Calculate the reaction enthalpy for the formation of anhydrous aluminum chloride, \(2 \mathrm{Al}(\mathrm{s})+3 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{AlCl}_{3}(\mathrm{~s})\), from the following data: $$ \begin{array}{ll} 2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{~g}) \\ & \Delta H^{\circ}=-1049 \mathrm{~kJ} \\ \mathrm{HCl}(\mathrm{g}) \longrightarrow \mathrm{HCl}(\mathrm{aq}) & \Delta H^{\circ}=-74.8 \mathrm{~kJ} \\ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HCl}(\mathrm{g}) & \Delta H^{\circ}=-185 \mathrm{~kJ} \\ \mathrm{AlCl}_{3}(\mathrm{~s}) \longrightarrow \mathrm{AlCl}_{3}(\mathrm{aq}) & \Delta H^{\circ}=-323 \mathrm{~kJ} \end{array} $$

Short Answer

Expert verified
-1304.9 kJ

Step by step solution

01

Identify the Known Reactions and Enthalpies

We are given four reactions with their corresponding reaction enthalpies. The first reaction describes the reaction of solid aluminum and aqueous hydrochloric acid forming aqueous aluminum chloride and hydrogen gas, with an enthalpy change of -1049 kJ. The second reaction describes the dissolution of gaseous hydrochloric acid into the aqueous phase, with an enthalpy change of -74.8 kJ. The third reaction describes the formation of gaseous hydrochloric acid from hydrogen gas and chlorine gas, with an enthalpy change of -185 kJ. The fourth reaction depicts the dissolution of solid aluminum chloride into the aqueous phase, with an enthalpy change of -323 kJ.
02

Write the Target Reaction

Our target reaction is the formation of anhydrous aluminum chloride from aluminum metal and chlorine gas, which can be represented as: \(2 \mathrm{Al}(\mathrm{s}) + 3 \mathrm{Cl}_2(\mathrm{g}) \rightarrow 2 \mathrm{AlCl}_3(\mathrm{s})\). We need to manipulate the given reactions to arrive at this target reaction.
03

Manipulate Given Reactions to Derive the Target Reaction

We need to reverse the fourth reaction to represent the formation of solid aluminum chloride from its aqueous form. This reverses the sign of the reaction enthalpy, so it becomes +323 kJ. The first reaction will be used as is, but we need to account for three moles of chlorine gas as in the target reaction, which means we need to multiply the third reaction by 3/2 to adjust the stoichiometry.
04

Calculate the Enthalpy Change for the Adjusted Reactions

For the reversed fourth reaction, the enthalpy change is +323 kJ. Since the third reaction is multiplied by 3/2, its enthalpy change becomes \(\frac{3}{2} \times -185 \mathrm{kJ} = -277.5 \mathrm{kJ}\).
05

Add the Enthalpies of Adjusted Reactions

Now, we add the enthalpies of all the adjusted reactions to find the total reaction enthalpy for the target reaction: \(-1049 \mathrm{kJ} (from the first reaction) + 3 \times (-74.8 \mathrm{kJ}) (since we need 6 moles of HCl for the first reaction, each with a -74.8 kJ change) + (-277.5 \mathrm{kJ}) + 323 \mathrm{kJ} = -1304.9 \mathrm{kJ}\).
06

Combine the Adjusted Reactions to Form the Target Reaction

When we combine all adjusted reactions, including the enthalpy changes, we will notice that all species except for the ones present in the target reaction will cancel out. Ultimately, we will be left with the target reaction and the combined reaction enthalpy.
07

Conclude with the Reaction Enthalpy for the Target Reaction

By combining the adjusted reactions, we find that the reaction enthalpy for the formation of anhydrous aluminum chloride is -1304.9 kJ. This is the total enthalpy change for the target reaction.

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Key Concepts

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

Thermochemistry
Thermochemistry is the study of heat and energy associated with chemical reactions and physical transformations. It is a sub-branch of thermodynamics, which deals more broadly with heat, work, and the properties of matter. One key principle of thermochemistry is that energy cannot be created or destroyed—it can only be changed from one form to another.

When a chemical reaction occurs, it can either release energy into the surrounding environment, known as an exothermic reaction, or absorb energy, known as an endothermic reaction. The amount of heat involved in a reaction at constant pressure is represented by the term enthalpy (\( \triangle H \)). Keeping track of enthalpy is crucial for predicting whether a reaction will occur spontaneously and for understanding how much energy is exchanged with the environment.

To calculate reaction enthalpies, scientists use specific measurement techniques like calorimetry. For educational and practical purposes, however, we often rely on tabulated standard enthalpies of formation and Hess's Law to find the enthalpy changes of reactions.
Enthalpy Change
Enthalpy change, noted as \( \triangle H \), is the difference in enthalpy between the products and the reactants in a chemical reaction. It reflects the heat absorbed or released during the reaction at constant pressure. Enthalpy change can be measured directly or calculated using known values, such as those from standard enthalpies of formation and Hess's Law.

If a reaction is exothermic, the enthalpy change is negative because the system releases heat. Conversely, an endothermic reaction results in a positive enthalpy change as the system absorbs heat from its surroundings. Chemical equations often include the enthalpy change to indicate the heat involved.

Understanding enthalpy changes can help predict the direction and extent of a reaction. It also allows for the design of safer and more efficient industrial processes, as controlling heat release or absorption is vital in large-scale chemical manufacturing.
Hess's Law
Hess's Law is a fundamental principle in thermochemistry that states the total enthalpy change of a chemical reaction is the same, regardless of the number of steps the reaction takes to go from reactants to products. This concept is based on the law of conservation of energy, which implies that enthalpy is a state function - its value is determined only by the current state of the system, not the path taken to reach that state.

Applying Hess's Law allows chemists to calculate the enthalpy change of complex reactions by breaking them down into simpler steps with known enthalpy changes. For instance, if a direct reaction's enthalpy change is difficult to measure or calculate, it can be found indirectly by adding or subtracting the enthalpy changes of multiple steps that lead to the same final products.

Understanding how to utilize Hess's Law is crucial because it provides a method for determining the energy changes in reactions that would otherwise be challenging to study due to their complexity or slow progression.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is the part of chemistry that involves the calculation of the amounts of substances involved in reactions. Stoichiometry is based on the law of conservation of mass, which states that during a chemical reaction, matter is neither created nor destroyed.

Stoichiometric calculations make use of the coefficients in a balanced chemical equation to relate the amounts of different substances. These relationships are crucial for determining the proportions of reactants needed and the amount of products that will be formed. For example, the stoichiometry of a reaction might show that two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of water.

Understanding stoichiometry is essential for controlling and scaling up reactions for industrial production, ensuring reactions go to completion, and calculating yields. It also forms the basis for reaction enthalpy calculations, as it informs the chemist of the proportionate amount of heat involved for the various quantities of reactants and products in a reaction.

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Most popular questions from this chapter

(a) At its boiling point, the vaporization of \(0.235 \mathrm{~mol} \mathrm{CH}_{4}(1)\) requires \(1.93 \mathrm{~kJ}\) of heat. What is the enthalpy of vaporization of methane? (b) An electric heater was immersed in a flask of boiling crhanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), and \(22.45 \mathrm{~g}\) of ethanol was vaporized when \(21.2 \mathrm{~kJ}\) of energy was supplied. What is the enthalpy of vaporization of ethanol?

Which gas do you expect to have the higher molar heat capacity, \(\mathrm{NO}\) or \(\mathrm{NO}_{2}\) ? Why?

(a) Calculate the work for a system that releases \(346 \mathrm{~kJ}\) of heat in a process for which the decrease in internal energy is \(125 \mathrm{~kJ}\). (b) Is work done on or by the system during this process?

Use the enthalpies of formation in Appendix \(2 \mathrm{~A}\) to calculate the standard enthalpy of the following reactions: (a) the replacement of deuterium by ordinary hydrogen in heavy water: \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{D}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{D}_{2}(\mathrm{~g})\) (b) the removal of sulfur from the hydrogen sulfide and sulfur dioxide in natural gas: \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+\mathrm{SO}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{~S}(\mathrm{~s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (c) the oxidation of ammonia: \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

The standard enthalpies of combustion of graphite and diamond are \(-393.51\) and \(-395.41 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\), respectively. Calculate the change in molar enthalpy for the graphite \(\rightarrow\) diamond transition.

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