) Dinitrogen Tetroxide partially decomposes according to the following equilibrium: N2O4 (g) 2NO2 (g) A 1.00-L flask is charged with 0.400 mol of N2O4. At equilibrium at 373 K, 0.0055 mol of N2O4 remains. Keq for this reaction is .

The value of for this reaction is 1.578.

Explanation:

Given: Initial moles of = 0.4 mol

Volume = 1.00 L

Therefore, initial concentration of is calculated as follows.

Now, ICE table for the given reaction equation is as follows.

Initial:               0.4               0

Change:           -x                +2x

Equib:  0.4 - x = 0.0055       2x

Hence, the value of x is calculated as follows.

0.4 - x = 0.0055

x = 0.4 - 0.0055

= 0.3945

Now, the is calculated as follows.

2x = =

Therefore, for the given reaction is calculated as follows.

Thus, we can conclude that for this reaction is 1.578.

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