Question

At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.100 M and [NO]=0.500 M.

N₂(g) + O₂ (g) <----> 2NO(g)

If more NO is added, bringing its concentration to 0.800 M, what will the final concentration of NO be after equilibrium is re-established?

_____M

Answer 1

At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.100 M and [NO]=0.500 M.

N₂(g) + O₂ (g) <----> 2NO(g)

If more NO is added, bringing its concentration to 0.800 M, what will the final concentration of NO be after equilibrium is re-established?

First, calculate Kc in equilbirium

Kc = [NO]^2 / [N2][O2] = 0.5^2 /(0.1*0.1) = 25

then;

initially:

[NO] = 0.800

[N2] = 0.1

[O2] = 0.1

in equilibrium

[NO] = 0.800 - x

[N2] = 0.1 + x

[O2] = 0.1 +x

Now, this will be reversed:

2NO = N2 + O2 Knew = 1/K = 1/25 = 0.04

0.04 = (0.1 +x)^2 /(0.800 - x)^2

sqrt(0.04) = (0.1 +x) /(0.800 - x)

0.2*0.80 -0.2x = 0.1 + x

(1.2)x = (0.2*0.8)- 0.1

x = 0.06 / 1.2 = 0.05

so

[NO] = 0.800 - 0.05 = 0.75 M

[N2] = 0.1 + 0.05 = 0.15

[O2] = 0.1 +0.05 = 0.15

proof:

Kc = [NO]^2 /([N2][O2])

Kc = (0.75^2) / (0.15^2) = 25

this must be correct