Calculate the equilibrium concentrations of N2O4 and NO2 at 25 ∘C in a vessel that contains an initial N2O4 concentration of 0.0655 M . The equilibrium constant Kc for the reaction N2O4(g)⇌2NO2(g) is 4.64×10−3 at 25 ∘C. Express your answers using four decimal places separated by a comma.

[N2O4] = 0.0573M

[NO2] = 0.0163M

Explanation:

The equilibrium of N2O4 is:

N2O4(g)⇌2NO2(g)

Where Kc is defined as:

Kc = 4.64x10⁻³ = [NO2]² / [N2O4]

When you add just N2O4, the reaction will occurs until  [NO2]² / [N2O4] = 4.64x10⁻³. Here, the system reaches equilibrium.

That means if 0.0655M N2O4 begin reaction, in equilibrium we will have:

[N2O4] = 0.0655M - X

[NO2] = 2X

Where X is defined as reaction coordinate

Replacing in Kc:

4.64x10⁻³ = [NO2]² / [N2O4]

4.64x10⁻³ = [2X]² / [0.0655-X]

3.0392x10⁻⁴ - 4.64x10⁻³X = 4X²

3.0392x10⁻⁴ - 4.64x10⁻³X - 4X² = 0

Solving for X:

X = -0.0093 → False solution. there is no negative concentrations

X = 0.008156M → Right solution.

Replacing X, equilibrium concentrations are:

[N2O4] = 0.0655M - X

[NO2] = 2X

the answer is 0.8m.

by: nicholas