The following reaction is in equilibrium: #N_(2(g)) + O_(2(g)) rightleftharpoons 2NO_((g))#, what effect does an increase in pressure have on the yield of #NO#?

It is important to remember that in equilibrium reactions involving gases, a change in pressure will only have a meaningful impact if the reaction involves a change in the moles of gas involved.

In other words, if and only if there are different amounts of gas molecules on the two sides of the reaction, altering the pressure will result in a change in the equilibrium's location.

The position of the equilibrium will not be significantly affected by changing the pressure if there are the same number of moles of gas on both sides of the equation.

This is how the equilibrium reaction appears in your situation.

Observe that there are two moles of gas on the side of the reactants and two moles of gas on the side of the products.

This indicates that there will be no change in equilibrium when the pressure is increased.

Consider Le Chatelier's Principle, which states that any change in the reaction's conditions will cause the system at equilibrium to respond in a way that offsets the change.

In your situation, applying more pressure would compel the equilibrium to change in a way that would lessen the pressure increase.

The side with fewer gas molecules would be preferred because this shift will lead to fewer molecules forming and a decrease in pressure, as pressure is created by collisions between the molecules of gas and the container walls.

Since there are the same number of gas molecules on both sides in this instance, raising the pressure won't cause the equilibrium to move.

Naturally, this means that the forward and reverse reactions will keep happening at the same pace.

Increasing pressure favors the side with fewer moles of gas. In this case, an increase in pressure will favor the formation of NO.