Question

Consider the following equilibrium:

Answer 1

Answer:

$Keq=\frac{[O_{2}]^{3} }{[H_{2}O]^{2} }$

Explanation:

The equilibrium constant (Keq) is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. In Keq we include gases and aqueous species, but not solids nor pure liquids because their concentration remains almost constant over time.

Let's consider the following reaction.

4 KO₂(s) + 2 H₂O(g) ⇄ 4 KOH(s) + 3 O₂(g)

Then, the equilibrium constant is:

$Keq=\frac{[O_{2}]^{3} }{[H_{2}O]^{2} }$

Answer 2

Its the last one Keq=[O2]3[H2O]2