In reduction CH3COCH3(g)⇌CH3CH3(g)+CO(g), if the initial pressure of CH3COCH3(g) is 150mm and at equilibrium the mole fraction of CO(g) is 13 then the value Kp is
A
50mm
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B
100mm
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C
33.3mm
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D
75mm
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Solution
The correct option is D75mm CH3COCH3(g)⇌CH3CH3(g)+CO(g) att=015000 att=teq150−xxx So total pressure at equilibrium = 150+x According to the question, x150+x=13⇒x=75mm Hence total pressure (Po)=150+75mm=225mm Now, Kp=(PCH3CH3)(PCO)PCH3COOH=13Po×13Po13Po ⇒Kp=13×225mm=75mm