At room temperature `2NO+O_(2) to 2NO_(2) to N_(2)O_(4)` reaction proceeds near to completion. The dimer, `N_(2)O_(4)`, solidified at 262K. A 250 mL flask and a 100 mL flask are separated by a stop-cock. At 300K, nitric oxide in the longer flask exerts a presure of 1.053 atm and and the smaller one contains oxygen at 0789 atm. the gases are mixed by opening the stopcock and after the end off the reaction the flasks are cooled to 220K. Neglecting vapour pressure of the dimer, find out pressure and composition of the gas remaining at 220 K (assume that the gases behave ideally).

Explanation

Number of moles of NO gas is 250 mL flask,
`(PV)/(RT)=(1.053xx0.25)/(0.0821xx300)=0.01` mol
and that of `O_(2)` in 100mL flask,
`=(PV)/(RT)=(0.789xx0.1)/(0.0821xx300)=3.2xx10^(-3)` mol.
reaction: `2NO+O_(2) to 2NO_(2)`
Thus 1 mol of `O_(2)` gas completely reacts with 2 mol of NO gas.
therefore, `3.2xx10^(-3)` mol of `O_(2)` gas will react with `2xx3.2xx10^(-3)=6.4xx10^(-3)mol` of NO.
After the completion of reaction number of moles of NO left in the reaction system`=(0.01-6.4xx10^(-3))`
`=3.6xx10^(-3)`mol
Total volume of the reaction system`=(0.25+0.1)L=0.35L`
If the pressure of NO gas left in the reaction system be P, then `Pxx0.35=3.6xx10^(-3)xx0.0821x220`
`therefore P=0.185` atm
`therefore`After the reaction, the pressure of NO gas is 0.185 atm.