An equilibrium mixture at 300 K contains `N _(2) O _(4) and NO _(2),` their partial pressures are 0.28 and 1.1 atmospheres respectively. If the volume of container is doubled, calculate the new equilibrium partial pressure of two gases.

Options

A. 0.095 atm, 0.64 tm
B. 0.12 atm, 0.86 atm
C. 0.06 atm, 0.47 atm
D. 0.18 atm, 0.63 atm

Answer

A

Explanation

`{:(N_(2)O_(4),hArr,2NO_(2)),(0.28,,11):}` pressure at equilibrium
`therefore K_(p) = ((P_(NO_(2))))/((P_(N_(2)O_(4)))) = ((11)^(2))/(0.28) = 4.32` atm. Now if the volume of container is doubled , i.e, pressure increase in mole are deomposition of `N_(2)O_(4)` is favoured.
`{:(NCO_(4),hArr,2NO_(2),),((0.28)/(2)P,,(11)/(2)2P,"New pressure at equm."):}`
Where reactant `N_(2)O` equivalent to pressure P is used in doing so
Again `K_(p) = ([(1.1)/(2)+2P]^(2))/([(0.28)/(2)-P]) = ([0.55 + 2P]^(2))/([0.14-P]) = 4.32, P = 0.45`
Now, `P_(N_(2)O_(4)) = 0.14 - 0.045 =0.095` atm , `P_(NO_(2))` at new eq `= 0.55 + 2 xx 0.64` atm