Question

Consider this equilibrium: 3Fe(s) + 4H2O(l) Fe3O4(s) + 4H2(g). What is the Keq expression for this equilibrium?

Answer 1

remember that in a Keq expression, you do products over reactants, and these can only be compounds in the gaseous or aqueous state.

for this equation

Keq= [H2]^4

Answer 2

Answer: The $K_{eq}$ for the given reaction is $K_{eq}=[H_2]^4$

Explanation:

The expression for the equilibrium constant is defined as the ratio of the concentrations of the products to the concentrations of the reactants each raised to power their stoichiometric ratios.

For the reaction:

$aA+bB\rightarrow cC+dD$

Expression for the equilibrium constant is:

$K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

For the reaction given in the equation:

$3Fe(s)+4H_2O(l)\rightarrow Fe_3O_4(s)+4H_2(g)$

The expression for the equilibrium constant is:

$K_{eq}=\frac{[H_2]^4[Fe_3O_4]}{[Fe]^3[H_2O]^4}$

The activity of solids and liquids are taken to be 1. Hence, the expression for the equilibrium constant is:

$K_{eq}=[H_2]^4$