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

Options

A. `K_(P_(1))` for the equilibrium is `0.4 atm^(-1)`
B. Partial pressure of `N_(2)O_(4)` at equilibrium is `1.6 atm`
C. Partial pressure of `N_(2)O_(3)` at equilibrium is `2atm`
D. Partial pressure of `NO` at equilibrium is `2.5 atm`

Answer

B, C

Explanation

`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"`