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kc and equilibria

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sorry had to put up another question,
im pretty sure that the higher the kc, the more the products are formed
and so I can cancel out C but im not sure about the others

cheers

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No. Q in that equation is the same thing, but I don't mean that equation. It is calculated in exactly the same way as K, but it does not matter if the concentrations of each species is at equilibrium. The point of the calculation is to determine whether or not Q = K, and thus whether or not the system is at equilibrium. Whether or not Q is less than, greater than, or equal to K will tell you if it's at equilibrium, or if not, which direction it will go to reach equilibrium.

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No. Q in that equation is the same thing, but I don't mean that equation. It is calculated in exactly the same way as K, but it does not matter if the concentrations of each species is at equilibrium. The point of the calculation is to determine whether or not Q = K, and thus whether or not the system is at equilibrium. Whether or not Q is less than, greater than, or equal to K will tell you if it's at equilibrium, or if not, which direction it will go to reach equilibrium.

 

thanks i get it now :)

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