## Psuedo 1st order rate law [ENDORSED]

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

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

### Psuedo 1st order rate law

What exactly is a pseudo 1st order rate law and how is it different from a normal 1st order rate law? It was mentioned in lecture on Friday and I don't fully understand the concept.

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

### Re: Psuedo 1st order rate law

It's considered a pseudo rate law because the if the normal rate law is Rate=k*A*B*C, we have to change the concentrations of B and C to see how A affects the rate. So under the assumption that B and C are in excess and are constants, we use the psuedo rate law Rate=kprime*A, where kprime=k*B*C. I think we use the pseudo rate law because it is easier to work with the equation Rate=kprime*A than Rate=k*A*B*C.

Charles Ang 1E
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

### Re: Psuedo 1st order rate law  [ENDORSED]

The post above me explains it well. The pseudo rate law assumes two of the variables are in excess allowing us to reduce the number of variables in our equation, simplifying the calculation.