`A(g) hArr B(g)+ C/2(g)`
Find relationship between `K`, alpha, equilibrium pressure P.

For A(g) hArr 2B(g) , equilibrium constant at total equilibrium pressure p_(1) is K_(p1)& for C(g)hArrD(g)+E(g) . equilibrium constant at total equilibrium pressure p_(2) is K_(p2) . If degree of dissciation of A & C are same, then the ratio K_(p1) //K_(p2) , if 2p_(1)=p_(2) , is :

In the dissociation of PCl_(5) as PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) If the degree of dissociation is alpha at equilibrium pressure P, then the equilibrium constant for the reaction is

Calculate the equilibrium constant K_(p) and K_(c ) for the reaction: CO(g)+1//2O_(2)(g) hArr CO_(2)(g) . Given that the partial pressure at equilibrium in a vessel at 3000 K are p_(CO)=0.4 atm, p_(CO_(2))=0.2 atm

A(s) harr M(s) + 1/2O_2(g) K_p of the reaction is 4. Find the partial pressure of O_2 at equilibrium

2AB_2(g)hArr 2AB(g)+B_2(g) Degree of dissociation of AB_2 is x. What will be equation for x in terms of K_p and equilibrium pressure P ?

The degree of dissociation 'alpha' of the reaction N_(2)O_(4)(g)hArr 2NO_(2)(g) Can be related of K_(p) as [Given : Total pressure at equilibrium = P]

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.

1 mol of A in 1 litre vessel maintained at constant T shows the equilibrium A(g) hArr B(g) +2C(g) " " K_(C_(1)) C(g) hArr 2D(g)+3B(g) " " K_(C_(2)) If the equilibrium pressure is 13/6 times of initial pressure and [C]_(eq)=4/9[A]_(eq) , Calculate K_(C_(1)) and K_(C_(2)) .

For the reaction, 2HI(g)toH_2(g)+I_2(g) What is the relationship between K_c and K_p at 25^@C ?