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.....0.26-2x
Kc = (NO2^2/(N2O4)
Substitute the new equilibrium line and solve for x, then evaluate NO2 and N2O4.
0.25+0.1 is not 0.26, but 0.35
To find the new equilibrium concentration of N2O4, we need to use the concept of the equilibrium constant and apply Le Chatelier's principle to determine how the system will react to the addition of NO2.
Let's first write the balanced equation for the reaction:
N2O4(g) ⇌ 2NO2(g)
The equilibrium constant expression for this reaction can be written as:
Kc = [NO2]^2 / [N2O4]
Given that the initial concentrations are [N2O4] = 0.55 M and [NO2] = 0.25 M, we can plug these values into the equilibrium constant expression to calculate the initial equilibrium constant (Kc_initial).
Kc_initial = (0.25 M)^2 / (0.55 M)
Kc_initial = 0.1136
Now, if we add 0.10 M NO2 to the reaction mixture, the concentration of NO2 will increase by 0.10 M. Therefore, the new concentration of NO2 will be:
[NO2] = 0.25 M + 0.10 M
[NO2] = 0.35 M
To determine the new equilibrium concentration of N2O4, we can rearrange the equilibrium constant expression and solve for [N2O4].
Kc = [NO2]^2 / [N2O4]
0.1136 = (0.35 M)^2 / [N2O4]
Rearranging the equation and solving for [N2O4]:
[N2O4] = (0.35 M)^2 / 0.1136
[N2O4] ≈ 1.078 M
Therefore, the new equilibrium concentration of N2O4 will be approximately 1.078 M.