Chemistry word problem

Question

Calculate the equilibrium concentrations of N2O4 and NO2 at 25 ∘C in a vessel that contains an initial N2O4 concentration of 0.0450 M . The equilibrium constant Kc for the reaction N2O4(g)⇌2NO2(g) is 4.64×10−3 at 25 ∘C.

Answer 1

N₂O₄(g)⇌2NO₂(g), so the law of mass action has it as 4.64×10⁻³ = [NO₂]²/[N₂O₄]. What are the units of the equilibrium constant? (M). This is an initial, change, and at equilibrium problem; solving for x in a quadratic.

________________________[N₂O₄]_______⇌_______2[NO₂]

Initial (M)_______________0.0450 _______________0

Change (M)_____________x_____________________2x

Equilibrium (M)_________0.0450-x_____________+2x

4.64×10⁻³ = [NO₂]²/[N₂O₄], so plug and chug.

4.64×10⁻³ = [+2x]²/[0.0450-x]

solve, 4x²+4.64×10⁻³x-(4.64×10⁻³)(0.0450) = 0

4x²+4.64×10⁻³x-2.088×10⁻⁴ = 0

The positive root is x = 0.0067

[N₂O₄] = [0.0450-x] = 0.0450-0.0067 = 0.0383 M

[NO₂] = [+2x] = 2* 0.0123 = 0.0246 M