Problem 65 - Introductory Chemistry Atoms First

Question

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) \(2 \mathrm{HF}(g) \rightleftarrows \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\) \(K_{\mathrm{eq}}=1 \times 10^{-95}\)

Step-by-Step Solution

Answer

Considering the equilibrium constants (K_eq), reaction (a) goes essentially to completion, as its K_eq value (3 x 10^81) is significantly larger than that of reaction (b). On the other hand, reaction (b) hardly proceeds or is strongly reactant-favored, given the very small K_eq value of 1 x 10^-95.

Step 1: Understand the question

The question asks: 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) \rightlefta

Step 2: Apply Chemistry knowledge

Using fundamental Chemistry concepts to analyze the problem.

Step 3: Formulate the answer

Considering the equilibrium constants (K_eq), reaction (a) goes essentially to completion, as its K_eq value (3 x 10^81) is significantly larger than that of reaction (b). On the other hand, reaction (b) hardly proceeds or is strongly reactant-favored, given the very small K_eq value of 1 x 10^-95.

Step 4: Conclude

The answer is: Considering the equilibrium constants (K_eq), reaction (a) goes essentially to completion, as its K_eq value (3 x 10^81) is significantly larger than that of reaction (b). On the other hand, reaction (b) hardly proceeds or is strongly reactant-favored, given the very small K_eq value of 1 x 10^-95.

Key Concepts

Equilibrium Constant (K_eq)

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.

Reaction 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.

Reactant-favored Reaction

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.