## Unique Average Rate --> Rate Law [ENDORSED]

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

Sydney Wu 2M
Posts: 50
Joined: Fri Jul 22, 2016 3:00 am

### Unique Average Rate --> Rate Law

Page 60 of the course reader says that the reaction rate is -(1/a)*(dR/dt) = k[R]n. Is there any specific derivation to come to this conclusion, or is this just the definition of a reaction rate?

Danny Nguyen 2H
Posts: 17
Joined: Fri Jul 22, 2016 3:00 am

### Re: Unique Average Rate --> Rate Law  [ENDORSED]

If you look on page 59 starting where it says "Think about this, if you want to write...", it basically starts from saying the rate of the reactants is equal and opposite of the rate of the products. So then we apply the problem above and the reaction for that particular one and from that we can generalize it for all reactions which ends up to be -(1/a)*(dR/dt) = k[R]n for the rate of the reaction in the slow step only.

Chem_Mod
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### Re: Unique Average Rate --> Rate Law

It's the general form of a rate law. What it says is that the change in concentration ([A]) per unit time (t) (d[A]/dt) is proportional to some constant k (rate constant) as a function of the concentration ([A]) raised to its order (n). If you're familiar with differential equations: the solution to that general form when solved is the differential rate law.