15.5 a /unique rate of reaction


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sofiakavanaugh
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Joined: Thu Jul 13, 2017 3:00 am
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15.5 a /unique rate of reaction

Postby sofiakavanaugh » Tue Feb 27, 2018 10:13 pm

Hi,

So this is a really simple question but I am just looking for a little clarification, on 15.5 it gives you the unique rate of the rxn: 0.44 m/Ls and it gives you the reaction C2H4(g) + 3O2(g) ----> 2CO2(g) + 2H20(g) and asks for the rate at which the O2 reacts. So in the answers you multiply the rate by three to get your answer which is very straightforward, and I can see it comes from the stoichiometric coefficient 3, and I guess my question more relates to what is the unique rate of a reaction and how would it (conceptually) be different than the rate at which O2 is used? I understand that different reactants will have different reaction rates but if that is true then what does the 0.44 m/Ls represent? I think I am just missing something in my understanding here.


Thanks!

Erik Khong 2E
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

Re: 15.5 a /unique rate of reaction

Postby Erik Khong 2E » Wed Feb 28, 2018 2:17 am

0.44 m/Ls represents the rate of C2H4's reaction. This would be the unique rate of reaction, and you can use it and the other reactants or products' coefficients to find their own rate of reactions. For example, for every one mole of C2H4 being reacted, 3 moles of O2 are being formed. So to find O2's rate of formation you would multiply C2H4's rate of reaction by 3, which gives you the answer of 1.32m/Ls.


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