Question

N2O4 decomposes into NO2. At a certain temperature, the equilibrium pressures of NO2 and N2O4 are 1.6 bar and 0.58 bar, respectively. If the volume of the container is doubled at constant temperature, what would be the partial pressures of the gases when equilibrium is re-established

Answer 1

The partial pressure of NO₂ is 0.958 bar and the partial pressure of N₂O₄ is 0.211 bar

Step-by-step explanation:

The reaction is:

2NO₂ = N₂O₄

The equilibrium constant is:

`K_(p) =\frac{P_{N_(2)O_(4) } }{P_{NO_(2) }^(2) } =(0.58)/(1.6^(2) ) =0.227`

When the volume is double, the pressure will be halved, and will be:

PNO₂ = 0.8 bar

PN₂O₄ = 0.29 bar

The ICE table for this exercise is:

2NO₂ = N₂O₄

I......... 0.8..........0.29

C....... -2x.......... +x

E-----0.8-2x....... 0.29+x

The equilibrium constant is:

`K_(p) =\frac{P_{N_(2)O_(4) } }{P_{NO_(2) }^(2) } \\0.23=(0.29+x)/((0.8-2x)^(2) ) \\x=-0.079`

The partial pressure of NO₂ is:

`p_{NO_(2) } =0.8-2x=0.8-(2*(-0.079))=0.958bar`

The partial pressure of N₂O₄ is:

`p_{N_(2)O_(4) } 0.29+x=0.29-0.079=0.211bar`