For the equilibrium
`AB(g) hArr A(g) +B(g)`,
`K_(p)` is equal to four times the total pressure. Calculate the number of moles of `A` formed.

For the equilibrium AB(g) hArr A(g)+B(g) . K_(p) is equal to four times the total pressure. Calculate the number moles of A formed if one mol of AB is taken initially.

For the equilibrium AB(g) hArr A(g)+B(g) at a given temperature, the pressure at which one-third of AB is dissociated is numerically equal to

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

For the equilibrium reaction NH_4Cl hArr NH_3(g)+HCl(g), K_P=81 "atm"^2 .The total pressure at equilibrium is 'X' times the pressure of NH_3 .The value of X will be

For the following equilibrium N_(2)O_(4)(g)hArr 2NO_(2)(g) K_(p) is found to be equal to K_(c) . This is attained when :

A sample of mixture of A(g), B (g) and C(g) under equilibrium has mean molecular mass equal to 80. The equilibrium is : A(g)hArr B(g)+C(g) If initially 4 mole of 'A' gas is present then total number of mole at equilibrium is : [M_(A)=100,M_(B)=60,M_(C)=40]

For the reactions, CO(g) +Cl_2( g) hArr COCl_2(g), " the " (K_P)/(K_c) is equal to

For the reaction, AB(g)hArr A(g)+B(g), AB" is "33% dissociated at a total pressure of 'p' Therefore, 'p' is related to K_(p) by one of the following options

For the reaction, AB(g)hArr A(g)+B(g), AB" is "33% dissociated at a total pressure of 'p' Therefore, 'p' is related to K_(p) by one of the following options

For the reaction is equilibrium : 2NOBr_((g))hArr2NO_((g))+Br_(2(g)) If P_(Br_(2)) is (P)/(9) at equilibrium and P is total pressure, prove that (K_(p))/(P) is equal to (1)/(81) .