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Dinitrogen tetraoxide partially decomposes according to the following equilibrium:
N2O4(g) = 2NO2(g)
A 1.00 L flask is charged with 0.0400 mol of N2O4. At equilibrium at 373K, 0.0055 mol of N2O4 remains. Keq for this reaction is ____________.
A. 0.87
B. 13
C. 2.2

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Answer:

Keq for this reaction is 0.87 (option a.)

Explanation:

The equilibrium is this:

N₂O₄  ⇄ 2N₂O₂

1 mol of N₂O₄ decomposes to 2 moles N₂O₂

                    N₂O₄  ⇄ 2N₂O₂

Initially       0.04m           -

React               x             2x

Initially I have 0.04 moles of N₂O₄. Some amount (x) reacts to decompose and make 2 moles of N₂O₂; according to reaction, ratio is 1:2.

So if x amount of compound, reacted. I will have the double, as product decomposed.

                    N₂O₄  ⇄ 2N₂O₂

Initially       0.04m           -

React               x             2x

Eq.              0.055m       2x

We have moles of N₂O₄ in equilibrium, so we have to know how many amount has reacted.

(0.04 - 0.0055) = 0.0345

The double of this, is what we have in equilibrium of N₂O₂.

0.0345 .2 = 0.069

Let's make Kc:

[N₂O₂]² / [N₂O₄]

0.069² / 0.0055 = 0.87

Keq for this reaction is 0.87.

A. 0.87

Chemical reaction:

N₂O₄  ⇄ 2N₂O₂

1 mol of N₂O₄ decomposes to 2 moles N₂O₂

                  N₂O₄  ⇄   2N₂O₂

Initially       0.04m           -

React               x             2x

Initially I have 0.04 moles of N₂O₄. Some amount (x) reacts to decompose and make 2 moles of N₂O₂; according to reaction, ratio is 1:2.

So if x amount of compound, reacted. I will have the double, as product decomposed.

                   N₂O₄  ⇄  2N₂O₂

Initially       0.04m           -

React               x             2x

At Eq.          0.055m       2x

The moles of N₂O₄ in equilibrium, so we have to know how many amount has reacted.

(0.04 - 0.0055) = 0.0345

The double of this, is what we have in equilibrium of N₂O₂.

0.0345 * 2 = 0.069

Calculation for kc:

[tex]k_c=\frac{[N_2O_2]^2}{[N_2O_4]} \\\\k_c=\frac{0.069^2}{0.0055}\\\\k_c=0.87[/tex]

Thus, the value of kc = 0.87. Hence, correct option is A.

Find more information about equilibrium constant here:

brainly.com/question/12270624

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