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Topic: The Equilibrium Law Expression....I AM RIGHT!!  (Read 5487 times)

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Offline Isomer

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The Equilibrium Law Expression....I AM RIGHT!!
« on: July 25, 2006, 08:00:29 PM »
In a class of 30, 10 of us think B is correct. The rest think C. What do you think?

At 900 K, the equilibrium constant for reaction this reaction

2 SO2 + O2 <----> 2 SO3 is 13.0.

The equilibrium concentrations are [SO2] = 0.361 mol/L
                                                [SO3] = 0.840 mol/L

Given the above values, the calculated equilibrium concentration of O2 is

B. 0.416 mol/L

C. 2.40 mol/L

Here is the equation I used.. 13 = (0.840) squared divided by unknown x (0.361) squared.

I hope you guys can understand that. Are the 10 of us right?

Offline Dan

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Re: The Equilibrium Law Expression....I AM RIGHT!!
« Reply #1 on: July 26, 2006, 05:46:57 AM »
13 = (0.840) squared divided by unknown x (0.361) squared.

Yeah,

K = [SO3]2 / [SO2]2[O2]
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Offline sdekivit

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Re: The Equilibrium Law Expression....I AM RIGHT!!
« Reply #2 on: July 26, 2006, 07:28:28 AM »
yeah, B is correct :)

Offline Isomer

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Re: The Equilibrium Law Expression....I AM RIGHT!!
« Reply #3 on: July 27, 2006, 05:06:03 PM »
My teacher still says C is correct! So we took it to another chem teacher who agreed with us!

Offline sdekivit

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Re: The Equilibrium Law Expression....I AM RIGHT!!
« Reply #4 on: July 27, 2006, 05:58:55 PM »
tell your teacher that in general for a homogenous sytem mA + nB <--> qC + rD the equilibruim fraction will be

Qc = ( [C]q[D]r ) / ( [A]mn ) = Kc

and let him read a good chemistry book. (omg what a n00b)

Offline stoneburner

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Re: The Equilibrium Law Expression....I AM RIGHT!!
« Reply #5 on: July 30, 2006, 02:14:32 PM »
Easy enough error to make - the question is, is the equilibrium constant products over reactants, or reactant over products? The teacher just confused, and got it backward, is all. (But should have recognized the error as soon as you questioned him about it).

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