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When NO and NO(2) are mixed, the followi...

When `NO` and `NO_(2)` are mixed, the following equilibria are readily obtained,
`2NO_(2) hArr N_(2)O_(4), K_(p)=6.8 atm^(-1)`
`NO+NO_(2) hArr N_(2)O_(3)`
In an experiment when `NO` and `NO_(2)` are mixed in the ratio of `1:2`, the final total pressure was `5.05` atm and the partial pressure of `N_(2)O_(4)` was `1.7` atm. Calculate
a. the equilibrium partial pressure of `NO`.
b. `K_(p)` for `NO+NO_(2) hArr N_(2)O_(3)`.

Text Solution

Verified by Experts

For I equilibrium `2NO_(2) hArr N_(2)O_(4)`
`K_(p)=(P'_(N_(2)O_(4)))/((P'_(NO_(2)))^(2))=6.8`
`P'_(N_(2)O_(4))=1.7 "atm"`
`:.` By equation (i), `P'_(NO_(2))=0.5 "atm"`
The equilibria is maintained using `NO` and `NO_(2)` in the ratio `1:2`
`{:("For II equilibrium",NO,+,NO_(2),hArr,N_(2)O_(3)),("Initial pressure",P,,2P,,0),("Pressure at equilibrium",(P-x),,(2P-x-3.4),,x):}`
`:' 3.4 "atm"` of `NO_(2)` is used for I equilibrium to have `P'_(N_(2)O_(4))=1.7 "atm"`
`{:("At equilibrium",,(P-x),0.5,x):}`
`( :' P'_(NO_(2))` is same for both the equilibrium since both reactions are at equilibrium at a time)
Total pressure at equilibrium (Given `5.05` atm)
`=P'_(NO)+P'_(NO_(2))+P'_(N_(2)O_(5))+P'_(N_(2)O_(4))`
`=p-x+0.5+x+1.7`
`:. 5.05=P+2.20`
`:. P=5.05-2.20`
`:. p=2.85 "atm"`
`2P-x-3.4=0.5`
`2xx2.85-x-3.4=0.5`
`:. x=5.70-3.90=1.80 "atm"`
`:. P'_(NO)=2.85-1.80=1.05 "atm"`
Now `K_(p)` for `NO+NO_(2) hArr N_(2)O_(3)`
`K_(p)=(P'_(N_(2)O_(3)))/(P'_(NO)xxP'_(NO_(2)))=(1.80)/(1.05xx0.5)=3.43 "atm"^(-1)`
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