`N_2 + 3H_2hArr2NH_3` 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 will be ?

N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) For the reaction intially 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 parial pressure of NH_(3) at equilibrium is :

N_(3) +3H_(2) hArr 2NH_(3) Starting with one mole of nitrogen and 3 moles of hydrogen, at equiliibrium 50% of each had reacted. If the equilibrium pressure is P , the partial pressure of hydrogen at equilibrium would be

If Ar is added to the equilibrium N_2(g) +3H_2(g) hArr 2NH_3 at constant volume , the equilibrium will

In a chemical reaction N_(2)+3H_(2) hArr 2NH_(3) , at equilibrium point