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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)` isA. `50 mm`B. `100 mm`C. `33.3 mm`D. `75 mm` |
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Answer» Correct Answer - A `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` |
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