Page 1 of 1

### Hw 7A.13

Posted: Sat Mar 14, 2020 2:01 pm
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.

### Re: Hw 7A.13

Posted: Sat Mar 14, 2020 2:08 pm
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.

### Re: Hw 7A.13

Posted: Sat Mar 14, 2020 2:09 pm
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

### Re: Hw 7A.13

Posted: Sat Mar 14, 2020 2:10 pm
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.

### Re: Hw 7A.13

Posted: Sat Mar 14, 2020 2:54 pm
Yes, because the power will be 1, making it a first order reaction.

### Re: Hw 7A.13

Posted: Sun Mar 15, 2020 5:00 am
so if both are first order independently, does this make the reaction an overall 2nd order reaction?

### Re: Hw 7A.13

Posted: Sun Mar 15, 2020 5:46 am
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.

### Re: Hw 7A.13

Posted: Sun Mar 15, 2020 7:32 am
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?

### Re: Hw 7A.13

Posted: Sun Mar 15, 2020 7:33 am
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?