11.39 6th edition

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CaminaB_1D
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Joined: Fri Sep 28, 2018 12:16 am

11.39 6th edition

Postby CaminaB_1D » Tue Jan 15, 2019 4:48 pm

Can someone please explain this problem:

Use the information in Table 11.2 to determine the value of K at 300 K for the reaction 2 BrCl(g) + H2(g) ∆ Br2(g) + 2 HCl(g)

Tyra Nguyen 4H
Posts: 74
Joined: Fri Sep 28, 2018 12:25 am

Re: 11.39 6th edition

Postby Tyra Nguyen 4H » Tue Jan 15, 2019 5:05 pm

There is a table earlier in the chapter which is needed for reference in this problem. The table will give you the equilibrium constant, K, at various temperatures for specific equilibrium reactions, such as H2(g)+Cl22HCl (g). Knowing the specific equilibrium constants of these referenced equilibrium reactions at certain temperatures, you should be able to find the equilibrium constant of the given reaction in the question at 300K.

Lynsea_Southwick_2K
Posts: 55
Joined: Fri Sep 28, 2018 12:25 am

Re: 11.39 6th edition

Postby Lynsea_Southwick_2K » Fri Jan 18, 2019 10:16 am

Since it is composed of two different equations on the table, how does that work?

Samantha Kwock 1D
Posts: 61
Joined: Fri Sep 28, 2018 12:24 am

Re: 11.39 6th edition

Postby Samantha Kwock 1D » Fri Jan 18, 2019 10:24 am

If a reaction is composed of two different reactions, multiply the K constants to determine the K constant of the third reaction.

Nina Do 4L
Posts: 61
Joined: Fri Sep 28, 2018 12:27 am

Re: 11.39 6th edition

Postby Nina Do 4L » Tue Jan 22, 2019 7:53 pm

Answer: Refer to the table referenced earlier in the chapter. It tells you that K equals 377 so you would set 377 equal to your K equation (products over reactions, not including solids or liquids), and then determine K from the table again from the other elements associated with the reaction. see that it is 4.0+10^31. So you would multiply the two K values to get 1.5*10^34.


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