Question

For the following reaction of N2O4, the equilibrium constant is 0.593 at a particular temperature.

N2O4(g) ⇌ 2 NO2(g)
If the initial concentration of N2O4 is 0.880M, what are the equilibrium concentrations?

Please show work!

Answer 1

"0.583" is the appropriate answer.

Step-by-step explanation:

Let,

The initial constant of
`N_2O_4` be "C".

Amount of
`N_2O_4` dissociated into
`NO_2` be "x".

now,

`N_2O_4 \rightleftharpoons 2NO_2`

Initial constant C -

Equilibrium constant C 2x

The Kc is given as:

⇒
`K_c = ([NO_2]^2)/([N_2O_4])`

`=((2x)^2)/(C-x)`

`0.593=(4x^2)/(0.88-x)`

`4x^2=0.593(0.88-x)`

`4x^2=0.512-0.593\ x`

`x=0.291`

hence,

The constant of
`NO_2` will be:

=
`2x`

=
`0.583`