5H.3

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

5H.3

Postby Chris Tai 1B » Wed Jan 08, 2020 11:44 pm

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?
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Christineg1G
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Joined: Fri Aug 09, 2019 12:15 am
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Re: 5H.3

Postby Christineg1G » Thu Jan 09, 2020 9:06 am

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

Postby Caitlyn Tran 2E » Thu Jan 09, 2020 9:45 am

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
Posts: 49
Joined: Wed Nov 20, 2019 12:21 am

Re: 5H.3

Postby 805329408 » Thu Jan 09, 2020 11:33 am

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

Postby 505316964 » Thu Jan 09, 2020 11:37 am

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.


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