Question

a) Starting with only NO(g), what will be the predominate specie(s) present at equilibrium? N2(g) + O2(g) ↔ 2NO(g) Kc = 4.6 x 10-31. b) If [N2] = [O2] = 8.6 x 10-2M and [NO] = 5.0 x 10-20, is the system in equilibrium? If not, what which way will it go to reach equilibrium? c) If, at equilibrium, [N2] = [O2] = 8.6 x 10-2M, what is the equilibrium concentration of NO?

Answers: a). N2 and O2

b). no, right shift

c). 5.8x10^-17 M please explain answers

Answer 1

a)

N2 + 02 ---> 2NO

the equilibrium constant is given by

Kc = [NO]^2 / [N2] [02] = 4.6 x 10-31

we can see that

Kc <<< O

so

reactants are the predominate species

that is

N2 and O2 are the predominate species

b)

consider the given reaction

N2 + 02 ---> 2 NO

the reaction quotient is given by

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

Q = [5 x 10-20]^2 / [8.6 x 10-2 ] 8.6 x 10-2]

Q = 3.38 x 10-37

we know that

if

Q = Kc , the reaction is at equilibrium

Q < Kc , the reaction will move towards products to reach equilibrium

Q > Kc , the reaction will move towars reactants to reach equilibrium

So

in this case

Q < Kc , so the reaction will move towards products ( right side ) to reach equilibrium

c)

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

so

4.6 x 10-31 = [NO]^2 / [8.6 x 10-2 ] [ 8.6 x 10-2]

[NO] = 5.833 x 10-17

so

the equilibrium concentration of NO is 5.833 x 10-17 M