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### Example in class

Posted: Fri Jan 18, 2019 5:02 pm
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?

### Re: Example in class

Posted: Fri Jan 18, 2019 6:27 pm
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.

### Re: Example in class

Posted: Fri Jan 18, 2019 8:06 pm
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 :)

### Re: Example in class

Posted: Fri Jan 18, 2019 10:39 pm
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.

### Re: Example in class

Posted: Sat Jan 19, 2019 12:39 am
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.

### Re: Example in class

Posted: Sat Jan 19, 2019 6:46 pm
for the example he gave, he used Q since the equation wasn't at equilibrium