Hw 7A.13


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kennedyp
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Hw 7A.13

Postby kennedyp » 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.

Chem_Mod
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Re: Hw 7A.13

Postby Chem_Mod » 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.

Joseph Saba
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Re: Hw 7A.13

Postby Joseph Saba » 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

Siddiq 1E
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Re: Hw 7A.13

Postby Siddiq 1E » 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.

Ayushi2011
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Re: Hw 7A.13

Postby Ayushi2011 » Sat Mar 14, 2020 2:54 pm

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

BeylemZ-1B
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Re: Hw 7A.13

Postby BeylemZ-1B » Sun Mar 15, 2020 5:00 am

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

Connor Chappell 2B
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Re: Hw 7A.13

Postby Connor Chappell 2B » 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.

BeylemZ-1B
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Re: Hw 7A.13

Postby BeylemZ-1B » 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?

BeylemZ-1B
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Re: Hw 7A.13

Postby BeylemZ-1B » 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?

Iman Gauhar 3E
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Re: Hw 7A.13

Postby Iman Gauhar 3E » Sun Mar 13, 2022 11:00 am

Hi! The way you would approach this problem is to find out the order of each individual reactant. Since the change in rate is proportional to the change in concentration, you know that both reactants are first-order. Hope this helps!


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