1.

Two equilibria are simultaneously existing in a vessel at 25^(@)C NO(g)+NO_(2)(g)hArrN_(2)O_(3)(g)K_(P_(1)) 2NO_(2)(g) hArrN_(2)O_(4)(g)K_(P_(2))=8 atm^(-1) Initially only NO(g) and NO_(2)(g) are present in 3:5 mole ratio. The total pressure at equilibrium is 5.5 atm and the partial pressure of NO_(2) at equilibrium is 0.5 atm The incorrect statement(s) regarding the above equilibria is/are

Answer»

`K_(P_(1))` for the equilibrium is `0.4 ATM^(-1)`
PARTIAL pressure of `N_(2)O_(4)` at equilibrium is `1.6 atm`
Partial pressure of `N_(2)O_(3)` at equilibrium is `2atm`
Partial pressure of `NO` at equilibrium is `2.5 atm`

Solution :`3P 5P 0`
`3P-X 5P-x-2y x`
`2NO_(2)(g)hArrN_(2)O_(4)(g)`
`5P-x-2y y`
`K_(P_(2))=(P_(N_(2)O_(4)))/((P_(NO_(2)))^(2)), 8=(P_(N_(2)O_(4)))/((0.5)^(2)`
`P_(N_(2)O_(4)=2` atm
`:.y=2atm`
`P_("TOTAL")=P_(NO)+P_(NO_(2))+P_(N_(2)O_(3))+P_(N_(2)O_(4)`
`5.5=3P-x+0.5+x+2`
`3P=3,P=1`
`5xx1-x-2xx2=0.5`atm
`K_(P_(1))=(P_(N_(2)O_(3)))/((P_(NO))(P_(NO_(2))))`
`=0.5/(2.5xx0.5)=0.4` atm
`P_(NO)=2.5 "atm" P_(NO_(2))=0.5 "atm" P_(N_(2)O_(3))=0.5"atm", P_(N_(2)O_(4))=2"atm"`


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