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 :

Ammonium hydrogen sulphide dissociated according to the equation, NH_4HS(s) hArr NH_3(g) +H_2S(g) . If the observed pressure of the mixture is 1.12 atm at 106^@C , what is the equilibrium constant K_p of the reactions ?

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

NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g) In the above reaction, if the pressure at equilibrium and at 300K is 100atm then what will be equilibrium constant K_(p) ?

NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g) In the above reaction, if the pressure at equilibrium and at 300K is 100atm then what will be equilibrium constant K_(p) ?

NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g) In the above reaction, if the pressure at equilibrium and at 300K is 100atm then what will be equilibrium constant K_(p) ?

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

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 .

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 .