In reaction:
`CH_(3)COCH_(3)(g)hArrCH_(3)CH_(3)(g)+CO(g),`
if the initial pressure of `CH_(3)COCH_(3)(g)` is `150 mm` and at equilibrium the mole fraction of `CO(g)` is `1/3`, then the value `K_(P)` is

Options

A. `50 mm`
B. `100 mm`
C. `33.3 mm`
D. `75 mm`

Answer

A

Explanation

`1` mol of `CH_(3)COCH_(3) =1` mol" of `CH_(3)CH_(3)`
`1` mol of `CO` = mol fraction of `CO=1/3`
Pressure of gas = mole fraction xx Initial pressure
`=1/3xx150=50 mm`
`:. K_(p)=((P_(CH_(3)CH_(3)))(P_(CO)))/((P_(CH_(3)COCH_(3))))=(50xx50)/50=50 mm`
Alternative method:
`CH_(3)COCH_(3)hArrCH_(3)CH_(3)+CO`
`{:("Initial",,1,0,0),("At equilibrium",,1-x,x,x):}`
Total "moles" `=1-x+x+x=1+x`
`:.` mole fraction of `CO=(x)/(1+x)=1/3 :. x=1/2`
`:.` mole of each gas `=1/2`
`:.` mole fraction of each gas `=(1/2)/(1/2+1/2+1/2)=1/3`
`:.` Pressure of each = mole fraction xx Initial pressure
`=1/3xx150=50 mm`
`:. K_(p)=((P_(CH_(3)CH_(3)))(P_(CO)))/((P_(CH_(3)COCH_(3))))=(50xx50)/50=50 mm`