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

Strategy: Use the initial equilibrium pressures to calculate the value of `K_p`. When the volume of the container is doubled, the pressure exerted by each gas is halved and the equilibrium shifts to the right (`Deltan_g` is `+ve`) until new equilibrium is established. Represent the new equilibrium pressures and substitute them into the `K_p` expression and solve for the new equilibrium pressures.
Solution:
`N_2O_4(g)=2NO_2(g)`
Equilibrium (atm) `0.28`
`K_p=(P_(NO_2)^2)/(P_(N_2O_4))=((1.1)^2)/(0.28)=(1.21)/(0.28)=4.32`
When the volume of the container is halved:
`{:(,N_(2)O_(4)(g)hArr2NO_(2)(g)),("New initial pressure (atm)"," 0.14 0.55"),("Change"," -x +2x"),("New equilibrium pressure (atm)",bar((0.14-x)(0.55+2x))):}`
Substitution of these values into `K_p` allows us to evaluate x
`K_p=4.32=((0.55+2x)^2)/((0.14-x))`
Solving the quardratic equation, we get
`x = 0.045`
`:. P_(N_2O_4)=0.14-x=0.14-0.045 = 0.095atm`
`P_(NO_2)=0.55+2x=0.55+2xx0.045`
`=0.64atm`