Help on 11.9: b and c

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Joanna Pham - 2D
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Joined: Fri Apr 06, 2018 11:04 am

Help on 11.9: b and c

Postby Joanna Pham - 2D » Mon Jan 07, 2019 2:54 pm

For this question, we are asked to balance then write the equilibrium expression Kc for the reactions

For part b, the equation is . I balanced the equation and got . Then I used the equation in the book, , inputted the corresponding elements and solved for Kc. I got
for my answer, but in the back of the book, there's no (RT)^4 in front. Everything else was correct, but I am unsure why there's no RT in the answer. Is there another way I am supposed to solve this?

Thanks!

Neil Hsu 2A
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Joined: Fri Sep 28, 2018 12:16 am

Re: Help on 11.9: b and c

Postby Neil Hsu 2A » Mon Jan 07, 2019 3:09 pm

I'm not sure where that equation came from, but I would just find the equilibrium expressions like how Lavelle taught us in class. Given a hypothetical chemical reaction aA + bB--> cC + dD, the equilibrium constant Kc = ([C]^c [D]^d) / ([A]^a [B]^b), or simply just Kc = [products] / [reactants]. So in the case of this question, you balanced the equation right, and Kc should equal [IF5]^2 / ( [F2]^5 [I2]).

Laura Gong 3H
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Re: Help on 11.9: b and c

Postby Laura Gong 3H » Mon Jan 07, 2019 3:23 pm

When the question asks you to write the Kc expression, it is just looking for the form, Kc=[P]/[R].
The equation K=KcRT^delta(n) is the equation used to convert Kc to Kp (or the equilibrium constant using partial pressures).
For some reason, it appears the textbook does not include the subscript p.

kamalkolluri
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Re: Help on 11.9: b and c

Postby kamalkolluri » Mon Jan 07, 2019 4:29 pm

I think we just need to use the general Kp and Kc. You don't need the complex equation, I'm pretty sure that's for later.

Kayla Vo 1B
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Joined: Fri Sep 28, 2018 12:26 am

Re: Help on 11.9: b and c

Postby Kayla Vo 1B » Tue Jan 08, 2019 4:56 pm

That equation from the book is used when you want to write Kp in terms of Kc. For this problem you can just use the formula given in lecture to find the equilibrium constant.


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