`H_(2)S`(gas) initially at a pressure of 10.0 atm and a temperature of 800K dissociates as `2H_(2)S`(g) `harr``2H_(2)`(g) + `S_(2)`(g) . At equilibrium, the partial pressure of `S_(2)` vapour is 0.020 atm Thus Kp is

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

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

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 ?

A rigid container holds an equal number of moles of N_(2) and H_(2) gas at a total pressure of 10.0 atm. The gases react according to the equation, ltBrgt N_(2)(g)+3H_(2)(g) hArr2NH_(3)(g) . If the total pressure of the gas decreases at a rate of 0.20atm.s^(-1) , what is the rate of change of the partial pressure of N_(2) in the container?

For NH_4HS(s)hArrNH_3(g)+H_2S(g) , if K_p = 64 atm^2 , equilibrium pressure of mixture is

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.