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

For N_2+3H_2⇌2NH_3 , one mole of N_2 and three moles of H_2 are at pressure of 4 atm. Equilibrium pressure is found to be 3 atm. Hence, K_p is:

For the equilibrium 2SO_(2)+O_(2)hArr 2SO_(3) we start with 2 moles of SO_(2) and 1 mole of O_(2) at 3 atm. When equilibrium is attained, pressure changes to 2.5atm. Hence, K_(p) is :

N_(2) + 3H_(2)hArr2NH_(3) 1mole N_(2) and 3 moles H_(2) are present at start in 1 L flask. At equilibrium NH_(3) formed required 100 mL of 5 M HC1 for neutralisation hence K_(c) is

In a container of constant volume at a particular temparature N_(2) and H_(2) are mixed in the molar ratio of 9:13 . The following two equilibria are found o be coexisting in the container N_(2)(g)+3H_(2)(g)harr2NH_(3)(g) N_(2)(g)+2H_(2)(g)harr N_(2)H_(4)(g) The total equiibrium pressure is found to be 305 atm while partial pressure of NH_(3)(g) and H_(2)(g) are 0.5 atm and 1 atm respectivly. Calculate of equilibrium constants of the two reactions given above.

For the reaction, N_(2) + 3H_(2) hArr 2NH_(3) in a vessel, equal moles of N_(2) and H_(2) are mixed to attain equilibrium.

For the reaction : NH_(2)COONH_(4)(g) hArr 2NH_(3)(g)+CO_(2)(g) equilibrium pressure was found to be 3 atm at 1000 K. K_(p) is :

If 1 mole of H_(2) , 2 moles of O_(2) and 3 moles of N_(2) are mixed in a vessel and total pressure was found to be 12 atm then the partial pressure exerted by N_(2) in the vessel will be

If 1 mole of H_(2) , 2 moles of O_(2) and 3 moles of N_(2) are mixed in a vessel and total pressure was found to be 12 atm then the partial pressure exerted by N_(2) in the vessel will be