contestada

Consider the following reversible reaction.
[tex]H_2O(g) \longleftrightarrow 2H_2(g) + O_2(g)[/tex].
What is the equilibrium constant expression for the given system?
a. [tex]k_{eq} = \frac{H_2O}{{H_2}{O_2}}.[/tex]
b. [tex]k_{eq} = \frac{{H_2O}^2}{{H_2}^2{O_2}^2}.[/tex]
c. [tex]k_{eq} = \frac{{H_2}{O_2}}{H_2O}.[/tex]
d. [tex]k_{eq} = \frac{{H_2}^2{O_2}^2}{{H_2O}^2}.[/tex]

Respuesta :

anfabba15

Answer:

No one is correct. The correct expression is:

Keq = [H₂]²  . [O₂]² / [H₂O]²

Explanation:

To build the Keq expression in a chemical equilibrium you must consider the molar concentrations of reactants / products, and they must be elevated to the stoichiometric coefficient.

The balance reaction is:

2 H₂O (g)  ⇄  2 H₂ (g)  +  O₂ (g)

Keq = [H₂]²  . [O₂]  / [H₂O]²

In opposite side: 2 H₂ (g)  +  O₂ (g)   ⇄  2 H₂O (g)

Keq =  [H₂O]² / [H₂]²  . [O₂]  

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