At room temperature the following reactions proceed nearly to completion : `2 NO + O_(2) to 2 NO_(2) to N_(2)O_(4)` . The dimer `N_(2) O_(4)` solidified 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 gases 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 of the gas remaining at 220 K. (Assume the gases to behave ideally)

Options

A. 0.682 atm
B. 0.383 atm
C. 0.221 atm
D. 0.536 atm

Answer

C

Explanation

To start with , we have
Amount of nitric oxide `= (pV)/(RT) = ((1.053 "atm") (0.250 L))/((0.0821 "atm" L K^(-1) mol^(-1)) (300 K)) = 0.01069` mol
Amount of oxygen = `(pV)/(RT) = ((0.789 "atm") (0.100 L))/((0.821 "atm" L K^(-1) mol^(-1)) (300 K)) = 3.023 xx 10^(-3)` mol
According to the reaction `2 NO + O_(2) to 2 NO_(3) to N_(2) O_(4)` , 1 mole of `O_(2)` reacts with 2 mol of NO.
Thus after the reaction , we will have , Amount of oxygen = 0
Amount of nitric acid oxide = `(0.01069 - 2 xx 0.0032)` mol `4.283 xx 10^(-2)` mol
Amount of `N_(2) O_(4)` formed `= 3.203 xx 10^(-3)` mol
Now on cooling `N_(2) O_(4) (g)` is condensed , Thus .only nitric oxide will be present in the entire volume of 0.350 L . Hence
Pressure of the gas = `(nRT)/(V) = ((4.283 xx 10^(-3) mol^(-1)) (0.0821 "atm" L K^(-1) mol^(-1)) (200 K))/((0.350 L)) = 0.221` atm