An equimolar mixture of CO_(2 (g)) and CF_(4 (g)) was taken in an empty flask at a particular temperature. These gases reacts as: CO_(2 (g)) + CF_(4 (g)) hArr 2 COF_(2 (g)) After this , mixture attains equilibrium and mole fraction of COF_(2 (g)) was found to be 0.2, then K_(P) for above reaction is :

An equimolar mixture of CO_(2 (g)) and CF_(4 (g)) was taken in an empt

1.	An equimolar mixture of CO_(2 (g)) and CF_(4 (g)) was taken in an empty flask at a particular temperature. These gases reacts as: CO_(2 (g)) + CF_(4 (g)) hArr 2 COF_(2 (g)) After this , mixture attains equilibrium and mole fraction of COF_(2 (g)) was found to be 0.2, then K_(P) for above reaction is :
Answer» 4 `(1)/(4)` `(1)/(2)` Can't be determined as total equilibrium PRESSURE in not given. Solution :`{:(,CO_(2 (g)),+,CF_(4 (g)),hArr,2 COF_(2 (g)),),("Moles initially",x,,x,,,),("Moles at",x - y,,x - y,,2y,),("equilibrium",,,,,,):}` `X_(COF_(2)) = (2y)/(2x - 2y + 2y) = (y)/(x) = 0.2` `y = 0.2 x` `n_(CO_(2)) = n_(CF_(4)) = x - 0.02x = 0.8 x` `n_(COF_(2)) = 0.4 x` Total Pressure not required as `(Delta N)_(g) = 0` `K_(P) = ((0.4x)^(2))/((0.8x) (0.8x)) = (0.4 xx 0.4)/(0.8 xx 0.8) = (1)/(4)`