Question 5H.3

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Megan Vu 1J
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Joined: Thu Jul 25, 2019 12:15 am

Question 5H.3

Postby Megan Vu 1J » Mon Jan 06, 2020 3:37 pm

I have a question on 5H.3, which states, "Use the information in Table 5G.2 to determine the value of K at 300 K for the reaction 2 BrCl(g) + H2(g) ->/<- Br2(g) + 2 HCl(g)."

I know that with the previous exercises with 5H, we had to use the K values listed on the table, but I am confused with this question because it utilizes BrCl as well as H2 which is different compared to the table. Does anyone have an idea to start on this question?

christabellej 1F
Posts: 109
Joined: Sat Aug 17, 2019 12:17 am

Re: Question 5H.3

Postby christabellej 1F » Mon Jan 06, 2020 3:53 pm

For this question, you could use the K values of 2 different equations, one that includes BrCl and another that has H2 in the equation. You can then find the overall K value of the equation in question by multiplying the K of the first equation with the K of the second equation.

Amy Pham 1D
Posts: 103
Joined: Fri Aug 09, 2019 12:15 am

Re: Question 5H.3

Postby Amy Pham 1D » Mon Jan 06, 2020 4:00 pm

The chemical equation given is the overall reaction, but we can break it down into two steps that make up the overall multistep reaction, both step reactions being given in the table. First is 2 BrCl (g) ->/<- Br_2 (g) + Cl_2 (g), and the second reaction is H_2 (g) + Cl_2 (g) ->/<- 2 HCl (g). The K listed for the first reaction is 377, and the K for the second reaction is 4.0*10^31. To find the overall K for the given reaction you would then multiply the two K's together, which gives 1.5*10^34.

Kishan Shah 2G
Posts: 132
Joined: Thu Jul 11, 2019 12:15 am

Re: Question 5H.3

Postby Kishan Shah 2G » Mon Jan 06, 2020 4:54 pm

This equation can be broken down into two smaller equation. By looking at the chart it becomes evident of what those two equations are. Then identify K values at the correct temperature for these reactions and multiply K values together for the overall K for the reaction.

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