I have a question about this one

2N2O(g) ⇋ O2(g) + 2N2(g), what happens to the equilibrium position if the pressure decreases?

Is going to shift to the right or to the left???

yes

Here is a statement to remember. If the pressure of a system in equilibrium

INCREASES the system will shift to the side with the fewer moles of gas (in order to occupy the smaller volume of course). I see 3 mols of gas on the right and 2 on the left..

Then if it Decrease the system will shift to the side with more moles of gas?

To determine the shift in the equilibrium position when the pressure decreases, we need to analyze the number of moles of gas on each side of the equation.

Looking at the balanced equation: 2N2O(g) ⇋ O2(g) + 2N2(g)

On the left-hand side, we have 2 moles of N2O.

On the right-hand side, we have 1 mole of O2 and 2 moles of N2.

Since there are more moles of gas on the right-hand side (3 moles) compared to the left-hand side (2 moles), we can conclude that the reaction produces more moles of gas when it shifts to the right.

When the pressure decreases, the system will try to reduce the excess pressure by shifting in the direction that produces fewer moles of gas. Therefore, the equilibrium position will shift to the left in order to reduce the total gas moles.

In summary, when the pressure decreases, the equilibrium position will shift to the left.

it is going to the left side.