11.39 6th Ed Multiplying K constants

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Sarah Bui 2L
Posts: 48
Joined: Thu Sep 27, 2018 11:29 pm

11.39 6th Ed Multiplying K constants

Postby Sarah Bui 2L » Sat Jan 12, 2019 4:46 pm

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).

I understand that this reaction is separated into two different reactions, but I'm not sure why we multiply the two k equilibrium constants.

Thank you!

Samantha Kwock 1D
Posts: 49
Joined: Thu Sep 27, 2018 11:24 pm

Re: 11.39 6th Ed Multiplying K constants

Postby Samantha Kwock 1D » Sat Jan 12, 2019 4:52 pm

If two reactions are added together to form another reaction, then their K constants are multiplied to give the new reaction's K constant.

Cynthia Ulloa
Posts: 32
Joined: Fri Feb 23, 2018 3:02 am

Re: 11.39 6th Ed Multiplying K constants

Postby Cynthia Ulloa » Sat Jan 12, 2019 9:34 pm

Samantha Kwock 1D wrote:If two reactions are added together to form another reaction, then their K constants are multiplied to give the new reaction's K constant.

can you explain why we would multiply the K constants when we add the reactions? Or is this just a property?

Samantha Kwock 1D
Posts: 49
Joined: Thu Sep 27, 2018 11:24 pm

Re: 11.39 6th Ed Multiplying K constants

Postby Samantha Kwock 1D » Fri Jan 18, 2019 10:17 am

For example, given three reactions:
(1) A + B -> C
(2) C -> A + D
(3) B -> D

The K constants for the reactions would be:
(1) K= [C]/[B]*[A]
(2) K= [A]*[D]/[C]
(3) K = [D]/[B]

If you multiply the two K constants together for reaction one and reaction two, you'll find that the [A] and [C] cancel out leaving the K constant for the 3rd reaction.
Hopefully this makes more sense!


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