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### AuthorTopic: Equilibrium  (Read 4426 times) !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="https://platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); (function() {var po = document.createElement("script"); po.type = "text/javascript"; po.async = true;po.src = "https://apis.google.com/js/plusone.js";var s = document.getElementsByTagName("script")[0]; s.parentNode.insertBefore(po, s);})();

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#### MitchTwitchita

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##### Equilibrium
« on: March 22, 2008, 06:11:50 PM »

Hey Guys, I'm having a really hard time with this one.

Consider the reaction H2(g) + I2(g) ---> 2HI(g).  Starting with a concentration of 0.040 M HI, calculate the concentrations of HI, H2, and I2 at equilibrium.  The Kc for the reaction is 54.3.

2HI             ---->            H2            +                I2

0.040 M                           0                               0
-x                                 +x                             +x
0.040 -x                          x                                x

Kc = [H2][I2] / [HI]^2
54.3 = x^2 / (0.040 - x)^2
7.37 = x / (0.040 - x)
0.2948 - 7.37x = x
x = 3.5 x 10^-2

[HI]0.040 - 2 x (3.5 x 10^-2) = 0.03
[H2] = 3.5 x 10^-2
[I2] = 3.5 x 10^-2

I have no idea where I'm going wrong.  The answers in the book are: [HI] = 0.031 M [H2] = 4.3 x 10^-3 [I2] = 4.3 x 10^-3
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#### Yggdrasil

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##### Re: Equilibrium
« Reply #1 on: March 22, 2008, 07:12:40 PM »

For HI, it should be your equilibrium concentration will be 0.040 - 2x because of the stoichiometry of the reaction.  It looks like you definitely know what you are doing, but just made a small oversight.
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#### MitchTwitchita

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##### Re: Equilibrium
« Reply #2 on: March 22, 2008, 07:30:14 PM »

Aaaaaah!  Thanks again Yggdrasil!
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