`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 :
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 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 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 :
Hydrogen (a moles ) and iodine (b moles ) react to give 2x moles of the HI at equilibrium . The total number of moles at equilibrium is
For N_(2)+3H_(2)hArr 2NH_(3) , 1 mole N_(2) and 3 mol H_(2) are at 4 atm. Equilibrium pressure is found to be 3 atm. Hence, K_(p) is
