## 15.15 Rate Law

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

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Kathleen Vidanes 1E
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Joined: Fri Sep 29, 2017 7:07 am

### 15.15 Rate Law

The question reads: In the reaction CH3Br(aq) OH (aq) S CH3OH(aq) Br (aq), when the OH concentration alone was doubled,
the rate doubled; when the CH3Br concentration alone was increased by a factor of 1.2, the rate increased by a factor of 1.2. Write the rate law for the reaction.

I'm still confused as to how the proportionality relates to the rate law (rate = k[CH3Br][OH-]). Can someone please explain this? Thank you.

Sarah Rutzick 1L
Posts: 50
Joined: Tue Oct 10, 2017 7:13 am

### Re: 15.15 Rate Law

When the OH- concentration doubles, the rate increased by the same factor, and when the CH3Br concentration increases by a factor or 1.2, the rate increased by the same factor as well. Therefore, both reactants form a 1:1 ratio with the rate law, yielding the rate above. If, for example, when the concentration of OH- doubled and the rate quadrupled, you would know that the rate law is rate = k[CH3Br][OH-]^2.

ClaireHW
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### Re: 15.15 Rate Law

For this problem, why aren't the products considered in the rate law?

(Claire Woolson Dis 1K)

Jessica Wakefield 1H
Posts: 58
Joined: Thu Jul 27, 2017 3:00 am

### Re: 15.15 Rate Law

if your concentration doubles and your rate doubles as well, the interaction is proportional and thus the order is considered first
in this problem they said that when concentration is increased by 1.2 the rate does as well and also when the other concentration is doubled the rate does as well, demonstrating this proportionality that would classify the reaction as first order

Ishita Monga 1B
Posts: 31
Joined: Thu Jul 27, 2017 3:00 am

### Re: 15.15 Rate Law

ClaireHW wrote:For this problem, why aren't the products considered in the rate law?

(Claire Woolson Dis 1K)

to make calculations easier we only consider the reactants when determining the rate law.

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