`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)` ?

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 ?

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 :

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

For NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g) , the observed, pressure for reaction mixture in equilibrium is 1.12 atm at 106^(@)C . What is the value of K_(p) for the reaction?

For NH_(4)HS(s)hArr NH_(3)(g)+H_(2)S(g) If K_(p)=64atm^(2) , equilibrium pressure of mixture is

On decomposition of NH_(4)HS , the following equilibrium is estabilished: NH_(4) HS(s)hArrNH_(3)(g) + H_(2)S(g) If the total pressure is P atm, then the equilibrium constant K_(p) is equal to

NH_(4)HS(s)hArrNH_(3)(g)+H_(2)S(g) The3 equilibrium pressure at 25^(@)C is 0.660 atm . What is K_(p) for the reaction ?