`AB_(3)(g)` is dissociation as `AB_(2)(g) hArr AB_(2)(g)+(1)/(2)B_(2)(g)`,
When the initial pressure of `AB_(2)` is 800 torr and the total pressure developed at equilibrium is 900 torr. What fraction of `AB_(3)(g)` is dissociated ?

AB_(3)(g) is dissociates as AB_(3)(g)hArrAB_(2)(g)+(1)/(2)B_(2)(g) When the initial pressure of AB_(3) is800 torr and the pressure developed at equilibrium is 900 torr , what fraction of AB_(3)(g) is dissociated?

AB_(3)(g) is dissociates as AB_(3)(g)hArrAB_(@)(g)+(1)/(2)B_(@)(g) When the initial pressure of AB_(3) is800 torr and the pressure developed at equilibrium is 900 torr , what fraction of AB_(3)(g) is dissociated?

AB_(3)(g) is dissociates as AB_(3)(g)hArrAB_(@)(g)+(1)/(2)B_(@)(g) When the initial pressure of AB_(3) is800 torr and the pressure developed at equilibrium is 900 torr , what fraction of AB_(3)(g) is dissociated?

AB_(2) dissociates as AB_(2)(g) hArr AB(g)+B(g) . If the initial pressure is 500 mm of Hg and the total pressure at equilibrium is 700 mm of Hg. Calculate K_(p) for the reaction.

AB_(2) dissociates as AB_(2)(g) hArr AB(g)+B(g) . If the initial pressure is 500 mm of Hg and the total pressure at equilibrium is 700 mm of Hg. Calculate K_(p) for the reaction.

AB_(2(g)) dissociates as AB_(2(g)) harr AB_(2(g))+B_((g)) . The initial pressure of AB_2 is 600 mm Hg and equilibrium pressure of the mixture of gases is 800 mm Hg. Then

AB_(2) dissociates as : AB_(2 (g)) hArr AB_((g)) + B_((g)) When the intial pressure of AB_(2) is 500 mm Hg , the total equilibrium pressure is 700 mm Hg . Calculate equilibrium constant for the reaction , assuming that the volume of the system remains unchanged .