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umdnjmed

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Suppose the reaction:
CO(g) + Cl2(g) ---> COCl2 (g)
has an equalibrium constant of 530. What is the equalibrium constant of the reaction.
2CO(g) + 2Cl2(g) ---> 2COCl2 (g)

Answer is 2.81 x 10^5

EK solution says that since everything doubled, means everything was squared in equalibrium expression therefore Keq is squared as well.

I understand the mathematics behind the answer but doesn't this answer suggest that the concentration of reactants and products affects the equilibrium constant? if i recall correctly, keq is independent of [R] and [P], or does this apply only to reactions which are already at equilibrium?
 

ssdfrb

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i just did this problem and i dont get their explanation either...


if you substitute numbers into the equilibrium constant you should get the same number for both reactions.
 

mehc012

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i just did this problem and i dont get their explanation either...


if you substitute numbers into the equilibrium constant you should get the same number for both reactions.

No, you'll still get the square of the original. Squares don't cancel like a constant; (xy)^2/z^2 does not equal xy/z. If you plug in numbers, you'll see the same thing:

(2*3/4) =/= (4*9/16)

You're multiplying the top and the bottom by different numbers, so you will get a different fraction.
 
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SN2ed

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Thread closed. These types of questions belong in the MCAT Study Q&A forum.
 
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