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### 5H.2

Posted: **Tue Jan 14, 2020 11:12 pm**

by **Rafsan Rana 1A**

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?

### Re: 5H.2

Posted: **Tue Jan 14, 2020 11:42 pm**

by **Justin Vayakone 1C**

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

### Re: 5H.2

Posted: **Tue Jan 14, 2020 11:47 pm**

by **Hui Qiao Wu 1I**

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.