`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

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_(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_(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_(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

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

A mixture of nitrogen and hydrogen in the ratio of 1: 3 reach equilibrium with ammonia, when 50 % of the mixture has reacted. If the total pressure is P , the partial pressure of ammonia in the equilibrium mixture was :

A mixture containing N_2 and H_2 in a mole ratio 1 : 3 is allowed to attain equilibrium when 50% of the mixture has reacted. If P is the pressure at equilibrium, then the partial pressure of NH_3 formed is

A mixture of nitrogen and hydrogen are initially in the molar ratio of 1:3 related equilibrium to form ammonia when 25% of the material had reacted. If the total pressure of the system is 28 atm , calculate the partial pressure of ammonia at the equilibrium.