In the reaction of: 2O4 (g) ⇌ 2O2(g) If the equilibrium constant is Kc = 0.513 and the initial concentration of 2O4 = 0.0500 . What are the equilibrium concentrations of 2O4 and O2 under these conditions?

[N₂O₄] = 0.00411 M

[NO₂] = 0.0918 M

Explanation:

In the reaction of: N₂O₄(g) ⇌ 2 NO₂(g) If the equilibrium constant is Kc = 0.513 and the initial concentration of N₂O₄ = 0.0500 M. What are the equilibrium concentrations of N₂O₄ and NO₂ under these conditions?

Step 1: Given data

Initial concentration of N₂O₄: 0.0500 M

Equilibrium constant (Kc): 0.513

Step 2: Make an ICE chart

        N₂O₄(g) ⇌ 2 NO₂(g)

I        0.0500           0

C           -x              +2x

E      0.0500-x        2x

Step 3: Find the value of "x"

The equilibrium constant is:

Kc = [NO₂]² / [N₂O₄]

0.513 = (2x)² / (0.0500-x)

4x² + 0.513 x -0.02565

We solve for x and we get x₁ = 0.0458942 and x₂ = −0.558894 (neglected).

Step 4: Calculate the concentrations at equilibrium

[N₂O₄] = 0.0500-x = 0.00411 M

[NO₂] = 2x = 0.0918 M

answer: mercury

explanation:

since electrons are negative the answer is neagtaive