average rate

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

Posts: 134
Joined: Sat Sep 14, 2019 12:17 am

average rate

Is there a negative in a unique rate law for the reactants because that is what is being used up? Or is the negative there for a different reason?

Ryan Yee 1J
Posts: 101
Joined: Sat Aug 17, 2019 12:16 am

Re: average rate

The rate law for a reaction aA -> bB is -(1/a)(d[A]/dt) = (1/b(d[B]/dt) the negative is there for the reactants to express that they are being used up.

Juliana Chopelas 1A
Posts: 50
Joined: Sat Aug 24, 2019 12:16 am

Re: average rate

Negative in a rate law is used to show when reactants are being all used up

Matthew Tsai 2H
Posts: 101
Joined: Wed Sep 18, 2019 12:20 am

Re: average rate

The negative just means that the reactant concentration is decreasing over time, which makes sense given that they are consumed as the reaction progresses.

Indy Bui 1l
Posts: 99
Joined: Sat Sep 07, 2019 12:19 am

Re: average rate

You are correct, the negative is used to show that the reactants are being used up.