Respuesta :
Answer: The value of [tex]K_{eq}[/tex] for this reaction is 1.578.
Explanation:
Given: Initial moles of [tex]N_{2}O_{4}[/tex] = 0.4 mol
Volume = 1.00 L
Therefore, initial concentration of [tex]N_{2}O_{4}[/tex] is calculated as follows.
[tex]Concentration = \frac{moles}{volume}\\= \frac{0.4}{1.0 L} mol\\= 0.4 M[/tex]
Now, ICE table for the given reaction equation is as follows.
[tex]N_{2}O_{4}(g) \rightleftharpoons 2NO_{2}[/tex]
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 [tex][NO_{2}][/tex] is calculated as follows.
2x = [tex][NO_{2}][/tex] = [tex]2 \times 0.3945 = 0.789[/tex]
Therefore, [tex]K_{eq}[/tex] for the given reaction is calculated as follows.
[tex]K_{eq} = \frac{[NO_{2}]^{2}}{[N_{2}O_{4}]}\\= \frac{(0.789)^{2}}{(0.3945)}\\= 1.578[/tex]
Thus, we can conclude that [tex]K_{eq}[/tex] for this reaction is 1.578.
