CHEMISTRY COMMUNITY

I am aware that change in pressure does not affect the equilibrium constant K, but when I googled it, it said when there is an increase in pressure, the reaction shifts towards the direction of lesser number of moles, and there is a decrease in pressure, then it shifts towards the side with more moles, which he did not mention in class.
Should we keep this in mind, or not worry about this?

I would probably just follow the way Dr. Lavelle taught us in lecture.

The information you googled was briefly mentioned in class, but Dr. Lavelle made it clear that this is only true if the pressure changes by also changing the volume. If the pressure changes by inserting an inert gas, then there is no equilibrium shift.

Just to clarify things, increasing the pressure (which can be done by decreasing the volume) or decreasing the pressure (which can be done by increasing the volume) does not result in a change in the equilibrium constant. It DOES however change the CONCENTRATION of the gasses as concentration is defined as n/V and changing V thus changes the concentration. This new concentration must then be plugged into the equilibrium equation to get a Q value which can then be compared to the K value to see which direction the reaction will shift. This method (of calculating the Q value and comparing it to the K value) is the "true" way of doing this problem as it really gets to the heart of the concept we are trying to teach. That being said, the trick of increasing the pressure makes the reaction shift to the side with more gas moles and decreasing the pressure makes the reaction shift to the side with more gas moles is a "trick" which DOES work. When doing these types of problems students most often use the trick method but it is a good idea to know where this comes from.