Example in class
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Example in class
When we were talking about that one example that had to do with pressure, why did we use Q instead of Kp to calculate the new pressure?
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Re: Example in class
I'm not in lecture 2 so I may not be correct, but if he used Q then it was probably because the rxn was not at equilibrium.
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Re: Example in class
We only use Kp when the reaction system is at equilibrium, other than that, we will always calculate Q and compare it with Kp (or sometimes Kc if the question is about concentrations) to see which direction (products or reactants) the reaction will move towards. Hope this helps :)
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Re: Example in class
I agree Q is used when the reaction is not at equilibrium. The Q value then will allow us to predict the direction the reaction will proceed when compared to Kc.
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Re: Example in class
Q looks at the system when it is not at equilibrium(when it ratio of products to reactants is equal to Kc). Q is compared to Kc to see whether the reaction proceeds to the right or the left. If Q<Kc then the reaction will "shift" to the right. If Q>K then the reaction will "shift" to the left. If Q=Kc, then the system is at equilibrium.
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