## Hw 7A.13

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

kennedyp
Posts: 56
Joined: Tue Nov 13, 2018 12:18 am

### Hw 7A.13

Can someone do a detailed walkthrough of how to do this problem? I keep getting the wrong answer :(
7A.13) In the reaction CH Br(aq) 1 OH2(aq) S CH OH(aq) 1 33
Br2(aq), when the OH2 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.

Chem_Mod
Posts: 18879
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 714 times

### Re: Hw 7A.13

When you increased each reactant separately, the rate of the reaction increased by the same factor. Therefore, both reactants are raised to the first power in the rate law.

Joseph Saba
Posts: 154
Joined: Thu Jul 11, 2019 12:16 am

### Re: Hw 7A.13

So the rate law would be based off of if each reactant is a first, second, or zero order reactant. Based from the info, you can infer that both reactants are first order because as you increase their concentration by a constant, the rate also increases by the same amount. therefore it would be k[CH3Br][OH-]=rate

Siddiq 1E
Posts: 106
Joined: Fri Aug 09, 2019 12:15 am

### Re: Hw 7A.13

kennedyp wrote:Can someone do a detailed walkthrough of how to do this problem? I keep getting the wrong answer :(
7A.13) In the reaction CH Br(aq) 1 OH2(aq) S CH OH(aq) 1 33
Br2(aq), when the OH2 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.

So if you increase one reactant by a certain factor and the rate is also increased by that factor, that means it is first order.

Ayushi2011
Posts: 101
Joined: Wed Feb 27, 2019 12:17 am

### Re: Hw 7A.13

Yes, because the power will be 1, making it a first order reaction.

BeylemZ-1B
Posts: 95
Joined: Thu Jul 25, 2019 12:17 am

### Re: Hw 7A.13

so if both are first order independently, does this make the reaction an overall 2nd order reaction?

Connor Chappell 2B
Posts: 59
Joined: Wed Feb 20, 2019 12:16 am

### Re: Hw 7A.13

yes, since when both concentrations of both reactants are doubled, this results in a doubling of the overall rate, both reactants are first order independently. Thus, when adding together n and m from k[A]^n[B]^m, 1+1=2, so it is overall a second-order reaction.

BeylemZ-1B
Posts: 95
Joined: Thu Jul 25, 2019 12:17 am

### Re: Hw 7A.13

in the homework there were some reactions (7A and 7B) that were larger than second order. We don't need to know much about these (like how to graph them), right?

BeylemZ-1B
Posts: 95
Joined: Thu Jul 25, 2019 12:17 am

### Re: Hw 7A.13

in the homework there were some reactions (7A and 7B) that were larger than second order. We don't need to know much about these (like how to graph them), right?