`N_(2)(g) +3H_(2)(g) hArr 2NH_(3)(g),`
In the reaction given above, the addition of small amount of an inert gas at constant pressure will shift the equilibrium towardss which side of

Addition of inert gas at constant pressure in equlibrium 2SO_2(g)+O_2(g) hArr 2SO_3(g) , will

Addition of inert gas at constant pressure in equlibrium 2SO_2(g)+O_2(g) hArr 2SO_3(g) , will

Addition of inert gas at constant volume to the equilibrium N_2(g)+3H_2(g)iff2NH_3(g) results.

Consider the equilibrium: N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)+ heat. Apply the i.e., chatelier's principle to explain the effect of pressure, temperature and addition of inert gas at constant volume on the equilibrium yield of NH_(3) .

Consider the reaction N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g) The equilibrium constant of the above reaction is K_(p) If pure ammonia is left to dissociated, the partial pressure of ammonia at equilibrium is given by : (Assume that P_(NH_(3)) lt lt P_(total) at equilibrium)

On addition of an inert gas at constant volume to the reaction : N_2 + 3H_2 hArr 2NH_3 at equilibrium:

On addition of an inert gas at constant volume to the reaction N_(2) + 3H_(2)hArr2NH_(3) at equilibrium

N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)) for the reaction initially the mole ratio was 1: 3 of N_(2). H_(2) . At equilibrium 50% of each has reacted. If the equilibrium pressure is p, the partial pressure of NH_(3) at equilibrium is