Question

Consider the following reversible reaction.


`H_2O(g) \longleftrightarrow 2H_2(g) + O_2(g)`.
What is the equilibrium constant expression for the given system?
a.
`k_(eq) = \frac{H_2O}{{H_2}{O_2}}.`
b.
`k_(eq) = \frac{{H_2O}^2}{{H_2}^2{O_2}^2}.`
c.
`k_(eq) = \frac{{H_2}{O_2}}{H_2O}.`
d.
`k_(eq) = \frac{{H_2}^2{O_2}^2}{{H_2O}^2}.`

Answer 1

No one is correct. The correct expression is:

Keq = [H₂]² . [O₂]² / [H₂O]²

Step-by-step 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]