Problem 76 - Chemistry An Atoms First Approach

Question

Write reactions for which the enthalpy change will be a. \(\Delta H_{\mathrm{f}}^{\circ}\) for solid aluminum oxide. b. the standard enthalpy of combustion of liquid ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\). c. the standard enthalpy of neutralization of sodium hydroxide solution by hydrochloric acid. d. \(\Delta H_{\mathrm{f}}^{\circ}\) for gaseous vinyl chloride, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}(g)\). e. the enthalpy of combustion of liquid benzene, \(C_{6} \mathrm{H}_{6}(l)\). f. the enthalpy of solution of solid ammonium bromide.

Step-by-Step Solution

Answer

a. \(4 \mathrm{Al}(s) + 3 \mathrm{O}_2(g) \rightarrow 2 \mathrm{Al}_2 \mathrm{O}_3(s)\) b. \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}(l) + 3 \mathrm{O}_2(g) \rightarrow 2 \mathrm{CO}_2(g) + 3 \mathrm{H}_2 \mathrm{O}(l) \) c. \(\mathrm{NaOH}(aq) + \mathrm{HCl}(aq) \rightarrow \mathrm{H}_2 \mathrm{O}(l) + \mathrm{NaCl}(aq) \) d. \(\mathrm{C}(s) + \frac{3}{2} \mathrm{H}_2(g) + \frac{1}{2} \mathrm{Cl}_2(g) \rightarrow \mathrm{C}_{2}\mathrm{H}_{3}\mathrm{Cl}(g) \) e. \(\mathrm{C}_6 \mathrm{H}_6(l) + \frac{15}{2} \mathrm{O}_2(g) \rightarrow 6 \mathrm{CO}_2(g) + 3 \mathrm{H}_2 \mathrm{O}(l) \) f. \(\mathrm{NH}_4\mathrm{Br}(s) \rightarrow \mathrm{NH}_4^+(aq) + \mathrm{Br}^-(aq) \)

Step 1: Write the balanced equation for the formation of aluminum oxide

Aluminum oxide (Al₂O₃) is formed when aluminum (Al) reacts with oxygen (O₂). To balance the equation, we need to have 2 aluminum atoms and 3 oxygen atoms on each side: \[4 \mathrm{Al}(s) + 3 \mathrm{O}_2(g) \rightarrow 2 \mathrm{Al}_2 \mathrm{O}_3(s)\] b. Standard enthalpy of combustion of liquid ethanol

Step 1: Write the balanced equation for the combustion of ethanol

In combustion reactions, the substance reacts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O). Ethanol (C₂H₅OH) combusts to form carbon dioxide and water: \[ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}(l) + 3 \mathrm{O}_2(g) \rightarrow 2 \mathrm{CO}_2(g) + 3 \mathrm{H}_2 \mathrm{O}(l) \] c. Standard enthalpy of neutralization of sodium hydroxide solution by hydrochloric acid

Step 1: Write the balanced equation for the neutralization of sodium hydroxide by hydrochloric acid

In a neutralization reaction, sodium hydroxide (NaOH) and hydrochloric acid (HCl) react to form water (H₂O) and sodium chloride (NaCl): \[ \mathrm{NaOH}(aq) + \mathrm{HCl}(aq) \rightarrow \mathrm{H}_2 \mathrm{O}(l) + \mathrm{NaCl}(aq) \] d. Formation of gaseous vinyl chloride

Step 1: Write the balanced equation for the formation of vinyl chloride

Vinyl chloride (C₂H₃Cl) is formed from the elements carbon (C), hydrogen (H₂), and chlorine (Cl₂): \[ \mathrm{C}(s) + \frac{3}{2} \mathrm{H}_2(g) + \frac{1}{2} \mathrm{Cl}_2(g) \rightarrow \mathrm{C}_{2}\mathrm{H}_{3}\mathrm{Cl}(g) \] e. Enthalpy of combustion of liquid benzene

Step 1: Write the balanced equation for the combustion of benzene

Benzene (C₆H₆) combusts to form carbon dioxide (CO₂) and water (H₂O): \[ \mathrm{C}_6 \mathrm{H}_6(l) + \frac{15}{2} \mathrm{O}_2(g) \rightarrow 6 \mathrm{CO}_2(g) + 3 \mathrm{H}_2 \mathrm{O}(l) \] f. Enthalpy of solution of solid ammonium bromide

Step 1: Write the balanced equation for the dissolution of ammonium bromide

Ammonium bromide (NH₄Br) dissolves in water (H₂O) to form ammonium ions (NH₄⁺) and bromide ions (Br⁻): \[ \mathrm{NH}_4\mathrm{Br}(s) \rightarrow \mathrm{NH}_4^+(aq) + \mathrm{Br}^-(aq) \]

Key Concepts

Enthalpy of Formation

Enthalpy of formation is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. The standard states are the most stable forms of the elements at 1 atm and a specified temperature, usually 25°C (298 K). When we consider the formation of aluminum oxide, the balanced chemical equation is:

• \[4 \mathrm{Al}(s) + 3 \mathrm{O}_2(g) \rightarrow 2 \mathrm{Al}_2 \mathrm{O}_3(s)\]

This equation is specifically written to show the formation of the compound from its elements. The standard state of aluminum is solid, and oxygen is diatomic gas, \(\mathrm{O}_2(g)\). Since the enthalpy of formation involves creating one mole of a substance, we often need to adjust the coefficients so that the product side of the equation reflects this standard condition properly. In this equation, we see the product is two moles of \(\mathrm{Al}_2 \mathrm{O}_3\); hence sometimes reporters might express enthalpy of formation per mole.

Enthalpy of Combustion

Enthalpy of combustion is the change in enthalpy when one mole of a substance completely reacts with oxygen under standard conditions. This reaction typically produces carbon dioxide \(\mathrm{CO}_2\) and water \(\mathrm{H}_2\mathrm{O}\). When examining the combustion of liquid ethanol, the balanced chemical reaction is:

• \[ \mathrm{C}_2\mathrm{H}_5\mathrm{OH}(l) + 3 \mathrm{O}_2(g) \rightarrow 2 \mathrm{CO}_2(g) + 3 \mathrm{H}_2 \mathrm{O}(l)\]

Ethanol combusts by reacting with oxygen to produce these products and releases heat energy. This is because combustion reactions fully oxidize the hydrocarbon, converting all carbon and hydrogen into their oxidized forms. The value of the enthalpy of combustion expresses the quantity of heat released when a mole of ethanol combusts, making it an important metric in fields like energy production and thermodynamics.

Enthalpy of Neutralization

Enthalpy of neutralization refers to the heat change that occurs when an acid and a base react to form water and a salt. For the reaction between sodium hydroxide \(\mathrm{NaOH}\) and hydrochloric acid \(\mathrm{HCl}\), the balanced equation is:

• \[ \mathrm{NaOH}(aq) + \mathrm{HCl}(aq) \rightarrow \mathrm{H}_2 \mathrm{O}(l) + \mathrm{NaCl}(aq)\]

During this reaction, the main event happening at the molecular level is the combination of hydrogen ions \(\mathrm{H}^+\) from the acid and hydroxide ions \(\mathrm{OH}^-\) from the base to form water. This process releases a certain amount of energy, which is characteristic for strong acids and bases. Typically, the enthalpy of neutralization for strong acid-base reactions is consistent, around \(-57\,\mathrm{kJ/mol}\), which indicates that this is an exothermic process as energy is released.

Enthalpy of Solution

Enthalpy of solution describes the heat change when one mole of a solute dissolves in a solvent, usually water, to form a solution. For ammonium bromide \( \mathrm{NH}_4\mathrm{Br}\), the dissolution can be represented by the equation:

• \[ \mathrm{NH}_4\mathrm{Br}(s) \rightarrow \mathrm{NH}_4^+(aq) + \mathrm{Br}^-(aq) \]

When the solid salt dissolves, it dissociates into its respective ions in the solution. This process may absorb or release energy depending on the nature of interactions broken and formed. If the energy required to break the lattice of the solid into individual ions is greater than the energy released upon solvation, the process will be endothermic. Conversely, if the solvation energy exceeds the lattice energy, the overall process will be exothermic. Understanding enthalpy of solution helps in predicting if more or less heat is needed or expelled and depends on the specific solute-solvent pair.