Question

1.575 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

the reaction

N2O3 = NO2 + NO

initially:

[N2O3] = mol/V = 1.575/2 = 0.7875 M

[NO2] = 0

[NO] = 0

in equilibrium

[N2O3] = 0.7875 - x

[NO2] = 0 + x

[NO] = 0 + x

and we know that

[NO2] = 0 + x = (0.3/2)

x = 0.15

substitute

[N2O3] = 0.7875 - 0.15 = 0.6375

[NO2] = 0 + 0.15

[NO] = 0 + 0.15

get Keq:

First, let us define the equilibrium constant for any species:

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

Kq = [NO2][NO2] / [N2O3]

K=(0.15*0.15)/0.6375

K = 0.0352