A 0.0300 mol sample of NO2(g) is placed in a rigid 1.00 L reaction vessel and allowed to reach equilibrium at a certain temperature. What is the value of Kc at this temperature if 0.00500 mol of N2O4(g) is present at equilibrium

12.5

Explanation:

Step 1: Write the balanced equation

2 NO₂(g) ⇄ N₂O₄(g)

Since the reaction takes place in a 1.00 L vessel, the initial concentration of NO₂ is 0.0300 M and the concentration at equilibrium of N₂O₄ is 0.00500 M.

Step 2: Make an ICE chart

        2 NO₂(g) ⇄ N₂O₄(g)

I          0.0300        0

C           -2x            +x

E     0.0300-2x        x

Step 3: Find the value of x

Since [N₂O₄]eq = 0.00500 M, x = 0.00500

Step 4: Calculate the concentrations at equilibrium

[NO₂] = 0.0300-2(0.00500) = 0.0200 M

[N₂O₄] = 0.00500 M

Step 5: Calculate the concentration equilibrium constant (Kc)

Kc = [N₂O₄]/[NO₂]²

Kc = 0.00500/0.0200²

Kc = 12.5

h let me think

c. 4.54 x 10–12 j happens to be correct