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N(2)O(3) is an unstable oxide of nitroge...

`N_(2)O_(3)` is an unstable oxide of nitrogen and it decomposes into NO (g) and `NO_(2)(g)`where `NO_(2)(g)` is further dimerise dimerise into `N_(2)O_(4)` as
`N_(2)O_(3)(g)hArrNO_(2)(g)+NO(g)" ",K_(p_(1)=2.5` bar
`2NO_(2)(g)hArrN_(2)O_(4)(g)" ": K_(P2)`
A flask is initially filled with pure `N_(2)O_(3)(g)` having pressure `2` bar and equilibria was established.
At equilibrium partial pressure of NO (g) was found to be `1.5` ber.
The equilibrium partial presure of `NO_(2)(g)` is:

A

a) `K_(P_(1))` for the equilibrium is `0.4 atm^(-1)`

B

b) Partial pressure of `N_(2)O_(4)` at equilibrium is `1.6 atm`

C

c) Partial pressure of `N_(2)O_(3)` at equilibrium is `2atm`

D

d) Partial pressure of `NO` at equilibrium is `2.5 atm`

Text Solution

Verified by Experts

The correct Answer is:
B, C
`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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