## Rate law factors [ENDORSED]

$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$

Martin_Sarafyan_2K
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Joined: Wed Sep 21, 2016 2:56 pm

### Rate law factors

So if rate laws are affected by energy, frequency, and orientation of collisions, meaning temperature, catalyst, pressure, and concentration, is there any other situation where the rate constant K and rate laws are affected? K isn't affected by concentration, right, because its only affected by temperature? However, increasing the number of catalysts doesn't affect K, why?

Chem_Mod
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### Re: Rate law factors  [ENDORSED]

The rate constant obeys the Arrhenius equation:

$k=A*exp(\frac{-E_a}{RT})$

Catalysts lower the activation energy needed for a reaction. Therefore, by providing or removing a catalyst, one is changing the activation energy which will have an affect on the rate. As for the number of catalysts - if you're asking specifically why increasing the number of catalysts may not increase the rate constant, it may just be that there are too many catalysts at work and the law of diminishing returns is applying.

BrianaBarr2A
Posts: 22
Joined: Wed Sep 21, 2016 3:00 pm

### Re: Rate law factors

I saw this in the chapter in the textbook and am curious about the answer/people's thoughts. It states, what would be the units of k for an overall order of 3/2 if the concentrations were expressed in grams per milliliter, g.mL-1?