Question

N2O4 ⇌ 2NO2

for the following reaction at 373 K, Kc = 0.36. If initial concentration of N2O4 is 0.1 mol dm^-3, what is the equilibrium concentration of NO2? (Is there a way to solve this without using quadratics?)

Answer 1

Okay, let's solve this step-by-step without using quadratics:

1) The equilibrium constant Kc = 0.36 means the equilibrium lies to the left. So there will be more N2O4 than NO2 at equilibrium.

2) The initial concentration of N2O4 is 0.1 mol dm^-3. Let's call this [N2O4]initial.

3) At equilibrium, the concentrations of N2O4 and NO2 will be [N2O4]equil and [NO2]equil respectively.

4) We know the equilibrium constant expression for this reaction is:

Kc = ([NO2]equil)^2 / [N2O4]equil

5) Setting this equal to 0.36 and plugging in 0.1 for [N2O4]initial, we get:

0.36 = ([NO2]equil)^2 / (0.1 - [NO2]equil)

6) Simplifying, we get:

0.036 = [NO2]equil^2

7) Taking the square root of both sides, we get:

[NO2]equil = 0.06 mol dm^-3

So the equilibrium concentration of NO2 is 0.06 mol dm^-3.

Let me know if you have any other questions! I can also provide a more step-by-step explanation if needed.