An equilibrium mixture at 300 K has `N_(2)O_(4)` and `NO_(2)` at 0.28 atm and 1.10 atm pressures. If the volume of container doubled at same temperature. Calculate the new equilibrium pressures for the gases.

Options

A. 0.095, 0.64 atm
B. 1.095, 2.64 atm
C. 1.250, 2.30 atm
D. 3.250, 1.50 atm

Answer

A

Explanation

`underset("0.28 atm")(N_(2)O_(4)) hArr underset("1.10 atm")(2NO_(2)) , K_(p) = (1.10^(2))/(0.28) =4.32 ` atm
On doubling the volume, pressure are halved and equilibrium favoured in forward direction.
`N_(2)O_(4) hArr 2NO_(2)`
`(0.14-p)" " ((1.10)/(2) +2p) = (0.55 +2p)`
`K_(p) = 4.32 = ((0.55 +2p)^(2))/(0.14-p)`
`:.p=(0.045`
`pN_(2)O_(4)=0.14 -0.045 =0.095` atm
`pNO_(2) =0.55 +2 xx 0.045 =0.64` atm.