Problem 67 - Chemistry The Molecular Science

Question

Complete each of these reactions by filling in the blanks. Predict whether each reaction is product-favored or reactant-favored, and explain your reasoning. (a) _________ \((\mathrm{aq})+\mathrm{HSO}_{4}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{HCN}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (b) \(\mathrm{H}_{2} \mathrm{~S}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\) ____________ (aq) (c) \(\mathrm{H}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{OH}^{-}(\mathrm{aq})+\) ______________ (g)

Step-by-Step Solution

Answer

(a) CN⁻, reactant-favored. (b) HS⁻, product-favored. (c) H₂, product-favored.

Step 1: Complete Reaction (a)

In reaction \((a)\), fill in the blank by determining which species would complete the acid-base reaction. The reaction shows an exchange involving \(\mathrm{HSO}_{4}^{-}\) and \(\mathrm{HCN}\). To balance this acid-base reaction, we recognize \(\mathrm{HCN}\) is a moderate acid and \(\mathrm{HSO}_{4}^{-}\) is a stronger acid due to its related acid \(\mathrm{H}_{2}\mathrm{SO}_{4}\). Therefore, the reactant should be \(\mathrm{CN}^{-}\) to form \(\mathrm{HCN}\). The completed reaction is: \(\mathrm{CN}^{-}( ext{aq}) + \mathrm{HSO}_{4}^{-}( ext{aq}) \rightleftharpoons \mathrm{HCN}( ext{aq}) + \mathrm{SO}_{4}^{2-}( ext{aq})\).

Step 2: Determine Direction for Reaction (a)

The equation formed in Step 1 could move in the forward or reverse direction. Comparing the acid strengths, \(\mathrm{HSO}_{4}^{-}\) is stronger than \(\mathrm{HCN}\), indicating the reaction tends to favor the side with weaker acids and bases to reach lower energy, more stable state. Thus, this reaction is reactant-favored.

Step 3: Complete Reaction (b)

In reaction \((b)\), \(\mathrm{H}_{2}S\) interacting with water would release \(\mathrm{H}^{+}\) and form \(\mathrm{HS}^{-}\). Thus, the blank is filled with \(\mathrm{HS}^{-}(\mathrm{aq})\). The completed reaction is: \(\mathrm{H}_{2}S(\text{aq}) + \mathrm{H}_{2}O(\text{l}) \rightleftharpoons \mathrm{H}_{3}O^{+} (\text{aq}) + \mathrm{HS}^{-}(\text{aq})\).

Step 4: Determine Direction for Reaction (b)

Consider the strengths: \(\mathrm{H}_{2}S\) is a weak acid and \(\mathrm{H}_{3}O^{+}\) is a stronger acid, suggesting that the reaction will favor the formation of weaker acids/bases as products. Therefore, this reaction is product-favored.

Step 5: Complete Reaction (c)

In reaction \((c)\), the hydride ion \(\mathrm{H}^{-}\) reacts with water, typically forming hydrogen gas and hydroxide ions. Hence, the blank will be \(\mathrm{H}_2(\text{g})\). The completed equation is: \(\mathrm{H}^{-}( ext{aq}) + \mathrm{H}_{2}O(\text{l}) \rightleftharpoons \mathrm{OH}^{-}( ext{aq}) + \mathrm{H}_{2}( ext{g})\).

Step 6: Determine Direction for Reaction (c)

\(\mathrm{H}^{-}\) is a strong base, much stronger than \(\mathrm{OH}^{-}\). Thus, the forward reaction forms a weaker base (\(\mathrm{OH}^{-}\)) which stabilizes the system, suggesting a product-favored reaction.

Key Concepts

Understanding Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that describes the state of a reaction where both the forward and reverse reactions occur at the same rate. This results in the concentrations of the reactants and products remaining constant over time. It’s essential to recognize that equilibrium doesn’t mean the reactants and products are equal in concentration, but rather that their rates of production are equal, achieving a balance.
A dynamic state is maintained in equilibrium reactions, meaning that the molecules are continuously reacting, but with no net change in concentration. The equilibrium position can lie towards the products or the reactants, depending on various factors like temperature, pressure, and the initial concentrations of the substances involved. This balance is typically represented by a double-headed arrow (↔) in the chemical equation, indicating the reversible nature of the reaction.
Factors affecting equilibrium include:

• Concentration of reactants or products
• Temperature
• Pressure (mainly for gases)
• Catalysts

Exploring Reaction Favorability

Reaction favorability refers to the tendency of a chemical reaction to proceed in one direction over the other. It is largely determined by the relative strengths of acids and bases involved in the reaction. In an acid-base reaction, the side of the reaction that is favored is often the one with the weaker acid and base, leading to a more stable molecular configuration.
For instance:

• In the first reaction \(\mathrm{HSO}_{4}^{-} + \mathrm{CN}^{-} \rightleftharpoons \mathrm{HCN} + \mathrm{SO}_{4}^{2-}\), reactant favorability is indicated since \mathrm{HSO}_{4}^{-}\ is a stronger acid compared to \mathrm{HCN}\.
• In contrast, the second reaction demonstrates product favorability as \mathrm{H}_2 ext{S}\, a weaker acid, transitions to form \mathrm{H}_3 ext{O}^{+}\, a stronger acid.

Understanding reaction favorability helps predict which direction a reaction will naturally proceed, allowing chemists to control and manipulate reactions efficiently.

Completing Chemical Equations

Completing chemical equations involves determining missing reactants or products to satisfy the stoichiometry of the reaction and the principle of conservation of mass. In acid-base reactions, this often means identifying the conjugate acid-base pairs and ensuring atom and charge balance.
Let’s look at the processes:

• **Identify the Species**: Determine which substances are missing in the equation. For example, understanding that in a hydride ion reaction \(\mathrm{H}^{-} + \mathrm{H}_2\text{O} \rightleftharpoons \mathrm{OH}^{-} + \mathrm{H}_2\), the missing product \mathrm{H}_2\ must be filled in.
• **Balance the Equation**: Ensure that the number of atoms on each side of the equation is equal. This adherence to the law of conservation of mass ensures that nothing is lost during the reaction process.
• **Analyze Acid-Base Strengths**: Use known pKa values or acid strength to predict reaction outcomes and adjustments.

Completing equations provides a comprehensive understanding of chemical processes and prepares students for more complex problem-solving in chemistry.