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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Twickel on June 09, 2012, 07:03:43 AM

Title: Writing rate laws
Post by: Twickel on June 09, 2012, 07:03:43 AM
Hi
I am having trouble understanding how to write rate laws.

O3 + Cl   :rarrow: ClO + O2
ClO + O --> Cl + O2

Overall O3 + O  :rarrow: 2O2

why is it d[ClO]/dt= k [O3][Cl]

Thanks
Title: Re: Writing rate laws
Post by: Borek on June 09, 2012, 07:39:11 AM
Think about stoichiometry.
Title: Re: Writing rate laws
Post by: Twickel on June 09, 2012, 08:24:13 AM
I actually want to know why we use those reactants products in the rate law.

II thought A+B --> C then its k[a]/dt = k[a]

Title: Re: Writing rate laws
Post by: Borek on June 09, 2012, 09:02:40 AM
I am not sure where the problem is, as what you wrote doesn't make much sense to me. You can put it any way you want, as concentration changes are related to each other by the reaction stoichiometry.

C + D  :rarrow: E

[tex]\frac{d[C]}{dt}=\frac{d[D]}{dt}=-\frac{d[E]}{dt}[/tex]

Or

F + 2G  :rarrow: 3H

[tex]\frac{d[F]}{dt}=\frac{d[G]}{2dt}=-\frac{d[H]}{3dt}[/tex]

(unless I am having a massive brain fart)
Title: Re: Writing rate laws
Post by: Twickel on June 09, 2012, 10:09:22 AM
Sorry, maybe I was not being clear ( as is usually the case)

Why in this specific example, have they put the ClO in the reaction rate and not O2?

Also shouldnt it be -d[ClO]/dt because the ClO is a product?

What does the rate law tell me about the reaction?
Title: Re: Writing rate laws
Post by: Twickel on June 10, 2012, 06:22:56 AM
Is it because the first equation is the rate limiting step, so we use that as the elementary reaction so its simple A+ B ---> C+D rate law with regards to Clo is d[Clo]/dt=k[O3][Cl]?

I thought we should not put ClO in the rate of reaction because we do not observe it, it is a reaction intermediate.
Title: Re: Writing rate laws
Post by: Borek on June 10, 2012, 12:56:00 PM
[tex]\frac{d[ClO]}{dt} = \frac{d[O_2]}{dt}[/tex]

You can use whichever you want, perhaps they had a reason to do it this way, no idea.

You are most likely about the sign thing.