## Integrated Rate Laws when a =/= 1

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

Joanne Guan 1B
Posts: 30
Joined: Sat Jul 22, 2017 3:01 am

### Integrated Rate Laws when a =/= 1

For a reaction like 2H2O2 --> 2H2O + O2, should we integrate a rate law using a = 2 or just use the ones given to us on the constant sheet? I know that the integrated first-order rate law for this would be ln[A] = ln[A]0 - 2kt, but I remember being told to just use the a = 1 rate law. Can someone clarify why?

Yashaswi Dis 1K
Posts: 56
Joined: Fri Sep 29, 2017 7:04 am

### Re: Integrated Rate Laws when a =/= 1

I am not sure if I will be able to exactly answer your question, but I think each order has it's own integrated rate law. To find the order, I suggest using the given concentrations and if time is also given, then plot each type of integrated rate law and see what the order is based off of that.

Yashaswi Dis 1K
Posts: 56
Joined: Fri Sep 29, 2017 7:04 am

### Re: Integrated Rate Laws when a =/= 1

I am not sure if I will be able to exactly answer your question, but I think each order has it's own integrated rate law. To find the order, I suggest using the given concentrations and if time is also given, then plot each type of integrated rate law and see what the order is based off of that.

404995677
Posts: 82
Joined: Fri Sep 29, 2017 7:07 am

### Re: Integrated Rate Laws when a =/= 1

Why is the integrated first-order rate law for that chemical equation ln[A] = ln[A]0 - 2kt?

Joanne Guan 1B
Posts: 30
Joined: Sat Jul 22, 2017 3:01 am

### Re: Integrated Rate Laws when a =/= 1

Because to find the first order integrated rate law, we integrate (-1/a)d[A]/dt = -k[A], where a is the coefficient of the reactant.