7th Edition 5H 3

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Cole Elsner 2J
Posts: 88
Joined: Fri Sep 28, 2018 12:25 am

7th Edition 5H 3

Postby Cole Elsner 2J » Mon Jan 14, 2019 2:21 pm

I'm a bit confused as to how I can calculate K for the combination of 2 reactions on table 5G.2. This question asks for the value of K at 300 K for
2BrCl+H2 <-->Br2 +HCl. I don't believe this was taught in lecture, but it is on outline 1 so I'm looking for some tips.

Thanks!

Jerome Mercado 2J
Posts: 60
Joined: Tue Nov 28, 2017 3:02 am

Re: 7th Edition 5H 3

Postby Jerome Mercado 2J » Mon Jan 14, 2019 2:25 pm

K should still be the quotient of concentration of products over the concentration of reactants. In this case there will be two concentrations of products on top and 2 concentrations of reactants on the bottom:

[Br2] [HCL] / [BrCl]^2 [H2]

Emmaraf 1K
Posts: 94
Joined: Sat Oct 20, 2018 12:16 am

Re: 7th Edition 5H 3

Postby Emmaraf 1K » Mon Jan 14, 2019 2:30 pm

Looking at the table, you have to divide the reaction into two parts so imagine the reaction happening in two steps to get to the final equation. The first step will have a K value as will the second step. Therefore, to find the final K value, multiply the Ks from the first and second step to get the final K of the reaction they gave you.


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