2Al(S)+3H2O(g)---->Al2O3(s)+6H2(g)
A.) what is the correct equilibrium expression?
Keq = (H2)^6/(H2O)^3
The equilibrium expression for the given chemical equation is written using the concentrations of products and reactants at equilibrium. In this case, the equation shows a reaction between aluminum (Al) with water (H2O) to form aluminum oxide (Al2O3) and hydrogen gas (H2).
To write the equilibrium expression, we consider the concentrations of the products (Al2O3 and H2) raised to their respective stoichiometric coefficients, divided by the concentrations of the reactants (Al and H2O) raised to their stoichiometric coefficients.
Based on the balanced chemical equation:
2Al(S) + 3H2O(g) → Al2O3(S) + 6H2(g)
The equilibrium expression can be written as:
K = [Al2O3]^a * [H2]^b / ([Al]^c * [H2O]^d)
Where a, b, c, and d represent the stoichiometric coefficients of Al2O3, H2, Al, and H2O, respectively.
By substituting the coefficient values into the expression, we get:
K = [Al2O3] * [H2]^6 / ([Al]^2 * [H2O]^3)
Therefore, the correct equilibrium expression for the given chemical equation is:
K = [Al2O3] * [H2]^6 / ([Al]^2 * [H2O]^3)