The dissociation equilibrium pressure of `NH_(4)HS_((s))` is `60 cm` at `27^(@)C`. The total pressure at equilibrium, when the `NH_(4)HS` is dissociated in pressence of `45 cm` of `NH_(3)` is :

For the reaction NH_(4)HS(s)rArrNH_(3)(g)+H_(2)S(g) ,K_(p)=0.09 . The total pressure at equilibrum is :-

The Kp of the reaction is NH_4HS(s)⇌NH_3(g)+H_2S( g) . If the total pressure at equilibrium is 42 atm.

The Kp of the reaction is NH_4HS(s)⇌NH_3(g)+H_2S( g) . If the total pressure at equilibrium is 30 atm.

The K_(p) of the reaction is NH_(4)HS_((s)) Leftrightarrow NH_(3(g))+H_(2)S_((g)) . If the total pressure at equilibrium is 30 atm.

For the following reaction NH_(4)HS_((s)) harr NH_(3(g))+H_(2)S_((g)) the total pressure at equilibrium is 30 atm. The value of K_(P) is

The equilibrium NH_(4)HS(s)hArr NH_(3)(g)+H_(2)S(g) , is followed to set - up at 127^(@)C in a closed vessel. The total pressure at equillibrium was 20 atm. The K_(C) for the reaction is

The equilibrium NH_(4)HS(s)hArr NH_(3)(g)+H_(2)S(g) , is followed to set - up at 127^(@)C in a closed vessel. The total pressure at equillibrium was 20 atm. The K_(C) for the reaction is

For the equilibrium of the reaction NH_(4) Cl(s) hArr NH_(3) (g) + HCl (g) Kp = 81 atm^(2) . Total pressure at equilibrium will be x times the pressure of NH_(3) . The value of x will be ........

At a given temperature, a closed vessel contains NH_3 gas and solid NH_4HS . The pressure of NH_3 gas in the vessel is 0.50 atm. On dissociation, NH_4HS produces NH_3 and H_2S gases. The total pressure in the flask at equilibrium is 0.84 atm. The equilibrium constant for the dissociation reaction (KP) of NH_4HS is-

Ammonium hydrogen sulphide dissociates according to the equation : NH_4 HS(s) hArr NH_3(g)+H_2S(g) The total pressure at equilibrium at 400 K is found to be 1 atm.The equilibrium constant K_p of the reaction is :