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#?
No effect whatsoever.
The important thing to be aware of here is that for equilibrium reactions that feature gases, changing the pressure will only produce a significant effect if the reaction features a change in the number of moles of gas.
Simply put, changing the pressure will cause a shift in the position of the equilibrium if and only if you have different numbers of moles of gas on the two sides of the reaction.
If you have the same number of moles of gas on both sides of the equation, then changing the pressure will not cause a significant effect on the position of the equilibrium.
In your case, the equilibrium reaction looks like this
#"N"_text(2(g]) + "O"_text(2(g]) rightleftharpoons 2"NO"_text((g])#
Notice that you have two moles of gas on the reactants' side and two moles of gas on the products' side.
This tells you that increasing the pressure will not cause the equilibrium to shift.
Think about Le Chatelier's Principle, which tells you that a system at equilibrium will react to any change in the conditions of the reaction in such a way as to counteract this change.
In your case, increasing the pressure would force the equilibrium to shift in such a way as to reduce this increase in pressure.
Since pressure is caused by collisions between the molecules of gas and the walls of the container, the side that features fewer molecules of gas would be favored, since such a shift will result in less molecules being formed, and thus in a decrease in pressure.
In this particular case, both sides feature the same number of gas molecules, so increasing the pressure will not result in a shift in the position of the equilibrium.
Of course, this implies both the forward reaction and the reverse reaction will continue to take place at the same rate.