Page 1 of 1

Hw 7A.13

Posted: Sat Mar 14, 2020 2:01 pm
by kennedyp
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
by Chem_Mod
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
by Joseph Saba
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
by Siddiq 1E
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
by Ayushi2011
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
by BeylemZ-1B
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
by Connor Chappell 2B
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
by BeylemZ-1B
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
by BeylemZ-1B
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?