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Question

In reduction
CH3COCH3(g)CH3CH3(g)+CO(g), if the initial pressure of CH3COCH3(g) is 150 mm and at equilibrium the mole fraction of CO(g) is 13 then the value Kp is

A
50 mm
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B
100 mm
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C
33.3 mm
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D
75 mm
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Solution

The correct option is D 75 mm
CH3COCH3(g)CH3CH3(g)+CO(g)
at t=0 150 0 0
at t=teq 150x x x
So total pressure at equilibrium = 150+x
According to the question,
x150+x=13x=75 mm
Hence total pressure (Po)=150+75 mm=225 mm
Now, Kp=(PCH3CH3)(PCO)PCH3COOH=13Po×13Po13Po
Kp=13×225 mm=75 mm

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