- Identify the physical state of each component: \(H_2O(g)\), \(C(s)\), \(H_2(g)\), and \(CO(g)\).
- Recall that pure solids \((s)\) and liquids \((l)\) are omitted from the equilibrium constant expression \(K_c\).
- Write the expression \(K_c = \frac{[H_2][CO]}{[H_2O]}\), which excludes the solid \(C\).
- Conclude that \([C]\) is the species that does not appear in the expression, corresponding to option \(\boxed{d}\).
Explanation
- Analyzing the Reaction and States of Matter
The chemical reaction provided is:
\[ H_2O(g) + C(s) \longleftrightarrow H_2(g) + CO(g) \]
To determine the equilibrium constant expression \(K_c\), we first identify the physical states of all reactants and products. In this reaction:
- \(H_2O\) is in the gaseous state \((g)\).
- \(C\) is in the solid state \((s)\).
- \(H_2\) is in the gaseous state \((g)\).
- \(CO\) is in the gaseous state \((g)\).
- Rules for Equilibrium Constant Expressions
In chemistry, the law of mass action states that for a reversible reaction at equilibrium, the equilibrium constant \(K_c\) is determined by the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants. However, there is a fundamental rule when dealing with heterogeneous equilibria: pure solids \((s)\) and pure liquids \((l)\) have constant concentrations and are therefore omitted from the equilibrium constant expression. Only gaseous species \((g)\) and species in aqueous solution \((aq)\) are included.
- Constructing the Expression and Identifying the Missing Species
Applying this rule, we exclude the solid carbon \((C)\) from the expression. The concentrations of the remaining gaseous species are included. Therefore, the equilibrium constant expression for the given reaction is:
\[ K_c = \frac{[H_2][CO]}{[H_2O]} \]
Comparing this to the provided options, we can see that \([H_2]\), \([CO]\), and \([H_2O]\) all appear in the expression, while \([C]\) does not.
- Conclusion
Based on the analysis, the species that is omitted from the equilibrium constant expression due to it being a pure solid is carbon \((C)\). Thus, the correct choice among the options is d. \([C]\).
Examples
In industrial chemical processes, such as the water-gas shift reaction, engineers and chemists use equilibrium constants to optimize product yields. By knowing which substances affect the equilibrium (gases and solutes) and which do not (pure solids), they can accurately model how changes in temperature or pressure will influence the production of gases like hydrogen \((H_2)\) and carbon monoxide \((CO)\). This distinction is vital for maintaining efficiency and safety in large-scale chemical reactors.