order rate laws

After you have identified the reactant(s) in large excess, making them negligible in terms of the rate law. Are there any fundamental differences when composing a pseudo-first-order rate law vs. a normal first-order rate law?

Parth Mungra
Posts: 72
Joined: Fri Sep 28, 2018 12:28 am

### Re: Pseudo 1st order rate laws

the k value for the psuedo system will incorporate something similar to k=k1k2/k2, and these k1 and k2 come from the molecules in large excess.

605168557
Posts: 65
Joined: Fri Sep 28, 2018 12:18 am

### Re: Pseudo 1st order rate laws

I think the main thing to remember for pseudo-first order rate law is just that N=1 because [B] and [C] are in large excess they remain essentially constant.