## 5H.3

Chris Tai 1B
Posts: 102
Joined: Sat Aug 24, 2019 12:16 am

### 5H.3

Use the information in Table 5G.2 to determine the value of K at 300 K for the reaction 2 BrCl(g) + 1 H2(g) <-> Br2(g) + 2 HCl(g).

Is there a way to determine the value of K given this table? If so, how?
Attachments Christineg1G
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Joined: Fri Aug 09, 2019 12:15 am
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### Re: 5H.3

You would use the values of K in the table to determine the value of K for the reaction. So you would use K=377 for the expression 2BrCl<->Br2+Cl2 and K=4.0x10^31 for the expression H2+Cl2<->2HCl. You would then multiply the values of K1 and K2 to get your answer.

Caitlyn Tran 2E
Posts: 100
Joined: Fri Aug 09, 2019 12:15 am

### Re: 5H.3

If this were a problem on a test, the K values would have to be given to you. The only time you would ever have to calculate K is if you could find the concentrations of all reactants and products or the partial pressures of all reactants and products to plug into the K expression. This is not the case here, which is why the textbook allows you to reference the table. Hope this helps!

805329408
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Joined: Wed Nov 20, 2019 12:21 am

### Re: 5H.3

Does anyone know what the very last column is used for?

505316964
Posts: 95
Joined: Thu Jul 11, 2019 12:17 am

### Re: 5H.3

When you manipulate an equation you also manipulate the K value, which is why you multiply the K values of the two equations to get your composite equation's K value.

For example if you multiply an equation by 2 you square the K value.