\(\mathrm{H}_{2}(\mathrm{g})\) can be prepared by passing steam over hot iron: \(3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+4 \mathrm{H}_{2}(\mathrm{g})\) (a) Write an expression for the thermodynamic equilibrium constant for this reaction. (b) Explain why the partial pressure of \(\mathrm{H}_{2}(\mathrm{g})\) is independent of the amounts of \(\mathrm{Fe}(\mathrm{s})\) and \(\mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})\) present. (c) Can we conclude that the production of \(\mathrm{H}_{2}(\mathrm{g})\) from \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) could be accomplished regardless of the proportions of \(\mathrm{Fe}(\mathrm{s})\) and \(\mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})\) present? Explain.

The equilibrium constant for the reaction, K, is defined by the concentrations of the products divided by the concentrations of the reactants, each raised to the power of their stoichiometric coefficients. Therefore, given the balanced chemical reaction is: \(3 \, Fe(s)+4 \, H_{2}O(g) \leftrightharpoons Fe_{3} O_{4}(s)+4 \, H_{2}(g)\) For gases and aqueous solutions, '[ ]' represents concentration in mol/L, but for solids and liquids, it represents activity, and their activity is always equal to 1. The equilibrium expression for the reaction becomes: \( K=[H_{2}]^{4}=[H_{2}O]^{0}=[Fe]^{0}=[Fe_{3}O_{4}]^{0}\ K=[H_{2}]^{4} \) Note: The activities of the solid components \(Fe_{3}O_{4}(s)\) and \(Fe_{s}\) are taken as 1 and hence are not considered in the expression for K.

Partial pressures are directly related to the concentrations of the gases involved, so explaining why the partial pressure of \(H_{2}(g)\) is independent of the amounts of \(Fe(s)\) and \(Fe_{3}O_{4}(s)\) present requires understanding the role of these solid substances in the reaction. It's key to note that the amounts of solids neither enter into the equilibrium constant (as we derived above) nor affect the partial pressures of the gases. This is because the activity of a pure solid or liquid phase in a reaction is constant (equal to 1), and therefore independent of the amount present.

The question asks whether the production of \(H_{2}(g)\) from \(H_{2}O(g)\) could be accomplished regardless of the proportions of \(Fe(s)\) and \(Fe_{3}O_{4}(s)\) present. Based on our previous discussion, the amounts of \(Fe(s)\) and \(Fe_{3}O_{4}(s)\) present do not affect the equilibrium constant or the partial pressure of \(H_{2}(g)\). However, while these amounts may not impact the equilibrium condition, they can affect the rate at which equilibrium is reached. Even though the concentrations of the solids do not appear in the equilibrium expression, these solids are catalysts that speed up the reaction rate by offering a surface for the reaction to occur. Therefore, while the production of \(H_{2}(g)\) can theoretically occur regardless of the proportions of solids present, in practice these proportions can affect how quickly that production happens.