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

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

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 a hypothetical reaction of kind. AB_(2)(g)+1/2 B_(2)(g) hArr AB_(3)(g), DeltaH=-x kJ More AB_(3) could be produceed at equilibrium by

In the following reaction, 3A (g)+B(g) hArr 2C(g) +D(g) , Initial moles of B is double at A . At equilibrium, moles of A and C are equal. Hence % dissociation is :

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

At temperature T, a compound AB_2(g) dissociation according to the reaction, 2AB _2(g) hArr 2AB(g) +B_2(g) with degree of dissociation, alpha , which is small compared to unity . Deduce the expression for alpha in terms of the equilibrium constant K_p and the total pressure P.