Question

1.191 mol N2O3 is put into a 2.00 L flask at 25°C where it decomposes into NO2(g) and NO(g). What is the equilibrium constant (to 4 decimal places) if the reaction mixture contains 0.300 mol NO2 at equilibrium?

Answer 1

K = 0.0505

Step-by-step explanation:

Based on the equilibrium:

N2O3 ⇄ NO2 + NO

K, equilibrium constant, is defined as:

K = [NO2] [NO] / [N2O3]

Where [] are the equilibrium concentration of each species in the mixture.

The initial molarity of N2O3 is:

1.191mol / 2.00L = 0.5955M

In equilibrium, 0.5955M of N2O3 reacts producing X Molar of NO2 and X Molar of NO:

[N2O3] = 0.5955M - X

[NO2] = X

[NO] = X

As equilibrium concentration of NO2 is 0.300mol/2.00L = 0.15M; X = 0.15M:

[N2O3] = 0.5955M - 0.15M = 0.4455M

[NO2] = 0.15M

[NO] = 0.15M

And K is:

K = [0.15M] [0.15M] / [0.4455M]

K = 0.0505