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 ........

For the equilibrium of the reaction NH_(4) Cl(s) hArr NH_(3) (g) + HCl

1.	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 ........
Answer» <P> Solution : `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)`