chemistry
posted by Jim .
Consider the following equation:
N2O4(g)+ 2NO2(g) Kc= 5.8 x 10^3
If the initial concentration of N2O4(g)= 0.040 M and the initial concentration of NO2(g) is 0 M, what is the equilibrium of concentration of N2O4(g)?

Did you make a typo, typing + instead of >

Yes, I meant to right >...I apologize for typing the question incorrectly.

If so, then
N2O4 ==> 2NO2
Set up ICE chart.
initial:
N2O4 = 0.040
NO2 = O
change:
NO2 = 2x
N2O4 = x
equilibrium:
NO2 = 2x
N2O4 = 0.040x
Set up Keq expression of
Keq = (NO2)^2/(N2O4), plug in the equilibrium values from above and solve for x. Finally, 0.040x will be the N2O4 at equilibrium.