## Rate Law

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

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

### Rate Law

I understand that aA's rate would be (-1/a)*(dA/dt). Does this mean 1/a*A's rate is (-a)*(dA/dt)? Thank you!

Danah Albaaj 1I
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

### Re: Rate Law

No, the answer would just be -1 because you still have to follow the formula and dividing the fractions will cancel out the value of a and leave you with -1.

Sandhya Rajkumar 1C
Posts: 50
Joined: Fri Jun 23, 2017 11:40 am

### Re: Rate Law

From my understanding of it, if your stoichiometric coefficient is now 1/a instead of a, you plug in 1/a to where you would plug in a. This leaves you with
(-1/(1/a))*(d[A]/dt) which simplifies to -a*(d[A]/dt)