Consider the reaction below that is at equilibrium. 2 NO(g) + O2 (g) <—> 2NO2 (g)

If the concentration of NO2 is 7.7 M, the concentration of NO is 1.9, and the K value is 2.9, what would be the concentration of O2? Units of concentration will be in M.

Concentration of N₂ is 4.8.10^{9}10

9

M.

Explanation: K_{c}K

c

is a constant of equilibrium and it is dependent of the concentrations of the reactants and the products of a balanced reaction. For

N2(g) + 2 O2(g) ⇄ 2 NO2(g)

K_{c}K

c

= \frac{[NO2]^{2} }{[N2][O2]^{2} }

[N2][O2]

2

[NO2]

2

From the question concentration of NO2 is twice of O2:

[NO2] = 2[O2]

Substituting this into K_{c}K

c

:

K_{c}K

c

= \frac{[2O2]^{2} }{[N2][O2]^{2} }

[N2][O2]

2

[2O2]

2

8.3.10^{-10}10

−10

= \frac{4O2^{2} }{[N2].O2^{2} }

[N2].O2

2

4O2

2

[N2] = \frac{4O2^{2} }{8.3.10^{-10}.O2^{2} }

8.3.10

−10

.O2

2

4O2

2

[N2] = \frac{4}{8.3.10^{-10} }

8.3.10

−10

4

[N2] = 4.8.10^{9}10

9

The concentration of N2 in the equilibrium is [N2] = 4.8.10^{9}10

9

M.

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)

thats how its supposed to be