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

Answer:

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:

$Kc = \frac{[NO_{2}]^{2}}{[N_{2}O_{4}]]}$

$0.490 = \frac{(0.04 + 2x)^{2}}{0.04 - x}$

$0.490*(0.04 - x) - (0.04 + 2x)^{2} = 0$

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!