Question

) 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 __________.

Answer 1

Answer: The value of
`K_(eq)` for this reaction is 1.578.

Step-by-step explanation:

Given: Initial moles of
`N_(2)O_(4)` = 0.4 mol

Volume = 1.00 L

Therefore, initial concentration of
`N_(2)O_(4)` is calculated as follows.

`Concentration = (moles)/(volume)\\= (0.4)/(1.0 L) mol\\= 0.4 M`

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

`N_(2)O_(4)(g) \rightleftharpoons 2NO_(2)`

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
`[NO_(2)]` is calculated as follows.

2x =
`[NO_(2)]` =
`2 * 0.3945 = 0.789`

Therefore,
`K_(eq)` for the given reaction is calculated as follows.

`K_(eq) = ([NO_(2)]^(2))/([N_(2)O_(4)])\\= ((0.789)^(2))/((0.3945))\\= 1.578`

Thus, we can conclude that
`K_(eq)` for this reaction is 1.578.