5H.1 and 5H.3?

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Andrew Pfeiffer 2E
Posts: 101
Joined: Sat Sep 28, 2019 12:16 am

5H.1 and 5H.3?

Postby Andrew Pfeiffer 2E » Sun Jan 19, 2020 3:35 pm

Can anyone explain the logic behind H.1 and H.3? Thanks y'all.

Johnathan Smith 1D
Posts: 108
Joined: Wed Sep 11, 2019 12:16 am

Re: 5H.1 and 5H.3?

Postby Johnathan Smith 1D » Sun Jan 19, 2020 5:33 pm

I don’t really know how to explain it but page 410 in the book talks about 5H.1

KSong_1J
Posts: 101
Joined: Thu Jul 11, 2019 12:17 am

Re: 5H.1 and 5H.3?

Postby KSong_1J » Sun Jan 19, 2020 9:27 pm

For 5H.1, when you modify a reaction, the K value is also affected. So for part a, since the reaction was flipped backwards, you take the reciprocal of K (1/41) to get 0.024; for Part b, since the reaction was multiplied by 1/2, you take the square root of K to get 6.4; for Part c, since the reaction was multiplied by 2, you square K to get 1681 (or 1.7x10^3).
For 5H.3, you use table 5G.2 to find the K values for 2BrCl —> Br2 + Cl2 and H2 + Cl2 —> 2HCl because these two equations make up the general equation we are trying to find the K value of. You just take the K values of these two equations and multiply them to get the answer.
I hope this helps and if you need more help I would read through section 5H because it explains it all! :)


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