The equilibrium will evolve to oppose the external stress, but note that #"oppose"# #!=# #"counteract"#.
We have,
#2NO_2(g) rightleftharpoonsN_2(g) + 2O_2(g); DeltaH=-124*kJ*mol^-1#
We could also write this another way in order to vizualize the effect of the stress:
#2NO_2(g) rightleftharpoonsN_2(g) + 2O_2(g) +Delta#, i.e. energy is released in the decomposition.
#(a)# If temperature is increased, we are in effect adding #Delta# to the system; the equilibrium should move to the LEFT.
#(b)# Decreasing the volume should have no effect on a reaction where there are EQUAL numbers of moles of gaseous products and gaseous reactants. Here, there are 3 moles of gaseous products, and 2 moles of gaseous reactants. In order to compensate for the increased pressure change, the equilibrium should move to the left hand side, the product side, in order to alleviate the external perturbation, the increased pressure.
#(c)# Addition of an inert gas should also have no effect, in that the concentrations, i.e. the partial pressures of the reactants and products remain constant; cf. #"Dalton's Law of Partial Pressures"#, #"in a gaseous mixture, the pressure exerted by a component is the"# #"same as the pressure it would exert if it ALONE occupied the"# #"container"#.
#(d)# Addition of dinitrogen gas increases the concentration, i.e. the partial pressure of #N_2#; the equilibrium should move to the left hand side as written in order to reestablish equilibrium concentrations.
