Question

If a reaction vessel at that temperature initially contains 0.0400 M NO2 and 0.0400 M N2O4, what is the concentration of NO2 at equilibrium

Answer 1

0.088 M.

Step-by-step explanation:

Assuming that the temperature is 429 K, we can calculate the concentration of NO₂. Kc(429K) = 0.490.

The reaction is:

N₂O₄(g) ⇄ 2NO₂(g)

0.04 0.04

0.04 - x 0.04 + 2x

The constant of the reaction is:

`Kc = ([NO_(2)]^(2))/([N_(2)O_(4)]])`

`0.490 = ((0.04 + 2x)^(2))/(0.04 - x)`

`0.490*(0.04 - x) - (0.04 + 2x)^(2) = 0`

By solving the above equation for x we have:

x = 0.024 = [NO₂]

Hence the concentration of NO₂ is:

`[NO_(2)] = 0.04 + 2*0.024 = 0.088 M`

I hope it helps you!