During lecture, the following chemical equilibrium between nitrogen dioxide and

Question

During lecture, the following chemical equilibrium between nitrogen dioxide and dinitrogen tetroxide
was presented and discussed.
2 NO2(g)<=> N2O4(g) Kp = 7.5 at 300.0 K.
(a) If the total pressure in the container, which has a volume of 100.0 mL, is 0.85 atm, what are the
partial pressures (in atm) of NO2(g) and N2O4(g) in this container at 300.0 K?
(b) Using your answer to (b), what are the equilibrium concentrations (in M) of NO2(g) and N2O4(g) in
this container at 300.0 K?
(c) Use your answer to (c) to evaluate Kc for this equilibrium. which relates Kp and Kc for gas-phase equilibrium.
(d) At low temperatures, close examination of the closed reaction vessel containing essentially all
dinitrogen tetroxide shows small amounts of a blue-green liquid that comes from the reaction of
dinitrogen tetroxide with trace amounts of water leading to dinitrogen trioxide (blue) and nitric acid
(colorless). Write the balanced chemical equation for this reaction.
(e) With greater amounts of water, dinitrogen tetroxide hydrolyzes into nitrous acid and nitric acid.
Write the balanced chemical equation for this process.

Explanation / Answer

(a) Kp = 7.5 = [N2O4]/[NO2]^2

[N2O4] + [NO2] = 0.85

[N2O4] = 0.85 - [NO2]

7.5 = (0.85 - [NO2])/[NO2]^2

7.5[NO2]^2 + [NO2] - 0.85 = 0

Partial pressure of [NO2] = 0.276 atm

partial pressure of [N2O4] = 0.85 - 0.276 = 0.574 atm

(b) let x be the change in pressure at equilibrium,

7.5 = (0.574 + x)/(0.276 - 2x)^2

30x^2 - 9.28x - 0.0027 = 0

x = 0.31

So equilibrium concentrations of

[N2O4] = 0.574 + 0.31 = 0.884 atm = 0.884 x 0.1/0.08206 x 300 x 0.1 = 0.036 M

[NO2] = none

(c) Kc = Kp/(RT)^dn

dn = 1-2 = -1

Kc = 7.5/(0.08206 x 300)^-1 = 184.63

(d) Chemical reaction with water occuring would be,

3N2O4 + H2O ------> 2HNO3 + 2N2O3   

(e) With great amount of water,

N2O4 + H2O ---> HNO3 + HNO2

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