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`AB_(2)` dissociates as `AB_(2)(g) hArr AB(g)+B(g)`. If the initial pressure is `500` mm of Hg and the total pressure at equilibrium is `700` mm of Hg. Calculate `K_(p)` for the reaction. |
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Answer» Correct Answer - A::C After dissociation, suppose the decrease in the pressure of `AB_(2)` at equilibrium is `p` mm. `{:(,AB_(2)(g),hArr,AB(g),+,B(g)),("Initial pressure",500 mm,,0,,0),("Pressure at equilibrium" ,(500-p) mm,,p mm,,p mm):}` `:.` Total pressure at equilibrium `=500-p+p+p` `=500+p mm` `500+p=700` (Given) or `p=200 mm` Hence, at equilibrium, `p_(AB_(2))=500-200=300 mm` `p_(AB)=200 mm, p_(B)=200 mm` `:. K_(p)=(p_(AB)xxp_(B))/p_(AB_(2))=(200xx200)/300=133.3 mm` |
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