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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, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}(\mathrm{g})\). e. the enthalpy of combustion of liquid benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\). f. the enthalpy of solution of solid ammonium bromide.

Short Answer

Expert verified
a. \(4 Al(s) + 3 O_2(g) \rightarrow 2 Al_2O_3(s)\) b. \(C_2H_5OH(l) + 3 O_2(g) \rightarrow 2 CO_2(g) + 3 H_2O(l)\) c. \(NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)\) d. \(2 C(s) + 3 H_2(g) + Cl_2(g) \rightarrow C_2H_3Cl(g)\) e. \(C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6 CO_2(g) + 3 H_2O(l)\) f. \(NH_4Br(s) \rightarrow NH_4^+(aq) + Br^-(aq)\)

Step by step solution

01

Reaction

The formation of solid aluminum oxide from its elements, aluminum and oxygen, can be represented by the following balanced equation: \(4 Al(s) + 3 O_2(g) \rightarrow 2 Al_2O_3(s)\) b. Enthalpy of Combustion of Ethanol:
02

Reaction

The combustion of liquid ethanol, in the presence of oxygen gas, results in the formation of carbon dioxide and water. The balanced equation is: \(C_2H_5OH(l) + 3 O_2(g) \rightarrow 2 CO_2(g) + 3 H_2O(l)\) c. Enthalpy of Neutralization of Sodium Hydroxide and Hydrochloric Acid:
03

Reaction

The neutralization reaction between sodium hydroxide and hydrochloric acid forms salt (sodium chloride) and water. The balanced equation is: \(NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)\) d. Formation Enthalpy of Vinyl Chloride:
04

Reaction

The formation of gaseous vinyl chloride from its elements, carbon, hydrogen, and chlorine, can be represented by the following balanced equation: \(2 C(s) + 3 H_2(g) + Cl_2(g) \rightarrow C_2H_3Cl(g)\) e. Enthalpy of Combustion of Benzene:
05

Reaction

The combustion of liquid benzene in the presence of oxygen gas results in the formation of carbon dioxide and water. The balanced equation for this reaction is: \(C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6 CO_2(g) + 3 H_2O(l)\) f. Enthalpy of Solution of Ammonium Bromide:
06

Reaction

The dissolution of solid ammonium bromide in water can be represented by the following equation: \(NH_4Br(s) \rightarrow NH_4^+(aq) + Br^-(aq)\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy of Combustion
The enthalpy of combustion refers to the heat change that occurs when one mole of a substance completely reacts with oxygen. This process often releases heat, making it an exothermic reaction. It's crucial in understanding how different fuels work, whether you're thinking about burning gasoline or lighting a candle.

In a typical combustion reaction, the reactants, usually a hydrocarbon and oxygen, form carbon dioxide and water. For instance, when ethanol combusts, it reacts with oxygen to produce carbon dioxide and water according to this balanced equation:

\[C_2H_5OH(l) + 3 O_2(g) \rightarrow 2 CO_2(g) + 3 H_2O(l)\]

This kind of information is particularly useful in both practical applications, like designing engines, and theoretical chemistry, where understanding energy changes is key.
  • Combustion reactions are typically exothermic, releasing energy as heat.
  • The enthalpy change is often measured in kilojoules per mole (kJ/mol).
Enthalpy of Formation
The enthalpy of formation, often denoted by \(\Delta H_{\rm f}^{\circ}\), is the heat change when one mole of a compound is formed from its elements in their standard states. This is a cornerstone concept in thermochemistry because it helps chemists predict the stability of compounds.

For example, the formation of solid aluminum oxide from aluminum and oxygen is a formation reaction, represented by:

\[4 Al(s) + 3 O_2(g) \rightarrow 2 Al_2O_3(s)\]

For another example, think about the formation of gaseous vinyl chloride, \(C_2H_3Cl(g)\), from carbon, hydrogen, and chlorine. It can be represented as:

\[2 C(s) + 3 H_2(g) + Cl_2(g) \rightarrow C_2H_3Cl(g)\]

Understanding the enthalpy of formation helps predict reaction behavior and is key in calculating other enthalpic values using Hess's Law. Insights drawn from these values are critical to industries and academic research alike.
Enthalpy of Neutralization
When an acid and a base react, they form water and a salt, releasing heat in the process. This is known as the enthalpy of neutralization. It is always exothermic, meaning it releases heat into the surroundings. Neutralization reactions are vital in both laboratory settings and many industrial processes.

Let's take a look at a common example involving sodium hydroxide and hydrochloric acid:

\[NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)\]

During this reaction, the hydroxide ions \((OH^-)\) from the sodium hydroxide combine with the hydrogen ions \((H^+)\) from the hydrochloric acid to form water \((H_2O)\). The energy change in neutralization reactions can help determine the heat of reaction and is often measured to create precise thermal models.
Enthalpy of Solution
The enthalpy of solution refers to the heat change that occurs when a solute dissolves in a solvent. This can either be endothermic, where it absorbs heat, or exothermic, where it releases heat. This concept is particularly important in predicting temperature changes and solution behaviors.

An example is the dissolution of solid ammonium bromide in water:

\[NH_4Br(s) \rightarrow NH_4^+(aq) + Br^-(aq)\]

Here, the enthalpy change reflects the energy required to break the ionic bonds in ammonium bromide and the energy released when water molecules surround the newly formed ions. Understanding this process is significant for fields like pharmaceuticals, where dissolution rates directly impact drug activity, or in designing energy-efficient industrial processes.
  • Enthalpy of solution can help determine solubility levels.
  • It also aids in predicting how temperature affects solubility.

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Most popular questions from this chapter

The enthalpy change for a reaction is a state function and it is an extensive property. Explain.

A biology experiment requires the preparation of a water bath at \(37.0^{\circ} \mathrm{C}\) (body temperature). The temperature of the cold tap water is \(22.0^{\circ} \mathrm{C}\), and the temperature of the hot tap water is \(55.0^{\circ} \mathrm{C}\). If a student starts with \(90.0 \mathrm{~g}\) cold water, what mass of hot water must be added to reach \(37.0^{\circ} \mathrm{C}\) ?

On Easter Sunday, April 3,1983, nitric acid spilled from a tank car near downtown Denver, Colorado. The spill was neutralized with sodium carbonate: \(2 \mathrm{HNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(s) \longrightarrow 2 \mathrm{NaNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) a. Calculate \(\Delta H^{\circ}\) for this reaction. Approximately \(2.0 \times\) \(10^{4}\) gal nitric acid was spilled. Assume that the acid was an aqueous solution containing \(70.0 \% \mathrm{HNO}_{3}\) by mass with a density of \(1.42 \mathrm{~g} / \mathrm{cm}^{3} .\) What mass of sodium carbonate was required for complete neutralization of the spill, and what quantity of heat was evolved? ( \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\left.\mathrm{NaNO}_{3}(a q)=-467 \mathrm{~kJ} / \mathrm{mol}\right)\) b. According to The Denver Post for April 4,1983 , authorities feared that dangerous air pollution might occur during the neutralization. Considering the magnitude of \(\Delta H^{\circ}\), what was their major concern?

The enthalpy change for the reaction $$ \mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ is \(-891 \mathrm{~kJ}\) for the reaction as written. a. What quantity of heat is released for each mole of water formed? b. What quantity of heat is released for each mole of oxygen reacted?

A balloon filled with \(39.1\) moles of helium has a volume of \(876 \mathrm{~L}\) at \(0.0^{\circ} \mathrm{C}\) and \(1.00\) atm pressure. The temperature of the balloon is increased to \(38.0^{\circ} \mathrm{C}\) as it expands to a volume of \(998 \mathrm{~L}\), the pressure remaining constant. Calculate \(q, w\), and \(\Delta E\) for the helium in the balloon. (The molar heat capacity for helium gas is \(20.8 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{mol}\).)

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