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

Posted: **Fri Mar 02, 2018 1:24 pm**

by **Joanne Guan 1B**

For a reaction like 2H_{2}O_{2} --> 2H_{2}O + O_{2}, 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?

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

Posted: **Fri Mar 02, 2018 11:23 pm**

by **Yashaswi Dis 1K**

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.

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

Posted: **Fri Mar 02, 2018 11:23 pm**

by **Yashaswi Dis 1K**

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.

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

Posted: **Sat Mar 03, 2018 12:10 pm**

by **404995677**

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

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

Posted: **Sun Mar 04, 2018 2:16 pm**

by **Joanne Guan 1B**

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.