Some solid NH_(4) HS " is placed in a flask containing " NH_(3) " at " 0*5 " atm. What would be the pressure of " NH_(3) and H_(2)S " when equilibrium is reached " ? NH_(4) HS (s) hArr NH_(3) (g) + H_(2) S (g), K_(p) = 0*11

Some solid NH_(4) HS " is placed in a flask containing " NH_(3) " at "

1.	Some solid NH_(4) HS " is placed in a flask containing " NH_(3) " at " 05 " atm. What would be the pressure of " NH_(3) and H_(2)S " when equilibrium is reached " ? NH_(4) HS (s) hArr NH_(3) (g) + H_(2) S (g), K_(p) = 011
Answer» <P> Solution :`NH_(3) and H_(2)S ` PRODUCED by the decomposition of `NH_(4)HS`will be same . Suppose at equilibrium each has pressure= p atm due to decompistion of ` NH_(4)HS` . Then ` p_(H_(2)S) = p atm ` ` p_(NH_(3) ) p + 05" "(:' NH_(3) " is already presentat" 05atm )` Applying law of chemical equilibrium to the given reaction `p_(NH_(3)) xx p_(H_(2)S) = K_(p)` `:.p xx(p + 05)= 011` or ` p^(2) + 05 p = 011` or`p^(2) + 05 p - 011 = 0 ` ` :. p= -05pm sqrt((05)^(2)-4 (-011))/2 = (05 pm sqrt (025 + 044))/2= -(05 pm sqrt(069))/2` ` ( -05 pm 083 )/2 = (033)/2 = 0165 ("Neglecting-ve value ")` ` :. p_(H_(2)S) = 0* 165 "atm " , p_(NH_(3))= 0*665 " atm "`