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Topic: Equilibrium Problem  (Read 3988 times)

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Offline LoveInQuanta

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Equilibrium Problem
« on: September 29, 2009, 10:10:59 PM »
"In an analysis of interhalogen activity, 0.490 mol ICl was placed in a 5.00 L flask and allowed to decompose at a high temperature.

2 ICl(g)  I2(g) + Cl2(g)

Calculate the equilibrium concentrations of I2, Cl2, and ICl. (Kc = 0.110 at this temperature.)"

All right, I am dad-blasted sure I am doing this problem right. I calculate the concentration of ICl (0.490 mol/5.00L = 0.098M), set up an ICE table (initial, change, equilibrium concentrations) thus:

           I        C        E
ICl    .098    -2x       .098-2x
I2       0        +x        x
Cl2      0       +x        x

And then set up the equation to find Qc: (x)(x)
                                                     ---------               = .110 (which is our given Kc; I believe I can set this equal
                                                      (.098-2x)squared             to Qc b/c we're dealing with conditions at equilibrium)

After a nasty quadratic equation, we have x = .10811, and thus that is the concentration of both I2 and Cl2, and for some reason this makes the conc. of ICl negative, which makes no sense at all.

Only it's NOT right. My online homework is telling me it's wrong, and I only have 2 more chances. But where did I go wrong? I have checked my math twice.

Offline LoveInQuanta

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Re: Equilibrium Problem
« Reply #1 on: September 29, 2009, 10:21:38 PM »
Update: I tried simplifying the algebraic expression another way, by taking the square root of both sides before getting into the quadratic nastiness, and I've got it. Too much math for one problem the other way  ::)

Offline AWK

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Re: Equilibrium Problem
« Reply #2 on: September 30, 2009, 03:27:18 AM »
Quadratic equation has two solutions, in this case with different signs. Only the positive one has a physical meaning.
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Offline Borek

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Re: Equilibrium Problem
« Reply #3 on: September 30, 2009, 04:24:43 AM »
Seems to me like 0.108 was neither positive nor negative solution of the equation.
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Offline LoveInQuanta

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Re: Equilibrium Problem
« Reply #4 on: October 01, 2009, 07:32:10 PM »
Borek,

It wasn't. I was getting tangled up in the math. I just took the square root of both sides to do away with the x^2 terms, and it became much easier to solve :)

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