Answer:
[tex]K_{eq} = [\text{CO}_{2}]\text{[H$_\mathbf{2}$O]}[/tex]
Explanation:
[tex]\rm 2NaHCO_{3}(s) \, \rightleftharpoons \,Na_{2}CO_{3}(s) + CO_{2}(g) + H_{2}O(g)[/tex]
The general formula for an equilibrium constant expression is
[tex]K_{eq} = \dfrac{[\text{Products}]}{[\text{Reactants}]}[/tex]
Solids are not included in the equilibrium constant expression.
Thus, for this reaction,
[tex]K_{eq} = [\textbf{CO}_{\mathbf{2}}]\textbf{[H$_\mathbf{2}$O]}[/tex]
