At equilibrium, `AB_3(g)` is dissociated `40%` in a mixture with `AB_(3)(g)` according to the following reaction, `AB_(3)(g) ↔A(g)+3B(g)`,the partial pressure of the `A(g)` will be (P: Total equilibrium pressure)

Calculate the percentage of AB_3 dissociated in the following reaction AB_3(g)⇌AB_2(g) +1/2B_2(g) Given initial pressure 800 mm and total pressure at equilibrium is 900 mm of Hg.

Solid ammonium carbamate dissociated according to the given reaction NH_2COONH_4(s) hArr 2NH_3(g) +CO(g) Total pressure of the gases in equilibrium is 5 atm. Hence K_p .

For the reaction A_((s)) + 2B_((g)) ƒhArr 3C_((g)) . At constant pressure on addition of inert gas, the equilibrium state

For the dissociation reaction N_(2)O_(4) (g)hArr 2NO_(2)(g) , the degree of dissociation (alpha) interms of K_(p) and total equilibrium pressure P is:

K_p for the equation A(s)⇋ B(g) + C(g) + D(s) is 9 atm^2 . Then the total pressure at equilibrium will be

At certain temperature compound AB_(2)(g) dissociates according to the reaction 2AB_(2)(g) hArr2AB (g)+B_(2)(g) With degree of dissociation alpha Which is small compared with unity, the expression of K_(p) in terms of alpha and initial pressure P is :

Consider the following equilibrium 2AB(g)hArrA_(2)(g)+B_(2)(g) The vapour density of the equilibrium mixture does not depend upon

An equilibrium PCl_5(g) dissociate 50% ,then the value of (P +K_p) is (where P = total pressure at equilibrium)

At equilibrium degree of dissociation of SO_(3) is 50% for the reaction 2SO_(3)(g)hArr 2SO_(2)(g)+O_(2)(g) . The equilibrium constant for the reaction is

3 moles of A and 4 moles of B are mixed together and allowed to come into equilibrium according to the following reaction. 3A(g) + 4B(g) rarr 2C(g) + 3D(g) When equilibrium is reached, there is 1 mole of C. The equilibrium constant of the reaction is