5H.2

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Rafsan Rana 1A
Posts: 55
Joined: Sat Aug 24, 2019 12:16 am

5H.2

Postby Rafsan Rana 1A » Tue Jan 14, 2020 11:12 pm

So for problem 5H.2, the book says that for the reaction, K = 2.5 * 10^10, however in the solutions manual they use K = 2.5 * 10^7 for the calculations. Can someone please explain? Am I missing something here?

Justin Vayakone 1C
Posts: 110
Joined: Sat Sep 07, 2019 12:19 am

Re: 5H.2

Postby Justin Vayakone 1C » Tue Jan 14, 2020 11:42 pm

My solutions manual only has odds and not even numbered problems. Could you post a picture of it so I could see?

Hui Qiao Wu 1I
Posts: 56
Joined: Fri Aug 30, 2019 12:16 am

Re: 5H.2

Postby Hui Qiao Wu 1I » Tue Jan 14, 2020 11:47 pm

My solution manual only includes answers for odd problems, so I don't know how to explain the 2.5 * 10^7. Plus, there were three reactions to the problem, which one were you talking about? But the problem is connecting back to the topics that if a reaction is reversed, then the K would be the inverse; if the reaction is halved, then you would square root the K; and if the reaction is doubled, then you would square the K.


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