11.39 6th edition

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204929947
Posts: 76
Joined: Fri Apr 06, 2018 11:03 am

11.39 6th edition

Postby 204929947 » Fri Feb 08, 2019 8:52 pm

Hey guys, I don't quite understand how to calculate K. It says to use Table 11.2 but it doesn't show the full equation.. can someone help me please.
Here is the question:

Use the information in Table 11.2 to determine the value of K at 300 K for the reaction 2 BrCl(g) + H2(g) --> Br2(g) + 2 HCl(g).

harshitasarambale4I
Posts: 58
Joined: Fri Sep 28, 2018 12:26 am

Re: 11.39 6th edition

Postby harshitasarambale4I » Sun Feb 10, 2019 11:39 am

In the table, use the K values in the second column for this question.

aisteles1G
Posts: 117
Joined: Fri Sep 28, 2018 12:15 am

Re: 11.39 6th edition

Postby aisteles1G » Sun Feb 10, 2019 11:44 am

Yea so for this one you have to combine 2 separate equations from the given table to make your final equation (kind of how you do with rxn enthalpies as well). Two things to note: when you multiply an equation through by a number, you raise the K value to the power of that number and when you add equations, you multiply the corresponding K values instead of adding them. From the given table use equations 2BrCL->Br2 + Cl2; K=377 and add it with H2 + Cl2 --> 2HCl; K=4*10^31. When you add them the Cl2 cancels out and youre left with the equation you need: 2BrCl + H2 --> Br2+2HCl, and for the K's you multiply them together to get the one for this new equation so 377*(4*10^31)=1.5*10^34. Hope this helps!


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