A nitrogen-hydrogen mixture initially in the moler ratio of `1:3` reached equilibrium to from ammonia when 25% of the `N_(2)and N_(2)` had reacterd .If the pressure of the system was 21 atm , the partial pressure of ammonia at the equilibrium was :

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.

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 gas mixture was prepared by taking equal mole of CO and N_(2) . If total pressure of the mixture was 1 atm, the partial pressure of N_(2) in the mixture is-

C(s)+CO_(2)(g) iff 2CO(g) . At equilibrium, 25% of the CO_(2) got converted into CO. If the equilibrium pressure is 12 atm, the partial pressure of CO_(2) at equilibrium is

Peqfor NH_(4)COONH_(2(s)) harr 2NH_(3(g))+ CO_(2(g)) at certain temperature is 0.9 atm. Then, partial pressure of Ammonia at equilibrium (in atm)

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