Chem 2
posted by Beth .
The following system is at equilibrium with [N2O4] = 0.55 M and [NO2] = 0.25 M. If an additional 0.10 M NO2 is added to the reaction mixture with no change in volume, what will be the new equilibrium concentration of N2O4?
N2O4(g) 2 NO2(g)

First Kc must determined.
.........N2O4 ==> 2NO2
Kc = (NO2)^2/(N2O4).
Substitute the equilibrium numbers and solve for Kc. Then set up an ICE cart for the remainder of the problem.
............NO2 ==> 2NO2
initial....0.55......0.25
add....................0.1
change......+x........2x
new equil.0.55+x.....0.262x
Kc = (NO2^2/(N2O4)
Substitute the new equilibrium line and solve for x, then evaluate NO2 and N2O4. 
I was hoping DrBob222 could have been more vague and unclear in his response.

0.25+0.1 is not 0.26, but 0.35