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)

Answer

( 0.221 atm )

Explanation

First we calculate partial pressure of NO and `O_(2)` in the combined system when no reaction takes place.
p V = constant `rArr p_(1)V_(1) = p_(2)V_(2)`
`rArr p_(2) (NO) = (1.053 xx 250)/(350) = 0.752` atm, `p_(2) (O_(2)) = (0.789 xx 100)/(350) = 0.225` atm
Now the reaction stoichiometry can be worked out using partial pressure because in a mixture. `p_(i) prop n_(1)`
`2NO + O_(2) rarr 2NO_(2) rarr N_(2)O_(4)`
`{:("Initial",0.752 atm,0.225 atm,0,0,),("Final",0.302,0,0,0.225 atm,):}`
Now, on cooling to 220 K, `N_(2)O_(4)` will solidify and only unreacted NO will be remaining in the flask.
`:'p prop T " " (p_(1))/(p_(2)) = (T_(1))/(T_(2)) rArr (0.302)/(p_(2)) = (300)/(220) rArr p_(2) (NO) = 0.221` atm