At room temperature, the following reaction proceeds nearly to completion:
`2NO+O_(2)to2NO_(2)toN_(2)O_(4)`
The dimer, `N_(2)O_(4)`, solidfies at `262 K`. A `250 mL` flask and a `100 mL` flask are separated by a stopcock. At `300 K`, the nitric oxide in the larger flask exerts a pressure of `1.053 atm` and the smaller one contains oxygen at `0.789 atm`. The gase are mixed by opening the stopcock and after the end of the reaction the flasks are cooled to `220 K`. Neglecting the vapour pressure of the dimer, find out the pressure and composition of the gas remaining at `220 K`. (Assume the gases to behave ideally)

Explanation

For `NO:V =250 mL, T =300K, P =1.53 atm`
`n_(NO) = (PV)/(RT) = (1.053 xx 250)/(0.0821 xx 300 xx 1000) = 1.069 xx 10^(-2)`
For `O_(2) V =100mL, T =300K, P = 0.789 atm`
`:. N_(O_(2)) = (0.789 xx 100)/(0.0821 xx 300 xx 1000) =0.32 xx 10^(-2)`
`{:(Now,2NO+,O_(2)rarr,N_(2)O_(4),),("Molebefore",1.069xx10^(-2),0.32xx10^(-2),0,),(reaction,,,,),("Moleafter",(1.069xx10^(-2)-2xx0.32xx10^(-2)),0,0.32xx10^(-2),),(reaction,=0.429xx10^(-2),,,):}`
Mole of NO left` = 4.29 xx 10^(-3)`
`P_(NO) xx V =nRT`
Given `T =220K, V =(250xx100)/(1000) = (3500)/(1000)` litre
`P_(NO) xx (350)/(1000) =4.29 xx 10^(-3) xx 0.0821 xx 220`
`:. P_(NO) "left" = 0.221` atm .