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

0.088 M.

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:

By solving the above equation for x we have:

x = 0.024 = [NO₂]      

Hence the concentration of NO₂ is:

I hope it helps you!

it is a good conductor

explanation: