At 25 degree C, 64.0 grams of oxygen and 44.0 grams of dinitrogen oxide were pla

Question

At 25 degree 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. N_2O(g) + O_2(g) NO_2(g) At equilibrium, the NO_2 concentration was 0.100 M. Set up an equilibrium table with all the relevant information inside the table Write the equilibrium expression. What is the equilibrium concentration of O_2? What is the equilibrium concentration of N_2O? What is the value of K_c for this reaction? Does this reaction favor the products or the reactants?

Explanation / Answer

Moles= mass/molar mass

Moles of O2= 64/32= 2M, Moles of N2O= 44/44=1M

Molarity= Moles/unit volume (L)

Molarity (M) : O2= 2/10=0.2, N2O= 1/10=0.1

Let x= drop in concentration of N2O to reach equilibrium.

The reaction is N2O(g) + 1.5O2(g) ----->2NO2

At equilibrium [N2O] =0.1-x, [O2] =0.2-1.5x and [NO2] =2x

Given [NO2]=2x=0.1, x=0.05

At equilibrium [N2O] =0.1-0.05=0.05M, [O2]=0.2-1.5*0.05 =0.125M, [NO2] =0.1M

KC= [NO2]2/ [N2O] [O2]1.5 = (0.1)2/ (0.05*(1.25)1.5= 0.143

Lower values of Kc favors the reactants. Higher value of KC favours the products. Here it is reactant favoured.

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