## Pseudo-First Order Reaction [ENDORSED]

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

Veritas Kim 2L
Posts: 30
Joined: Fri Sep 29, 2017 7:05 am

### Pseudo-First Order Reaction

What is pseudo-first order reaction and how is that related to the second-order reaction?

Sarah_Stay_1D
Posts: 57
Joined: Sat Jul 22, 2017 3:00 am
Been upvoted: 1 time

### Re: Pseudo-First Order Reaction

Veritas Kim 2L wrote:What is pseudo-first order reaction and how is that related to the second-order reaction?

You can use a pseudo-first order reaction if you have more than one reactant. You would make the concentration of one of you reactants extremely high, and make the concentration of you second reaction low. Since the concentration of one reactant is so high, it remains effectively constant and therefore we can treat it as a constant. The constant for the reactant in high concentration is included in the constant k, and renamed k'. This means you now only have one reactant changing so it is in theory a first order reaction.

Kaileigh Yang 2I
Posts: 49
Joined: Sat Jul 22, 2017 3:00 am

### Re: Pseudo-First Order Reaction

I guess it's kind of related to the second order reaction because of the fact that you have more than one reactant.

Michael Cheng 1C
Posts: 51
Joined: Thu Jul 13, 2017 3:00 am

### Re: Pseudo-First Order Reaction

what is the difference between k' and k?

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

### Re: Pseudo-First Order Reaction  [ENDORSED]

From what I remember from lecture, K is the rate constant for a normal reaction. K' is the rate constant for the pseudo-reaction where the other reactants are in very high concentrations. In other words, K' is the rate constant for the experimental reaction with just the one reactant. In addition, k and k' can be mathematically related too. Using the example reaction in lecture, k= k'/([B]o^M) * ([C]o^L).