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?

Question

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)

DrBob222 · Answer

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.....0.26-2x

Kc = (NO2^2/(N2O4)
Substitute the new equilibrium line and solve for x, then evaluate NO2 and N2O4.

Meow Kapow · Answer

I was hoping DrBob222 could have been more vague and unclear in his response.

Dunno · Answer

0.25+0.1 is not 0.26, but 0.35