H_(2)S_((g))intially at a pressure of 10 atm anda temperature of 800K, dissociates as 2H_(2)S_((g)) hArr 2H_(2) +S_(2(g)) At equilibrium, the partial pressure of S, vapour is 0.02 atm. Thus, K_(p) is

H_(2)S_((g))intially at a pressure of 10 atm anda temperature of 800K,

1.	H_(2)S_((g))intially at a pressure of 10 atm anda temperature of 800K, dissociates as 2H_(2)S_((g)) hArr 2H_(2) +S_(2(g)) At equilibrium, the partial pressure of S, vapour is 0.02 atm. Thus, K_(p) is
Answer» `3.23 xx 10^(-7)` `6.45 xx 10^(-7)` `1.55 xx 10^(6)` `6.2 xx 10^(-7)` Solution : x=0.020 atm `PH_(2)S=10-2x=10-0.02 xx 2=9.96` atm `PH_(2)=x=0.02` `K_(P)=(PS_(2) xx PH_(2)^(2))/(P^(2)H_(2)S)=(0.02 xx (0.04)^(2))/((9.96)^(2))` `= 3.23 xx 10^(-7)` atm