At 25°C, 64.0 grams of oxygen and 44.0 grams of dinitrogen oxide were placed in

Question

At 25°C, 64.0 grams of oxygen and 44.0 grams of dinitrogen oxide were placed in a 10.0 liter container where the following equilibrium was established.

At equilibrium, the NO2 concentration was 0.100 M.

a) Set up an equilibrium table with all the relevant information inside the table.

b) Write the equilibrium expresion.

c) What is the equilibrium concentration of O2?

d) What is the equilibrium concentration of N2O?

e) What is the value of Kc for this reaction?

f) Does this reaction favor the products or the reactants?

Explanation / Answer

moles of O2 = 64 / 16 = 4 mol

moles of N2O = 44 / 44 = 1 mol

a)

2 N2O(g) +    3 O2(g) <-------------->    4 NO2(g)

1                      4                                    0    ---------------> initial

-2x                   -3x                                          +4x      -----------> change

1-2x                    4-3x                                     4x   -------------> equilibrium

b)

equilibrium expression :

Kc = [NO2]^4 / [N2O]^2[O2]^3

equilibrium concentration of NO2 = 0.100 M

4x = 0.1

x = 0.025

c)

equilibrium concentration of O2 = 4 - 3x = 4 - 3 x 0.025

equilibrium concentration of O2 = 3.925 M

d)

equilibrium concentration of N2O = 1 - 2x 0.025

                                                    = 0.950 M

e)

Kc = [0.1]^4 / [0.950]^2[3.925]^3

Kc = 1.83 x 10^-6

f)

this is reactants favoured

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