Question

Consider the reaction 3O2(g) 2O3(g) At 298 K, the equilibrium concentration of O2 is 1.6 x 10^-2 M, What is the equilibrium constant of the reaction at this temperature?

Answer 1

Answer: A. 2.0 x 10^-50

Explanation: That’s what I got on Edge.

Answer 2

Kc = 105062.5 at 298K

Step-by-step explanation:

First of all, we state the equilibrium:

3O₂(g) ⇄ 2O₃(g)

We know data about equilibrium concentration of oxygen. We suppose 1 mol of oxygen at the begining. During the reaction, x moles have reacted.

As ratio is 2:3, we can determine how many moles of ozone have been produced.

(x . 2)/3

So we have the final concentration of oxygen, so, let's find out x

1 - x = 0.016 moles

x = 0.984 moles

Then, the [O₃] in equilibrium will be (0.984 . 2) /3 = 0.656.

We supose a volume of 1 L, so we have molar concentration to determine Kc. Let's state the expression for it:

Kc = [O₃]² / [O₂]³

Kc = 0.656² / 0.016³ → 105062.5