On the basis of \(K_{\text {eq }}\) values, which reaction goes essentially to completion? How would you describe the other reaction? (a) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \leftrightarrows 2 \mathrm{H}_{2} \mathrm{O}(g)\); \(K_{\mathrm{eq}}=3 \times 10^{81}\) (b) \(2 \mathrm{HF}(g) \rightleftarrows \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\) \(K_{\mathrm{eq}}=1 \times 10^{-95}\)

Understanding the equilibrium constant, denoted as \( K_{eq} \), is crucial in determining the extent of a chemical reaction. This constant is a numerical value derived from the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their respective stoichiometric coefficients in the balanced equation.

In the realm of chemistry, a larger \( K_{eq} \) value, which means the constant is much greater than 1, indicates a product-favored reaction; conversely, a \( K_{eq} \) value significantly less than 1 suggests a reaction that is reactant-favored. This numeric approach allows chemists to predict how far a reaction will proceed under given conditions without needing to perform the experiment physically.

When we talk about a reaction 'going to completion,' we generally mean that nearly all the reactants are transformed into products. This scenario is represented by an exceptionally high \( K_{eq} \), as seen with reaction (a) in the exercise where \( K_{eq} \) equals \( 3 \times 10^{81} \). This value is an indicator that the reaction strongly favors product formation and hence, proceeds practically to completion.

The term 'reaction completion' is used to describe a condition where the reactants are almost entirely converted into products, leaving a negligible amount of the initial substances. This does not necessarily mean that 100% conversion is achieved, but rather that for practical purposes, the reactants have been exhausted.

A reaction that goes to completion is one where the equilibrium lies so far to the right that the backward reaction is almost non-existent. For instance, in the exercise, reaction (a) with a \( K_{eq} \) of \( 3 \times 10^{81} \) signifies that the production of water from hydrogen and oxygen gases is essentially complete in the given conditions. In such cases, the reaction vessel would contain almost exclusively water vapor with minuscule amounts of hydrogen and oxygen gases.

On the flip side, a 'reactant-favored' reaction is characterized by a low \( K_{eq} \) value, suggesting that at equilibrium, the concentration of reactants is much higher than that of the products. These reactions do not proceed to a significant extent and are sometimes referred to as reactions 'lying to the left.'

For example, reaction (b) in our exercise, with \( K_{eq} \) being \( 1 \times 10^{-95} \), clearly indicates that the equilibrium heavily favors the reactants, hydrogen fluoride in this case. This kind of reaction might seem to hardly proceed at all because the products, hydrogen and fluorine gases, are found in extremely low concentrations at equilibrium. It is important to note that while the forward reaction does occur, the reverse reaction is predominant, essentially re-forming the reactants from the products.