An equilibrium mixture at `300 K` contains `N_(2)O_(4)` and `NO_(2)` at `0.28` and `1.1 atm`, respectively. If the volume of container is doubles, calculate the new equilibrium pressure of two gases.

Options

A. 0.064 atm and 0.095 atm
B. 0.64 atm and 0.095 atm
C. 0.095 atm and 0.632 atm
D. 0.095 atm and 0.64 atm

Answer

D

Explanation

`underset(0.28)(N_(2)O_(4)(g)) hArr underset(1.1)(2NO_(2)(g))`
Pressure at equilibrium `K_(p)= (p_(NO_(2))^(2))/(p_(N_(2)O_(4))) = ((1.1)^(2))/(0.28)=4.32` atm
If volume of the container is doubled, the pressure will reduced to half `N_(2)O_(4) hArr 2N_(2)O`
New Pressure, `[(0.028)/(2)-p] [(1.1)/(2) + 2p]`
`K_(P)= ([(1.1)/(2)+ 2p]^(2))/([(0.28)/(2)-p])=4.32`
On solving, we get p= 0.045
`therefore p_(N_(2)O_(4))= 0.14- 0.045 = 0.095` atm
`p_(NO_(2)) = 0.55 + (2 xx 0.045)= 0.64` atm