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 atm`
B. `0.12 atm, 0.86 atm`
C. `0.06 atm 0.47 atm`
D. `0.18 atm, 0.63 atm 24`

Answer

A

Explanation

`underset(0.28)(N _(2) O _(4)) hArr underset(0.28) (2 NO _(2))`pressure at equilibrium
`therefore K _(P) = (( P _(NO _(2))))/((P _(N _(2 ) O _(4)))) = ((1.1) ^(2))/(0. 28) = 4.32` atm. Now if the volume of container is doubled, i.e., pressure decreases and will become half the reaction will proceed in the firection where the reactio shows an increase in mole i.e., decompostition of `N _(2)O_(4)` is favoured.
`N _(2) O _(4) hArr 2NO _(2)`
`(028)/(2) - P (11)/(2) + 2P` New pressure at equm.
Where reactant `N _(2) O _(4)` equivalent to pressure P is used up in domg so
Again `K _(P) = [((11)/(2) + 2P )/((0. 28)/(2) - P)] = ([0.55 + 2P]^(2))/([0.14 - P])= 4.32 , P = 0. 0 45 `
Now `P _(N_(2) O _(2))= 0. 14 - 0.045 = 0.095 atm, P _(NO _(2))` at new eq `= 0. 55 + 2 xx 0. 0 45 , P _(NO) = 0. 64 atm`