Question

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Dinitrogen tetroxide partially decomposes according to the following equilibrium: N204(g) → 2NO2(g) A 1.000-L flask is charged with 2.60 x 10-2 mol of N204. At equilibrium, 2.01 x 10 2 mol of N204 remains. Keg for this reaction is O 0.361 O 1.53 x 10-4 O 0.615 6.94 x 10-3

Answer 1

Option D is correct.

EXPLANATION -

The decomposition reaction is given as

N204(9) = 2N02(9)

The equilibrium constant for this reaction will be given by

Keq = [NO2(9) [N204(9)]

Initially, there is 2.60 * 10^-2 mol N₂O₄ . No NO₂ is present at that time.

Now, 1 molecule of N₂O₄ produces 2 molecules of NO₂.

So 1 mol N₂O₄ produces 2 mol NO₂.

Let x mol of  N₂O₄ decomposes during the reaction. It will produce 2x mol of NO₂.

     

N2049 = 2N0 (9)

t=0 2.6 * 10^-2 mol 0 mol

At equilibrium   (2.6 * 10^-2 - x) mol 2x mol

According to the question, 2.01 * 10^-2 mol of N₂O₄ remains after the reaction.

So  (2.6 * 10^-2 - x) mol = 2.01 * 10^-2 mol

So x= 0.59 * 10^-2 mol

So number of moles of NO₂ formed = 2x

= 2 * 0.59 * 10^-2 mol

= 1.18 * 10^-2 mol

Molar concentration of N₂O₄ = (number of moles of N₂O₄ )/ (volume of the container)

2.01 x 10-2 mol 1.000 L

= 2.01 * 10^-2 M [1 M = 1 mol L^-1]

similarly, Molar concentration of NO₂  

1.18 x 10-2 mol 1.000 L

= 1.18 * 10^-2 M

So the equilibrium constant is

Keq = [NO2(9) [N204(9)]

Keq= [1.18 x 10-22 [2.01 x 10-21

= 6.94 * 10^-3

Option D is correct.