Question

Consider the reaction: N2(g)+O2(g)<=>2NO(g). Suppose a sample of air contains initially N2]=0.80M and [O2]=0.20M. Calculate the equilibrium concentration of all reactants and products at equilibrium if Kc= 1.0x10-5 and if Kc=1.20

Answer 1

a) If Kc = 1 x 10^-5

N2(g) + O2(g) <=> 2NO(g)

0.8...........0.2.................0

-X...........-X.................+2X

0.8-X......0.2-X.............2X

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

=> 1 x 10^-5 = (2X)^2 / (0.8 - X) (0.2 - X)

Since Kc << 1, we neglect X with respect to 0.2 and 0.8

=> 10^-5 = 4X^2 / (0.8) x (0.2)

=> X^2 = 4 x 10^-7

=> X = 6.325 x 10^-4 M

Therefore at equilibrium,

[N2] = 0.8 - X = 0.7994 M

[O2] = 0.2 - X = 0.1994 M

[NO] = 2X = 1.265 x 10^-3 M

b)

Kc = 1.2

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

=> 1.2 = (2X)^2 / (0.8 - X) (0.2 - X)

=> 0.192 - 1.2X + 1.2X^2 = 4X^2

=> 2.8X^2 + 1.2X - 0.192 = 0

Solving for X, we get

X = 0.124 M

Therefore at equilibrium,

[N2] = 0.8 - X = 0.676 M

[O2] = 0.2 - X = 0.076 M

[NO] = 2X = 0.248 M