Question

At 100 oC, Kc=0.36. If a 1.00-L flask initially contains 0.100 M N2O4, what will be the equilibrium concentration of NO2? N2O4(g)⇌2NO2(g)

Answer 1

`[NO_2]=0.12M`

Step-by-step explanation:

Hello,

In this case, for the undergoing chemical reaction at equilibrium, the law of mass action for the equilibrium statement turns out:

`Kc=([NO_2]^2)/([N_2O_4])`

Next, by introducing the reaction extent
`x` (ICE methodology) we can write:

`Kc=((2x)^2)/([N_2O_4]_0-x)`

Whereas the equilibrium constant is 0.36 and the initial concentration of dinitrogen tetroxide is 0.100 M:

`0.36=((2x)^2)/(0.100M-x)`

Therefore, solving such quadratic equation we obtain two values of
`x`:

`x_1=-0.15M\\x_2=0.06M`

Clearly, the solution is 0.06M since the reaction extent must not be negative, thereby, the equilibrium concentration nitrogen dioxide turns out:

`[NO_2]=2*0.06M`

`[NO_2]=0.12M`

Best regards.