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An equimolar mixture of CO(2 (g)) and CF...

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 :

A

4

B

`(1)/(4)`

C

`(1)/(2)`

D

Can't be determined as total equilibrium pressure in not given.

Text Solution

Verified by Experts

The correct Answer is:
B
`{:(,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)`
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