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.

Explanation

`{:(,N_2O_4(g) hArr, 2NO_2(g)),("Pressure at equilibrium",0.28,1.1):}`
`K_p=(p(NO_2)^2)/(p(N_2O_4))=(1.1)^2/(0.28)`=4.32 atm
If volume of the container is doubled , the pressure will be reduced to half.
`{:(,N_2O_4 hArr, 2NO_2),("New pressure", (0.28/2-p),(1.1/2+2p)):}`
`K_p=((1/1/2+2p)^2)/(0.28/2-p)`=4.32
Solving p=0.045
`therefore p(N_2O_4)` =0.14-0.045=0.095 atm
`p(NO_2)` =0.55+2(0.045)=0.64 atm