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For the equilibrium of the reaction NH(4...

For the equilibrium of the reaction `NH_(4) Cl(s) hArr NH_(3)` (g) + HCl (g) Kp = `81 atm^(2)`. Total pressure at equilibrium will be x times the pressure of `NH_(3)`. The value of x will be ........

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The correct Answer is:
2

`K_(P)=PNH_(3).PHCl=P^(2)NH_(3)` (`:. PNH_(3)=PHCl`)
`PNH_(3)=sqrt(K_(P))=9` atm = PHCl
Total pressure = `PHN_(3)+PHCl=18` atm
`=2 xx PNH_(3)`
`=PNH_(3)=9` atm = x
Total pressure of equilibrium
`=PNH_(3)+PHCl=9+9=18` atm
`=2 xx "partial pressure of" NH_(3)`
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